本来是8天代码放一起的，但是太麻烦了，还是分成了2个博客放。

day1

F div1 爬爬爬山 (最短路)

  1. //F
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef long long LL;
  11.     typedef unsigned long long uLL;
  12.     typedef pair<int, LL> pii;
  13. }
  15. using namespace lh;
  17. const int MXN = 1e6 + 7;
  18. const int INF = 0x3f3f3f3f;
  19. const int MOD = 1000000007;
  21. int n, m, k, NODE;
  22. LL height[MXN];
  23. int other[MXN];
  24. vector<pii> mp[MXN];
  25. LL dis[MXN];
  26. struct lp {
  27.     int v;
  28.     LL w;
  29.     friend bool operator <(const lp &a, const lp &b) {
  30.         return a.w > b.w;
  31.     }
  32. }A, B;
  33. void add_edge(int u,int v,LL w) {
  34.     mp[u].push_back({v, w});
  35. }
  36. void dij() {
  37.     for(int i = 2; i <= NODE; ++i) dis[i] = 1e18;
  38.     dis[1] = 0;
  39.     priority_queue<lp> Q;
  40.     Q.push({1, 0});
  41.     while(!Q.empty()) {
  42.         A = Q.top(); Q.pop();
  43.         int u = A.v;
  44.         if(dis[u] < A.w) continue;
  45.         int len = mp[u].size();
  46.         for(int i = 0; i < len; ++i) {
  47.             int v = mp[u][i].fi;
  48.             if(dis[v] > dis[u] + mp[u][i].se) {
  49.                 dis[v] = dis[u] + mp[u][i].se;
  50.                 Q.push({v, dis[v]});
  51.             }
  52.         }
  53.     }
  54.     printf("%lld\n", dis[n]);
  55. }
  56. int main() {
  57.     scanf("%d%d%d", &n, &m, &k);
  58.     NODE = n;
  59.     for(int i = 1; i <= n; ++i) other[i] = -1;
  60.     for(int i = 1; i <= n; ++i) {
  61.         scanf("%lld", &height[i]);
  62.         if(i > 1 && height[i] - height[1] > k) {
  63.             other[i] = ++ NODE;
  64.             add_edge(other[i], i, (height[i]-height[1]-k)*(height[i]-height[1]-k));
  65.             //add_edge(i, other[i], (height[i]-height[1])*(height[i]-height[1]));
  66.             //printf("%d %d\n", i, (height[i]-height[1])*(height[i]-height[1]));
  67.         }
  68.     }
  69.     LL c;
  70.     for(int i = 0, a, b; i < m; ++i) {
  71.         scanf("%d%d%lld", &a, &b, &c);
  72.         if(other[b] == -1) add_edge(a, b, c);
  73.         else add_edge(a, other[b], c);
  74.         if(other[a] == -1) add_edge(b, a, c);
  75.         else add_edge(b, other[a], c);
  76.     }
  77.     dij();
  78.     return 0;
  79. }

B div2 吃豆豆 (dp)

  1. //B
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef long long LL;
  11.     typedef unsigned long long uLL;
  12.     typedef pair<int, LL> pii;
  13. }
  15. using namespace lh;
  17. const int MXN = 1e6 + 7;
  18. const int INF = 0x3f3f3f3f;
  19. const int MOD = 1000000007;
  21. int n, m, C;
  22. int sx,sy,ex,ey;
  23. int ar[15][15];
  24. int dp[15][15][10352];
  25. int dir[5][2]={1,0,0,1,-1,0,0,-1,0,0};
  26. int main() {
  27.     scanf("%d%d%d", &n, &m, &C);
  28.     for(int i = 1; i <= n; ++i) {
  29.         for(int j = 1; j <= m; ++j) {
  30.             scanf("%d", &ar[i][j]);
  31.         }
  32.     }
  33.     scanf("%d%d%d%d", &sx, &sy, &ex, &ey);
  34.     for(int k = 0; k <= 10351; ++k) {
  35.         for(int i = 1; i <= n; ++i) {
  36.             for(int j = 1; j <= m; ++j) {
  37.                 dp[i][j][k] = -INF;
  38.             }
  39.         }
  40.     }
  41.     dp[sx][sy][0] = 0;
  42.     for(int k = 1; k <= 10351; ++k) {
  43.         for(int i = 1; i <= n; ++i) {
  44.             for(int j = 1; j <= m; ++j) {
  45.                 for(int h = 0; h < 5; ++h) {
  46.                     int px = i + dir[h][0], py = j + dir[h][1];
  47.                     if(px < 1 || px > n || py < 1 || py > m) continue;
  48.                     if(dp[px][py][k-1] == -INF) continue;
  49.                     dp[i][j][k] = max(dp[i][j][k], dp[px][py][k-1]+((k%ar[i][j])==0));
  50.                 }
  51.             }
  52.         }
  53.     }
  54.     int ans = INF;
  55.     for(int i = 1; i <= 10351; ++i) {
  56.         if(dp[ex][ey][i] >= C) {
  57.             ans = i;
  58.             break;
  59.         }
  60.     }
  61.     printf("%d\n", ans);
  62.     return 0;
  63. }

J div2 夺宝奇兵(暴力)

  1. //J
  2. /*
  3. 把所有人宝物从大到小排个序，列出一个柱状图来
  4. 枚举其他人最高的宝物的数量为mid，所有高于mid的宝物我们肯定要买的
  5. 如果这时获得宝物已经多于mid了，就可以return了
  6. 不然就从剩下所有的宝物里面选出价值最小的几个宝物凑出mid+1个宝物出来
  7. */
  8. #include<bits/stdc++.h>
  9. #define fi first
  10. #define se second
  11. #define iis std::ios::sync_with_stdio(false)
  12. #define pb push_back
  13. namespace lh {
  14. #define o2(x) (x)*(x)
  15.     using namespace std;
  16.     typedef long long LL;
  17.     typedef unsigned long long uLL;
  18.     typedef pair<int, LL> pii;
  19. }
  21. using namespace lh;
  22. const LL INF = 1e18;
  23. const int N = 1e4 + 100;
  24. vector<LL> c[N];
  25. int n, m;
  26. LL x;
  27. LL solve(int mid){
  28.     LL tmp = 0;
  29.     int now = 0;
  30.     std::vector<LL> vs;
  31.     for(int i = 2; i <= n; ++i) {
  32.         int siz = c[i].size();
  33.         if(siz >= mid) {
  34.             for(int j = 0; j < siz - mid; ++j) tmp += c[i][j], now++;
  35.             for(int j = siz - mid; j < siz; ++j) vs.push_back(c[i][j]);
  36.         }else {
  37.             for(int j = 0; j < siz; ++j) vs.push_back(c[i][j]);
  38.         }
  39.     }
  40.     if(now > mid) return tmp;
  41.     if(now + vs.size() <= mid) return INF;
  42.     sort(vs.begin(), vs.end());
  43.     for(int i = 0; i <= mid - now && i < vs.size(); ++i) {
  44.         tmp += vs[i];
  45.     }
  46.     return tmp;
  47. }
  48. int main() {
  49.     scanf("%d%d", &n, &m);
  50.     for (int i = 1, ar; i <= m; ++i){
  51.         scanf("%lld%d", &x, &ar);
  52.         ++ ar; c[ar].pb(x);
  53.     }
  54.     ++ n;
  55.     for (int i = 2; i <= n; ++i){
  56.         sort(c[i].begin(), c[i].end());
  57.     }
  58.     LL tmp = 1e18;
  59.     for (int i = 1; i <= m; ++i){
  60.         tmp = min(tmp, solve(i));
  61.     }
  62.     printf("%lld\n", tmp);
  63.     return 0;
  64. }

J div1 夺宝奇兵 (权值线段树)

  1. //太菜了，后缀和都不会写了，找了一万年的bug
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define lson rt<<1
  6. #define rson rt<<1|1
  7. #define iis std::ios::sync_with_stdio(false)
  8. #define pb push_back
  9. namespace lh {
  10. #define o2(x) (x)*(x)
  11.     using namespace std;
  12.     typedef long long LL;
  13.     typedef unsigned long long uLL;
  14.     typedef pair<int, int> pii;
  15. }
  17. using namespace lh;
  19. const LL INF = 1e18;
  20. const int MXN = 4e5 + 7;
  22. int n, m;
  23. std::vector<int> shut[MXN];
  24. std::vector<pii > mp[MXN];
  25. pii ar[MXN];
  26. LL suf[MXN], SUM[MXN<<2];
  27. int num[MXN], NUM[MXN<<2];
  28. LL min_sum;
  29. int min_num;
  30. bool cmp(const pii &a, const pii &b) {
  31.     if(a.fi != b.fi) return a.fi > b.fi;
  32.     return a.se > b.se;
  33. }
  34. void push_up(int rt) {
  35.     NUM[rt] = NUM[lson] + NUM[rson];
  36.     SUM[rt] = SUM[lson] + SUM[rson];
  37. }
  38. void build(int l,int r,int rt) {
  39.     if(l == r) {
  40.         NUM[rt] = 1;
  41.         SUM[rt] = ar[l].fi;
  42.         return ;
  43.     }
  44.     int mid = (l + r) >> 1;
  45.     build(l, mid, lson), build(mid+1, r, rson);
  46.     push_up(rt);
  47. }
  48. void update(int p, int v, int l, int r, int rt) {
  49.     if(l == r) {
  50.         NUM[rt] = 0;
  51.         SUM[rt] = 0;
  52.         return ;
  53.     }
  54.     int mid = (l + r) >> 1;
  55.     if(p <= mid) update(p, v, l, mid, lson);
  56.     else update(p, v, mid+1, r, rson);
  57.     push_up(rt);
  58. }
  59. LL query(int nd, int l, int r, int rt) {
  60.     //printf("%d %d %d %d %lld\n", nd, l, r, NUM[rt], SUM[rt]);
  61.     if(l == r || nd == NUM[rt]) {
  62.         return SUM[rt];
  63.     }
  64.     int mid = (l + r) >> 1;
  65.     if(NUM[lson] >= nd) return query(nd, l, mid, lson);
  66.     else {
  67.         return SUM[lson] + query(nd - NUM[lson], mid+1, r, rson);
  68.     }
  69. }
  70. LL solve(int mid) {
  71.     LL ans = min_sum;//min_sum 和 suf[mid+1]
  72.     if(min_num > mid) return ans;//min_num 和 num[mid+1]
  73.     if(NUM[1] + min_num <= mid) return INF;
  74.     int nd = min(mid + 1 - min_num, NUM[1]);
  75.     ans += query(nd, 1, m, 1);
  76.     return ans;
  77. }
  79. int main() {
  80.     scanf("%d%d", &n, &m);
  81.     int x, a;
  82.     LL ans = 0;
  83.     for(int i = 1; i <= m; ++i) {
  84.         scanf("%d%d", &x, &a);
  85.         ar[i] = {x, a};
  86.         ans = ans + x;
  87.     }
  88.     sort(ar + 1, ar + 1 + m);
  89.     for(int i = 1; i <= m; ++i) {
  90.         mp[ar[i].se].push_back({ar[i].fi, i});
  91.     }
  92.     build(1, m, 1);
  93.     for(int i = 1; i <= n; ++i) {
  94.         sort(mp[i].begin(), mp[i].end(), cmp);
  95.         for(int j = mp[i].size(); j >= 1; --j) {
  96.             shut[j].push_back(mp[i][j-1].se);
  97.         }
  98.     }
  99.     //printf("[%lld]\n", query(4, 1, m, 1));
  100.     for(int i = m; i >= 1; --i) {
  101.         ans = min(ans, solve(i));
  102.         for(auto x: shut[i]) {
  103.             ++ min_num;
  104.             min_sum += ar[x].fi;
  105.             update(x, -1, 1, m, 1);
  106.         }
  107.     }
  108.     printf("%lld\n", ans);
  109.     return 0;
  110. }

C div1 拆拆拆数

  1. #include<bits/stdc++.h>
  2. #define fi first
  3. #define se second
  4. #define lson rt<<1
  5. #define rson rt<<1|1
  6. #define iis std::ios::sync_with_stdio(false)
  7. #define pb push_back
  8. namespace lh {
  9. #define o2(x) (x)*(x)
  10.     using namespace std;
  11.     typedef long long LL;
  12.     typedef unsigned long long uLL;
  13.     typedef pair<int, int> pii;
  14. }
  16. using namespace lh;
  18. const int MXN = 4e5 + 7;
  19. //一个偶数可以拆成一个奇数和一个奇素数之和
  20. int n, m;
  21. LL a, b;
  22. LL pp[20] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59};
  23. LL gcd(LL a, LL b) {
  24.     return b==0?a:gcd(b,a%b);
  25. }
  26. int main() {
  27.     scanf("%d", &n);
  28.     m = 17;
  29.     while(n --) {
  30.         scanf("%lld%lld", &a, &b);
  31.         if(gcd(a, b) == 1) {
  32.             printf("1\n%lld %lld\n", a, b);
  33.             continue ;
  34.         }
  35.         int flag = 0;
  36.         if(b % 2 == 0) swap(a, b), flag = 1;
  37.         if(a % 2 == 0) {
  38.             for(int i = 0; i < m; ++i) {
  39.                 if(gcd(b-2, pp[i]) == 1 && gcd(a-pp[i],2) == 1) {
  40.                     if(flag == 0) printf("2\n%lld %lld\n%lld %lld\n",pp[i],b-2,a-pp[i],2LL);
  41.                     else printf("2\n%lld %lld\n%lld %lld\n",b-2,pp[i],2LL,a-pp[i]);
  42.                     break;
  43.                 }
  44.             }
  45.         }else {
  46.             if(flag == 0) printf("2\n%lld %lld\n%lld %lld\n", 2LL, b-2, a-2, 2LL);
  47.             else printf("2\n%lld %lld\n%lld %lld\n", b-2, 2LL, 2LL, a-2);
  48.         }
  49.     }
  50.     return 0;
  51. }
  52. //这个也行
  53. using namespace lh;
  55. const int MXN = 4e5 + 7;
  57. int n, m;
  58. LL a, b;
  59. LL pp[20] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59};
  60. LL gcd(LL a, LL b) {
  61.     return b==0?a:gcd(b,a%b);
  62. }
  63. int main() {
  64. #ifndef ONLINE_JUDGE
  65.     freopen("E://ADpan//in.in", "r", stdin);
  66.     //freopen("E://ADpan//out.out", "w", stdout);
  67. #endif
  68.     scanf("%d", &n);
  69.     m = 17;
  70.     while(n --) {
  71.         scanf("%lld%lld", &a, &b);
  72.         if(gcd(a, b) == 1) {
  73.             printf("1\n%lld %lld\n", a, b);
  74.             continue ;
  75.         }else if(a % 2 == 1 && b % 2 == 1) {
  76.             printf("2\n%lld %lld\n%lld %lld\n", 2LL, b-2, a-2, 2LL);
  77.             continue;
  78.         }
  79.         printf("2\n");
  80.         int flag = 1;
  81.         for(LL i = 2; i <= 100 && flag; ++i) {
  82.             for(LL j = 2; j <= 100 && flag; ++j) {
  83.                 if(gcd(i,j) == 1 && gcd(a-i,b-j) == 1) {
  84.                     printf("%lld %lld\n%lld %lld\n", i,j,a-i,b-j);
  85.                     flag = 0;
  86.                 }else if(gcd(i,b-j) == 1 && gcd(a-i,j) == 1) {
  87.                     flag = 0;
  88.                     printf("%lld %lld\n%lld %lld\n", i,b-j,a-i,j);
  89.                 }
  90.             }
  91.         }
  92.     }
  93.     return 0;
  94. }

E div1 流流流动(冰雹猜想 ，树形dp)

  1. //这玩意会连成一个森林
  2. #include<bits/stdc++.h>
  3. namespace lh {
  4. #define o2(x) (x)*(x)
  5.     using namespace std;
  6.     typedef long long LL;
  7.     typedef unsigned long long uLL;
  8.     typedef pair<int, int> pii;
  9. }
  11. using namespace lh;
  13. const int MXN = 4e5 + 7;
  15. int n, m;
  16. std::vector<int> son[MXN];
  17. int f[MXN], d[MXN], vis[MXN];
  18. LL dp[MXN][2];
  19. //dp[i][0]表示不选i, dp[i][1]表示选i
  20. void dfs(int u,int ba) {
  21.     dp[u][1] += f[u];
  22.     vis[u] = 1;
  23.     for(auto v: son[u]) {
  24.         if(v == ba) continue;
  25.         dfs(v, u);
  26.         dp[u][0] += max(dp[v][0], dp[v][1]);
  27.         dp[u][1] += max(dp[v][0], dp[v][1]-d[min(u, v)]);
  28.     }
  29. }
  30. int main() {
  31.     scanf("%d", &n);
  32.     for(int i = 1; i <= n; ++i) scanf("%d", &f[i]);
  33.     for(int i = 1; i <= n; ++i) scanf("%d", &d[i]);
  34.     for(int i = 2; i <= n; ++i) {
  35.         if(i % 2 == 0) {
  36.             son[i].push_back(i/2);
  37.             son[i/2].push_back(i);
  38.         }else if(i*3+1 <= n){
  39.             son[i].push_back(i*3+1);
  40.             son[i*3+1].push_back(i);
  41.         }
  42.     }
  43.     LL ans = 0;
  44.     for(int i = 1; i <= n; ++i) {
  45.         if(vis[i] == 0) {
  46.             dfs(i, i);
  47.             ans += max(dp[i][0], dp[i][1]);
  48.         }
  49.     }
  50.     printf("%lld\n", ans);
  51.     return 0;
  52. }

I div2 起起落落 (dp)

\(dp[i]=\sum_{j=1}^{j<i}[p[j]>p[i]]\times dp[j]\times \sum_{h=j+1}^{h<i}[p[h]<p[j]]\)

  1. /*
  2. 题意：
  3. 给你一个长度为n的排列，问有多少个长度为(2*m+1)奇数且大于3的子序列满足一下条件：
  4. 1.下标序列递增（这个没啥重要的
  5. 2.(1<=i<=m) p[2*i-1]>p[2*i+1]>p[2*i]
  7. dp[i]表示以i结尾符合条件的序列的个数
  8. dp[i]=\sum_{j=1}^{j<i}[p[j]>p[i]]\times dp[j]\times \sum_{h=j+1}^{h<i}[p[h]<p[j]]
  9. 这种转移方法构造出来的方案中序列的长度肯定是奇数的
  10. */
  11. #include<bits/stdc++.h>
  12. #define fi first
  13. #define se second
  14. #define lson rt<<1
  15. #define rson rt<<1|1
  16. #define iis std::ios::sync_with_stdio(false)
  17. #define pb push_back
  18. namespace lh {
  19. #define o2(x) (x)*(x)
  20.     using namespace std;
  21.     typedef long long LL;
  22.     typedef unsigned long long uLL;
  23.     typedef pair<int, int> pii;
  24. }
  26. using namespace lh;
  28. const int mod = 1e9 +7;
  29. const int MXN = 4e5 + 7;
  31. int n, m;
  32. int p[MXN];
  33. LL dp[MXN], suf[MXN];
  34. int main() {
  35.     scanf("%d", &n);
  36.     for(int i = 1; i <= n; ++i) scanf("%d", &p[i]);
  37.     LL ans = 0;
  38.     for(int i = 3, cnt; i <= n; ++i) {
  39.         cnt = 0;
  40.         for(int j = i-1; j >= 1; --j) {
  41.             if(p[j] < p[i]) {
  42.                 ++ cnt;
  43.                 continue;
  44.             }
  45.             dp[i] = (dp[i]+ (1+dp[j])*cnt%mod) % mod;//加1是要加上取单独j这一个元素和cnt和i相组合的情况
  46.         }
  47.         if(i >= 3) ans = (ans + dp[i]) % mod;
  48.     }
  49.     printf("%lld\n", ans);
  50.     return 0;
  51. }

I div1 起起落落 (权值线段树优化dp)

  1. 待填坑

day2

A div2 Erase Nodes(贪心)

  1. //A
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. #define pb push_back
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef unsigned long long LL;
  11.     typedef pair<int, LL> pii;
  12. }
  14. using namespace lh;
  16. const int MXN = 1e4 + 100;
  17. int n;
  18. int ar[MXN];
  19. LL dp[20];
  20. int get(int x) {
  21.     int num = 0;
  22.     while(x) {
  23.         num ++;
  24.         x /= 10;
  25.     }
  26.     return num;
  27. }
  28. int main() {
  29.     dp[0] = 1;
  30.     for(int i = 1; i <= 10; ++i) {
  31.         dp[i] = dp[i-1] * 10;
  32.     }
  33.     int cas = 0;
  34.     int tim; scanf("%d", &tim);
  35.     while(tim --) {
  36.         scanf("%d", &n);
  37.         int cnt = 0;
  38.         for(int i = 1; i <= n; ++i) {
  39.             scanf("%d", &ar[i]);
  40.             if(ar[i] == 1000000000) cnt++;
  41.         }
  42.         int mmax = ar[n], wei = get(ar[n]);
  43.         LL ans = 0, tmp = 0;
  44.         int flag = 0;
  45.         for(int i = n-1; i >= 1; --i) {
  46.             ans = ar[i];
  47.             ans *= dp[wei];
  48.             ans += mmax;
  49.             tmp = max(tmp, ans);
  50.             if(ar[i] > mmax) {
  51.                 mmax = ar[i];
  52.                 wei = get(ar[i]);
  53.             }
  54.         }
  55.         printf("Case #%d: %llu\n", ++cas, tmp);
  56.     }
  57.     return 0;
  58. }

B div2 Erase Numbers III(贪心)

  1. //B
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. #define pb push_back
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef unsigned long long LL;
  11.     typedef pair<int, LL> pii;
  12. }
  14. using namespace lh;
  15. const int INF = 0x3f3f3f3f;
  16. const int MXN = 1e4 + 100;
  17. int n;
  18. int ar[MXN], len[MXN], br[MXN], is[MXN], len2[MXN], id[MXN];
  19. LL dp[20];
  20. int get(int x) {
  21.     int num = 0;
  22.     while(x) {
  23.         num ++;
  24.         x /= 10;
  25.     }
  26.     return num;
  27. }
  28. string toString(int x) {
  29.     string ans;
  30.     char c;
  31.     while(x) {
  32.         c = x%10+'0';
  33.         ans = c + ans;
  34.         x /= 10;
  35.     }
  36.     return ans;
  37. }
  38. int A[20], B[20];
  39. bool mycmp(int x, int y) {//x是否比y字典序小
  40.     A[0] = B[0] = 0;
  41.     while(x) {
  42.         A[++A[0]] = x%10;
  43.         x /= 10;
  44.     }
  45.     while(y) {
  46.         B[++B[0]] = y%10;
  47.         y /= 10;
  48.     }
  49.     reverse(A+1, A+A[0]+1);
  50.     reverse(B+1, B+B[0]+1);
  51.     int tmp = min(A[0], B[0]);
  52.     for(int i = 1; i <= tmp; ++i) {
  53.         if(A[i] > B[i]) return 0;
  54.         else if(A[i] < B[i]) return 1;
  55.     }
  56.     return 1;
  57. }
  58. int main() {
  59.     int cas = 0;
  60.     int tim; scanf("%d", &tim);
  61.     while(tim --) {
  62.         scanf("%d", &n);
  63.         int mmin = INF;
  64.         for(int i = 1; i <= n; ++i) is[i] = 0;
  65.         for(int i = 1; i <= n; ++i) {
  66.             scanf("%d", &ar[i]);
  67.             len[i] = get(ar[i]);
  68.             br[i] = len[i];
  69.             mmin = min(mmin, len[i]);
  70.         }
  71.         sort(br + 1, br + 1 + n);
  72.         int flag = 0;
  73.         if(br[1] == br[2]) flag = 1;
  74.         printf("Case #%d: ", ++cas);
  75.         int pos = -1;
  76.         if(flag == 1) {
  77.             std::vector<int> my;
  78.             for(int i = 1; i < n; ++i) {
  79.                 if(len[i] == mmin && mycmp(ar[i], ar[i+1])) {
  80.                     my.push_back(i);
  81.                     is[i] = 1;
  82.                     //printf("*%d\n", i);
  83.                     break;
  84.                 }
  85.             }
  86.             int cnt2 = 0;
  87.             for(int i = 1; i <= n; ++i) {
  88.                 if(is[i] == 0) br[++cnt2] = ar[i], len2[cnt2] = len[i], id[cnt2] = i;
  89.             }
  90.             for(int i = 1; i < cnt2; ++i) {
  91.                 if(len2[i] == mmin && mycmp(br[i], br[i+1])) {
  92.                     my.push_back(id[i]);
  93.                     is[id[i]] = 1;
  94.                     //printf("**%d\n", id[i]);
  95.                     break;
  96.                 }
  97.             }
  98.             for(int i = n; i >= 1 && my.size() < 2; --i) {
  99.                 if(is[i] == 0 && len[i] == mmin) my.push_back(i);
  100.             }
  101.             string ans1, ans2;
  102.             for(int i = 1; i <= n; ++i) {
  103.                 if(i == my[0] || i == my[1]) continue;
  104.                 ans1 += toString(ar[i]);
  105.             }
  106.             //cout<<ans1<<"\n";
  107.             my.clear();
  108.             for(int i = 1; i + 2 <= n; ++i) {
  109.                 if(len[i] + len[i+1] == mmin + mmin) {
  110.                     if(mycmp(ar[i], ar[i+2])) {
  111.                         my.push_back(i);
  112.                         break;
  113.                     }
  114.                 }
  115.             }
  116.             int hh = 0;
  117.             if(my.size() == 0) {
  118.                 for(int i = n; i >= 2; --i) {
  119.                     if(len[i] + len[i-1] == mmin + mmin) {
  120.                         my.push_back(i);
  121.                         break;
  122.                     }
  123.                 }
  124.                 if(my.size() == 1) {
  125.                     hh = 1;
  126.                     my.push_back(my[0]-1);
  127.                     for(int i = 1; i <= n; ++i) {
  128.                         if(i == my[0] || i == my[1]) continue;
  129.                         ans2 += toString(ar[i]);
  130.                     }
  131.                 }
  132.             }
  133.             if(hh) ans1 = max(ans1, ans2);
  134.             cout<<ans1<<"\n";
  135.             continue;
  136.         }
  137.         pos = -1;
  138.         int fuck = br[2];
  139.         for(int i = 1, nex; i < n; ++i) {
  140.             nex = ar[i+1];
  141.             if(len[i+1] == mmin && i + 2 <= n) nex = ar[i+2];
  142.             if(len[i] == fuck) {
  143.                 if(mycmp(ar[i], nex)) {
  144.                     pos = i;
  145.                     break;
  146.                 }
  147.             }
  148.         }
  149.         if(pos == -1) {
  150.             for(int i = n; i >= 1; --i) {
  151.                 if(len[i] == mmin) continue;
  152.                 if(len[i] != fuck) continue;
  153.                 pos = i;
  154.                 break;
  155.             }
  156.         }
  157.         for(int i = 1; i <= n; ++i) {
  158.             if(i == pos || len[i] == mmin) {
  160.             }else printf("%d", ar[i]);
  161.         }
  162.         printf("\n");
  163.     }
  164.     return 0;
  165. }//求ac

H div1&2 Cosmic Cleaner(计算几何)

球缺体积

用一个平面截去一个球所得部分叫球缺

球缺面积\(=2\times πh\)

球缺体积\(=π\times h^2\times (R−\frac{h}{3})\)

球缺质心：匀质球缺的质心位于它的中轴线上，并且与底面的距离为

\(C=\frac{(4R-h)\times h}{12R-4h}=\frac{(d^2+2h^2)\times h}{3d^2+4h^2}\)

其中，\(h\)为球缺的高，\(R\)为大圆半径，\(d\)为球缺的底面直径

https://blog.csdn.net/enterprise_/article/details/81624174

https://paste.ubuntu.com/p/5qZKXYFsjQ/

  1. //H
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. #define pb push_back
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef unsigned long long LL;
  11.     typedef pair<int, LL> pii;
  12. }
  14. using namespace lh;
  15. const int INF = 0x3f3f3f3f;
  16. const double pi = acos(-1.0);
  17. const int MXN = 1e4 + 100;
  18. int n;
  19. int A, B, C, r1;
  20. struct lp {
  21.     int a, b, c, r2;
  22. }cw[MXN];
  23. double tiji(lp ar) {
  24.     double d = sqrt(o2(ar.a-A)+o2(ar.b-B)+o2(ar.c-C));
  25.     double r2 = ar.r2;
  26.     double X = r1-(r1*r1-r2*r2+d*d)/(2*d);
  27.     double Y = r2-(r2*r2-r1*r1+d*d)/(2*d);
  28.     double ans = pi/3*X*X*(3*r1-X)+pi/3*Y*Y*(3*r2-Y);
  29.     return ans;
  30. }
  31. int main() {
  32.     int cas = 0;
  33.     int tim; scanf("%d", &tim);
  34.     while(tim --) {
  35.         scanf("%d", &n);
  36.         int a, b, c, r2;
  37.         double d;
  38.         for(int i = 1; i <= n; ++i) {
  39.             scanf("%d%d%d%d", &a, &b, &c, &r2);
  40.             cw[i] = {a, b, c, r2};
  41.         }
  42.         scanf("%d%d%d%d", &A, &B, &C, &r1);
  43.         double ans = 0;
  44.         for(int i = 1; i <= n; ++i) {
  45.             double d = sqrt(o2(cw[i].a-A)+o2(cw[i].b-B)+o2(cw[i].c-C));
  46.             if(d > r1 + cw[i].r2) continue;
  47.             if(d + cw[i].r2 <= r1) {
  48.                 ans += 4.0*pi/3*cw[i].r2*cw[i].r2*cw[i].r2;
  49.             }else ans += tiji(cw[i]);
  50.         }
  51.         printf("Case #%d: %.10f\n", ++cas, ans);
  52.     }
  53.     return 0;
  54. }
  55. /*X = r1-(r1^2-r2^2+d^2)/2d
  56. pi/3*(X)^2*(3r1-(X))
  57. Y = r2-(r2^2-r1^2+d^2)/2d
  58. +pi/3*(Y)^2*(3r2-(Y))*/

K div1 Sticks 搜索

  1. //12分6写的有点丑https://paste.ubuntu.com/p/kHBHNCGgnh/
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. #define pb push_back
  7. namespace lh {
  8. #define o2(x) (x)*(x)
  9.     using namespace std;
  10.     typedef long long LL;
  11.     typedef pair<int, int> pii;
  12. }
  14. using namespace lh;
  15. const int INF = 0x3f3f3f3f;
  16. const int MXN = 1e4 + 100;
  17. int n;
  18. LL ar[MXN];
  19. int is[13][13][13], g[13];
  20. struct lp {
  21.     LL x, y, z;
  22. }ans[5], tmp[5];
  23. int ANS;
  24. void dfs(int cur, int num) {
  25.     if(cur/3 + num <= ANS) return ;
  26.     if(cur == 3) {
  27.         if(ar[g[0]]+ar[g[1]] > ar[g[2]]) {
  28.             num ++;
  29.             tmp[num].x = ar[g[0]]; tmp[num].y = ar[g[1]]; tmp[num].z = ar[g[2]];
  30.         }
  31.         if(num > ANS) {
  32.             ANS = num;
  33.             for(int i = 1; i <= num; ++i) ans[i] = tmp[i];
  34.         }
  35.         return ;
  36.     }
  37.     int d[13];
  38.     for(int i = 0; i < cur; ++i) d[i] = g[i];
  39.     int k = 0, ret;
  40.     //tmp[num+1].x = ar[d[0]];
  41.     for(int i = 1; i < cur; ++i) {
  42.         //tmp[num+1].y = ar[d[i]];
  43.         for(int j = i + 1; j < cur; ++j) {
  44.             //tmp[num+1].z = ar[d[j]];
  45.             ret = num + is[d[0]][d[i]][d[j]];
  46.             if(is[d[0]][d[i]][d[j]]) {
  47.                 tmp[num+1].x = ar[d[0]];
  48.                 tmp[num+1].y = ar[d[i]];
  49.                 tmp[num+1].z = ar[d[j]];
  50.             }
  51.             k = 0;
  52.             for(int t = 1; t < cur; ++t) {
  53.                 if(t != i && t != j) g[k++] = d[t];
  54.             }
  55.             dfs(cur-3, ret);
  56.             if(ANS == 4) return ;
  57.         }
  58.     }
  59. }
  60. int main() {
  61.     int cas = 0;
  62.     int tim; scanf("%d", &tim);
  63.     while(tim --) {
  64.         for(int i = 0; i < 12; ++i) scanf("%lld", &ar[i]), g[i] = i;
  65.         sort(ar, ar + 12);
  66.         for(int i = 0; i < 12; ++i) {
  67.             for(int j = i + 1; j < 12; ++j) {
  68.                 for(int k = j + 1; k < 12; ++k) {
  69.                     is[i][j][k] = ((ar[i] + ar[j]) > ar[k]);
  70.                 }
  71.             }
  72.         }
  73.         ANS = 0;
  74.         dfs(12, 0);
  75.         printf("Case #%d: %d\n", ++cas, ANS);
  76.         for(int i = 1; i <= ANS; ++i) {
  77.             printf("%lld %lld %lld\n", ans[i].x, ans[i].y, ans[i].z);
  78.         }
  79.     }
  80.     return 0;
  81. }

day3

F div2 小清新数论(莫比乌斯反演)

莫比乌斯反演

  1. //F
  2. #include<bits/stdc++.h>
  3. using namespace std;
  4. typedef long long LL;
  6. const int MXN = 1e7+ 7;
  7. const int mod = 998244353;
  9. LL n, m;
  10. int noprime[MXN], pp[MXN], pcnt;
  11. int phi[MXN], mu[MXN];
  12. LL pre_mu[MXN];
  13. void init_prime() {
  14.     noprime[0] = noprime[1] = 1;
  15.     mu[1] = 1; phi[1] = 1;
  16.     for(int i = 2; i < MXN; ++i) {
  17.         if(!noprime[i]) pp[pcnt++] = i, phi[i] = i-1, mu[i] = -1;
  18.         for(int j = 0; j < pcnt && pp[j] * i < MXN; ++j) {
  19.             noprime[pp[j]*i] = 1;
  20.             phi[pp[j]*i] = (pp[j]-1)*phi[i];
  21.             mu[pp[j]*i] = -mu[i];
  22.             if(i % pp[j] == 0) {
  23.                 phi[pp[j]*i] = pp[j]*phi[i];
  24.                 mu[pp[j]*i] = 0;
  25.                 break;
  26.             }
  27.         }
  28.     }
  29.     for(int i = 1; i < MXN; ++i) pre_mu[i] = pre_mu[i-1] + mu[i];
  30. }
  32. LL solve(LL n) {
  33.     LL ans = 0, tmp;
  34.     for(LL L = 1, R; L <= n; L = R + 1) {
  35.         R = n/(n/L);
  36.         tmp = pre_mu[R]-pre_mu[L-1];
  37.         ans = (ans + tmp*(n/L)%mod*(n/L)%mod)%mod;
  38.     }
  39.     return (ans+mod)%mod;
  40. }
  41. int main() {
  42.     init_prime();
  43.     scanf("%lld", &n);
  44.     LL ans = 0, tmp;
  45.     for(LL L = 1, R; L <= n; L = R + 1) {
  46.         R = n/(n/L);
  47.         tmp = pre_mu[R]-pre_mu[L-1];
  48.         ans = (ans + tmp*solve(n/L)%mod)%mod;
  49.     }
  50.     printf("%lld\n", (ans+mod)%mod);
  51.     return 0;
  52. }

F div1 小清新数论(杜教筛)

杜教筛

题解

code1:利用欧拉函数

  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. typedef long long LL;
  5. const int MXN = 1e7+ 7;
  6. const LL mod = 998244353;
  7. const LL inf = 1e18;
  8. LL n, m, inv2;
  9. int noprime[MXN], pp[MXN], pcnt;
  10. int phi[MXN], mu[MXN];
  11. LL pre_mu[MXN], pre_phi[MXN];
  12. unordered_map<LL, LL> mp1, mp2;
  13. void init_prime() {
  14.     noprime[0] = noprime[1] = 1;
  15.     mu[1] = 1; phi[1] = 1;
  16.     for(int i = 2; i < MXN; ++i) {
  17.         if(!noprime[i]) pp[pcnt++] = i, phi[i] = i-1, mu[i] = -1;
  18.         for(int j = 0; j < pcnt && pp[j] * i < MXN; ++j) {
  19.             noprime[pp[j]*i] = 1;
  20.             phi[pp[j]*i] = (pp[j]-1)*phi[i];
  21.             mu[pp[j]*i] = -mu[i];
  22.             if(i % pp[j] == 0) {
  23.                 phi[pp[j]*i] = pp[j]*phi[i];
  24.                 mu[pp[j]*i] = 0;
  25.                 break;
  26.             }
  27.         }
  28.     }
  29.     for(int i = 1; i < MXN; ++i) {
  30.         pre_mu[i] = pre_mu[i-1] + mu[i];
  31.         pre_phi[i] = (pre_phi[i-1] + phi[i])%mod;
  32.     }
  33. }
  34. LL solve_mu(LL n) {
  35.     if(n < MXN) return pre_mu[n];
  36.     if(mp1[n] == inf) return 0;
  37.     if(mp1[n]) return mp1[n];
  38.     LL ans = 1;
  39.     for(LL L = 2, R; L <= n; L = R + 1) {
  40.         R = n/(n/L);
  41.         ans -= (R-L+1LL)%mod*solve_mu(n/L);
  42.     }
  43.     mp1[n] = ans;
  44.     if(mp1[n] == 0) mp1[n] = inf;
  45.     return ans;
  46. }
  48. LL solve_phi(LL n) {
  49.     if(n < MXN) return pre_phi[n];
  50.     if(mp2[n]) return mp2[n];
  51.     LL ans = n%mod*(n+1)%mod*inv2%mod;
  52.     for(LL L = 2, R; L <= n; L = R + 1) {
  53.         R = n/(n/L);
  54.         ans = (ans - (R-L+1LL)%mod*solve_phi(n/L)%mod)%mod;
  55.     }
  56.     ans = (ans + mod) % mod;
  57.     mp2[n] = ans;
  58.     return ans;
  59. }
  60. int main() {
  61.     inv2 = 499122177;
  62.     init_prime();
  63.     scanf("%lld", &n);
  64.     LL ans = 0;
  65.     for(LL L = 1, R; L <= n; L = R + 1) {
  66.         R = n/(n/L);
  67.         ans = (ans + (solve_mu(R)-solve_mu(L-1))%mod*(solve_phi(n/L)*2LL%mod-1LL)%mod+mod)%mod;
  68.     }
  69.     printf("%lld\n", (ans+mod)%mod);
  70.     return 0;
  71. }

code2:利用莫比乌斯反演和迪利克雷卷积

  1. #include<bits/stdc++.h>
  2. #define fi first
  3. #define se second
  4. #define iis std::ios::sync_with_stdio(false);cin.tie(0)
  5. #define pb push_back
  6. #define o2(x) (x)*(x)
  7. using namespace std;
  8. typedef long long LL;
  9. typedef pair<int, int> pii;
  10. typedef pair<LL, LL> pll;
  12. const int INF = 0x3f3f3f3f;
  13. const LL MOD = 998244353;
  14. const int MXN = 5e6 + 6;
  16. LL n, q, m;
  17. int noprime[MXN], pp[MXN/2], pcnt;
  18. int mu[MXN], phi[MXN];
  19. int pre_mu[MXN];
  20. LL mumu[MXN];
  21. unordered_map<LL,LL>mp1,mp2;
  23. void init_rime() {
  24.     noprime[0] = noprime[1] = 1;
  25.     mu[1] = 1;
  26.     for(int i = 2; i < MXN; ++i) {
  27.         if(!noprime[i]) pp[pcnt++] = i, mu[i]=-1;
  28.         for(int j = 0; j < pcnt && i*pp[j] < MXN; ++j) {
  29.             noprime[i*pp[j]] = 1;
  30.             mu[i*pp[j]] = -mu[i];
  31.             if(i % pp[j] == 0) {
  32.                 mu[i*pp[j]] = 0;
  33.                 break;
  34.             }
  35.         }
  36.     }
  37.     for(int i = 1; i < MXN; ++i) pre_mu[i] = pre_mu[i-1] + mu[i], mumu[i] = mu[i];
  38.     for(int i = 2; i < MXN; ++i) {
  39.         for(int j = i; j < MXN; j += i) {
  40.             mumu[j] += mu[i]*mu[j/i];
  41.             if(mumu[j] >= MOD) mumu[j] %= MOD;
  42.         }
  43.     }
  44.     for(int i = 2; i < MXN; ++i) mumu[i] = (mumu[i]+mumu[i-1]+MOD)%MOD;
  45. }
  47. LL solve_u(LL n) {
  48.     if(n < MXN) return pre_mu[n];
  49.     if(mp1.count(n)) return mp1[n];
  50.     LL ans = 1;
  51.     for(LL L = 2, R; L <= n; L = R + 1) {
  52.         R = n/(n/L);
  53.         ans = (ans - (R-L+1)%MOD*solve_u(n/L)%MOD+MOD)%MOD;
  54.     }
  55.     mp1[n] = ans;
  56.     return ans;
  57. }
  58. LL solve_uu(LL n) {
  59.     if(n < MXN) return mumu[n];
  60.     if(mp2.count(n)) return mp2[n];
  61.     LL ans = 0;
  62.     for(LL L = 1, R; L <= n; L = R + 1) {
  63.         R = n/(n/L);
  64.         ans = (ans + (solve_u(R)-solve_u(L-1)+MOD)%MOD*solve_u(n/L)%MOD);
  65.         if(ans >= MOD) ans %= MOD;
  66.     }
  67.     mp2[n] = (ans+MOD)%MOD;
  68.     return ans;
  69. }
  70. int main(int argc, char const *argv[]) {
  71.     init_rime();
  72.     scanf("%lld", &n);
  73.     LL ans = 0;
  74.     for(LL L = 1, R; L <= n; L = R + 1) {
  75.         R = n/(n/L);
  76.         ans = (ans + (n/L)%MOD*(n/L)%MOD*(solve_uu(R)-solve_uu(L-1)+MOD)%MOD)%MOD;
  77.     }
  78.     printf("%lld\n", (ans+MOD)%MOD);
  79.     return 0;
  80. }
  81. /*
  82. void init_rime() {
  83.     noprime[0] = noprime[1] = 1;
  84.     mu[1] = 1;mumu[1] = 1;
  85.     for(int i = 2; i < MXN; ++i) {
  86.         if(!noprime[i]) pp[pcnt++] = i, mu[i]=-1, mumu[i]=-2;
  87.         for(int j = 0; j < pcnt && i*pp[j] < MXN; ++j) {
  88.             noprime[i*pp[j]] = 1;
  89.             mu[i*pp[j]] = -mu[i];
  90.             mumu[i*pp[j]] = mumu[i]*mumu[pp[j]];
  91.             if(i % pp[j] == 0) {
  92.                 mu[i*pp[j]] = 0;
  93.                 if((i/pp[j])%pp[j]) mumu[i*pp[j]] = mumu[i/pp[j]];
  94.                 else mumu[i*pp[j]] = 0;
  95.                 break;
  96.             }
  97.         }
  98.     }
  99.     for(int i = 1; i < MXN; ++i) pre_mu[i] = pre_mu[i-1] + mu[i];
  100.     for(int i = 2; i < MXN; ++i) mumu[i] = (mumu[i]+mumu[i-1]+MOD)%MOD;
  101. }
  102. */

H div2 涂鸦(计数)

  1. //H
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define lson rt<<1
  6. #define rson rt<<1|1
  7. #define iis std::ios::sync_with_stdio(false)
  8. #define pb push_back
  9. namespace lh {
  10. #define o2(x) (x)*(x)
  11.     using namespace std;
  12.     typedef long long LL;
  13.     typedef unsigned long long uLL;
  14.     typedef pair<int, LL> pii;
  15. }
  17. using namespace lh;
  19. const int mod = 998244353;
  20. const int INF = 0x3f3f3f3f;
  21. const int MXN = 2e5 + 7;
  22. const int N = 1e3 + 7;
  24. int n, m;
  25. int ar[N][N];
  26. LL dp[1005][1005];
  27. LL ksm(LL a, LL b) {
  28.     LL res = 1;
  29.     while(b) {
  30.         if(b & 1) res = res *a % mod;
  31.         a = a * a % mod;
  32.         b >>= 1;
  33.     }
  34.     return res;
  35. }
  36. int da[N][N], di[N][N], dj[N][N],dij[N][N];
  37. void ADD(LL &a) {
  38.     if(a >= mod) a %= mod;
  39. }
  40. int lowbit(int x) {
  41.     return x&(-x);
  42. }
  43. //sumxy = (x+1)(y+1)sigma(d[i][j])-(y+1)sigma(i*d[i][j])-(x+1)sigma(j*d[i][j])+sigma(i*j*d[i][j])
  44. void add(int x, int y, int z){
  45.   for(int i=x;i<=n;i+=lowbit(i)){
  46.     for(int j=y;j<=m;j+=lowbit(j)){
  47.       da[i][j] += z; di[i][j] += z*x; dj[i][j] += z*y; dij[i][j] += z*x*y;
  48.     }
  49.   }
  50. }
  51. void update(int xa,int ya,int xb,int yb,int z){
  52.   add(xa,ya,z);add(xa,yb+1,-z);add(xb+1,ya,-z);add(xb+1,yb+1,z);
  53. }
  54. int query(int x, int y){
  55.   int res = 0;
  56.   for(int i = x; i>0; i -= lowbit(i)){
  57.     for(int j = y; j>0; j -= lowbit(j)){
  58.       res += (x+1)*(y+1)*da[i][j] - (y+1)*di[i][j] - (x+1)*dj[i][j] + dij[i][j];
  59.     }
  60.   }
  61.   return res;
  62. }
  63. int ask(int xa,int ya,int xb,int yb){
  64.   return query(xb,yb)-query(xb,ya-1)-query(xa-1,yb)+query(xa-1,ya-1);
  65. }
  67. void init(){
  68.   for(int i = 1; i <= n; ++i){
  69.     for(int j = 1; j <= m; ++j){
  70.       LL tmp = dp[i][j]-dp[i-1][j]-dp[i][j-1]+dp[i-1][j-1];
  71.       tmp = (tmp%mod + mod)%mod;
  72.       add(i,j,tmp);
  73.     }
  74.   }
  75. }
  76. int main(int argc, char const *argv[]) {
  77.     int q;
  78.     scanf("%d%d%d", &n, &m, &q);
  79.     LL tmp, l, r;
  80.     LL two = ksm(2, mod-2);
  81.     for(int i = 1; i <= n; ++i) {
  82.         scanf("%lld%lld", &l, &r);
  83.         tmp = (r-l+1)*(r-l+2)%mod;
  84.         tmp = ksm(tmp, mod - 2);
  85.         for(int j = l; j <= r; ++j) {
  86.             dp[i][j] = (j-l+1)*(r-j+1)%mod*tmp%mod*2%mod;
  87.         }
  88.     }
  89.     LL ANS = 0;
  90.     //init();
  91.     //printf("%lld %d\n", ANS, ask(1, 1, n, m));
  92.     int x1, y1, x2, y2;
  93.     while(q--) {
  94.         scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
  95.         if(x1+y1 > x2+y2) {
  96.             swap(x1,x2);
  97.             swap(y1,y2);
  98.         }
  99.         //printf("%d %d %d %d\n", x1,y1,x2,y2);
  100.         update(x1,y1,x2,y2,1);
  101.     }
  102.     ANS = 0;
  103.     for(int i = 1; i <= n; ++i) {
  104.         for(int j = 1; j <= m; ++j) {
  105.             if(ask(i,j,i,j)) continue;
  106.             ANS += dp[i][j];
  107.             ADD(ANS);
  108.         }
  109.     }
  110.     printf("%lld\n", ANS);
  111.     return 0;
  112. }

I div1&2 (并查集)

  1. //I石头剪刀布
  2. //不能路径压缩，必须按秩合并
  3. #include<bits/stdc++.h>
  4. #define clr(a, b) memset(a,b,sizeof((a)))
  5. #define lson rt<<1
  6. #define rson rt<<1|1
  7. using namespace std;
  8. typedef long long LL;
  10. const int MXN = 1e6 + 7;
  11. const LL MOD = 998244353;
  12. int n, m, Q;
  13. int fa[MXN], rk[MXN];
  14. LL all[MXN], ke[MXN];//场数，客场数
  15. LL ALL, KE;
  16. LL ksm(LL a, LL b) {
  17.     LL res = 1;
  18.     while(b) {
  19.         if(b&1) res = res * a % MOD;
  20.         a = a * a % MOD;
  21.         b >>= 1;
  22.     }
  23.     return res;
  24. }
  25. int Fi(int x) {
  26.     ALL = all[x]; KE = ke[x];
  27.     if(x == fa[x]) return x;
  28.     int tmp = fa[x];
  29.     while(tmp != fa[tmp]) {
  30.         ALL += all[tmp]; KE += ke[tmp];
  31.         tmp = fa[tmp];
  32.     }
  33.     ALL += all[tmp]; KE += ke[tmp];
  34.     return tmp;
  35. }
  36. int Find(int x) {
  37.     int ans = x;
  38.     if(x != fa[x]) {
  39.         int t = fa[x];
  40.         ans = Find(fa[x]);
  41.         ALL += all[t];
  42.         KE += ke[t];
  43.     }
  44.     return ans;
  45. }
  46. void UNION(int x,int y) {
  47.     int pa = Find(x), pb = Find(y);
  48.     if(pa == pb) return ;
  49.     ++ all[pa]; ++ all[pb];
  50.     ++ ke[pb];
  51.     if(rk[pa] < rk[pb]) swap(pa, pb);
  52.     ++ rk[pa];
  53.     all[pb] -= all[pa];
  54.     ke[pb] -= ke[pa];
  55.     fa[pb] = pa;
  56. }
  57. int main() {
  58.     scanf("%d%d", &n, &m);
  59.     for(int i = 1; i <= n; ++i) fa[i] = i;
  60.     while(m --) {
  61.         int opt, x, y;
  62.         scanf("%d%d", &opt, &x);
  63.         if(opt == 1) {
  64.             scanf("%d", &y);
  65.             UNION(x, y);
  66.         }else {
  67.             ALL = all[x]; KE = ke[x];
  68.             Fi(x);//Find(x);
  69.             LL a = ALL - KE, b = KE, c = n - a - b;
  70.             printf("%lld\n", ksm(3, c) * ksm(2, a) % MOD);
  71.         }
  72.     }
  73.     return 0;
  74. }

E div1&2 精简改良 生成树dp

迷之复杂度

  1. #include<bits/stdc++.h>
  2. #define fi first
  3. #define se second
  4. #define iis std::ios::sync_with_stdio(false)
  5. #define eb emplace_back
  6. #define o2(x) (x)*(x)
  7. using namespace std;
  8. typedef long long LL;
  9. typedef pair<int, int> pii;
  10. inline LL read(){
  11.   LL x=0;int f=0;char ch=getchar();
  12.   while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
  13.   while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
  14.   return x=f?-x:x;
  15. }
  16. inline void write(LL x) {
  17.     if(x==0){putchar('0'),putchar('\n');return;}
  18.     if(x < 0) {putchar('-');x=-x;}
  19.     static char s[23];int l = 0;
  20.     while(x!=0)s[l++]=x%10+48,x/=10;
  21.     while(l)putchar(s[--l]);
  22.     putchar('\n');
  23. }
  24. const int INF = 0x3f3f3f3f;
  25. const LL mod = 998244353;
  26. const int MXN = 1e3 + 7;
  27. int n, m;
  28. std::vector<int> vs[(1<<14)+5];
  29. LL dp[(1<<14)+5][20];
  30. LL dis[20][20], N[(1<<14)+5];
  31. int sta;
  32. //for (int x = S; x; x = (x-1)&S))枚举子集，复杂度2^m，会漏掉空集
  34. int main(int argc, char const *argv[]) {
  35. #ifndef ONLINE_JUDGE
  36.     freopen("E://ADpan//in.in", "r", stdin);
  37.     //freopen("E://ADpan//out.out", "w", stdout);
  38. #endif
  39.     n = read(), m = read();
  40.     //scanf("%d%d", &n, &m);
  41.     memset(dis, -1, sizeof(dis));
  42.     memset(dp, -1, sizeof(dp));
  43.     for(int i = 0, a, b, c; i < m; ++i) {
  44.         a = read(), b = read(), c = read();
  45.         //scanf("%d%d%d", &a, &b, &c);
  46.         dis[a][b] = dis[b][a] = c;
  47.     }
  48.     sta = 1 << n;
  49.     for(int i = 1; i < sta; ++i) {
  50.         for(int j = 0; j < n; ++j) if((i>>j)&1) vs[i].eb(j+1);
  51.         N[i] = vs[i].size();
  52.     }
  53.     LL tmp, ans = 0;
  54.     for(int i = 1, j = 1; i < sta; i <<= 1, ++j) dp[i][j] = 0;
  55.     for(int t = 1; t < sta; ++t) {
  56.         for(int old = (t-1)&t; old; old = (old-1)&t) {
  57.             for(auto i: vs[t]) {
  58.                 if(dp[t-old][i]==-1) continue;
  59.                 for(auto j: vs[old]) {
  60.                     if(dis[i][j]==-1||dp[old][j]==-1) continue;
  61.                     tmp = dp[old][j]+dp[t-old][i]+dis[i][j]*N[old]*(n-N[old]);
  62.                     dp[t][i] = (dp[t][i]>tmp?dp[t][i]:tmp);
  63.                 }
  64.             }
  65.         }
  66.     }
  67.     for(int i = 1; i <= n; ++i) ans = (ans>dp[sta-1][i]?ans:dp[sta-1][i]);
  68.     write(ans);
  69.     //printf("%lld\n", ans);
  70.     return 0;
  71. }

day4

I div2 咆咆咆哮 (贪心)

  1. //I
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. namespace lh {
  7. #define o2(x) (x)*(x)
  8.     using namespace std;
  9.     typedef long long LL;
  10.     typedef unsigned long long uLL;
  11.     typedef pair<int, int> pii;
  12.     typedef pair<int, LL> piL;
  13. }
  15. using namespace lh;
  17. const int MXN = 1e6 + 5;
  18. const double eps = 1e-8;
  20. int n;
  21. LL dp[1005][1005];
  22. struct lp {
  23.     LL a, b;
  24. }cw[1005], tmp[1005];
  25. bool cmp(const lp &A, const lp &B) {
  26.     if(A.a != B.a) return A.a > B.a;
  27.     return A.b < B.b;
  28. }
  29. bool cmp1(const lp &A, const lp &B) {
  30.     return (A.b-A.a) > (B.b-B.a);
  31. }
  32. int main(){
  33.     scanf("%d", &n);
  34.     LL ans = 0;
  35.     for(int i = 1; i <= n; ++i) {
  36.         scanf("%lld%lld", &cw[i].a, &cw[i].b);
  37.         ans += cw[i].a;
  38.     }
  39.     for(int t = 1; t < n; ++t) {
  40.         for(int i = 1; i <= n; ++i) {
  41.             tmp[i].a = cw[i].a;
  42.             tmp[i].b = cw[i].b * (n-t);
  43.         }
  44.         sort(tmp + 1, tmp + 1 + n, cmp1);
  45.         LL ret = 0;
  46.         for(int i = 1; i <= n; ++i) {
  47.             if(i <= t) ret += tmp[i].b;
  48.             else ret += tmp[i].a;
  49.         }
  50.         ans = max(ans, ret);
  51.     }
  52.     printf("%lld\n", ans);
  53.     return 0;
  54. }

I div1 咆咆咆哮 (奇怪的三分+贪心)

  1. //蜜汁三分
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. #define eb emplace_back
  7. #define o2(x) (x)*(x)
  8. using namespace std;
  9. typedef long long LL;
  10. typedef pair<int, int> pii;
  11. inline LL read(){
  12.   LL x=0;int f=0;char ch=getchar();
  13.   while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
  14.   while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
  15.   return x=f?-x:x;
  16. }
  17. inline void write(LL x) {
  18.     if(x==0){putchar('0'),putchar('\n');return;}
  19.     if(x < 0) {putchar('-');x=-x;}
  20.     static char s[23];int l = 0;
  21.     while(x!=0)s[l++]=x%10+48,x/=10;
  22.     while(l)putchar(s[--l]);
  23.     putchar('\n');
  24. }
  25. const int INF = 0x3f3f3f3f;
  26. const LL mod = 998244353;
  27. const int MXN = 1e5 + 7;
  29. int n;
  30. LL dp[1005][1005];
  31. struct lp {
  32.     LL a, b;
  33. }cw[1005], tmp[1005];
  34. bool cmp(const lp &A, const lp &B) {
  35.     if(A.a != B.a) return A.a > B.a;
  36.     return A.b < B.b;
  37. }
  38. bool cmp1(const lp &A, const lp &B) {
  39.     return (A.b-A.a) < (B.b-B.a);
  40. }
  41. LL check(int t) {//t个b
  42.     LL ans = 0;
  43.     priority_queue<LL> Q;
  44.     for(int i = 1; i <= n; ++i) {
  45.         ans += cw[i].a;
  46.         Q.push(cw[i].b*(n-t)-cw[i].a);
  47.     }
  48.     while(t --) {
  49.         ans += Q.top(); Q.pop();
  50.     }
  51.     return ans;
  52. }
  53. int main(){
  54.     n = read();
  55.     LL ans = 0;
  56.     for(int i = 1; i <= n; ++i) {
  57.         cw[i].a = read(), cw[i].b = read();
  58.         ans += cw[i].a;
  59.     }
  60.     int L = 0, R = n, mid1, mid2;
  61.     LL res1, res2;
  62.     while(L <= R) {
  63.         mid1 = (2*L+R)/3;mid2 = (L+2*R)/3;
  64.         //mid1 = (L+R)/2;mid2 = (mid1+R)/2;
  65.         res1 = check(mid1);
  66.         res2 = check(mid2);
  67.         if(res1 > res2) R = mid2-1, ans = max(ans, res1);
  68.         else L = mid1+1, ans = max(ans, res2);
  69.     }
  70.     write(ans);
  71.     return 0;
  72. }

C div2 (规律)

  1. //C
  2. 相邻点度数均大于1就不行或者链长为3

K div2 两条路径(树形dp)

  1. //K
  2. #include<bits/stdc++.h>
  3. #define fi first
  4. #define se second
  5. #define iis std::ios::sync_with_stdio(false)
  6. namespace lh {
  7. #define o2(x) (x)*(x)
  8.     using namespace std;
  9.     typedef long long LL;
  10.     typedef unsigned long long uLL;
  11.     typedef pair<int, int> pii;
  12.     typedef pair<int, LL> piL;
  13. }
  15. using namespace lh;
  17. const int MXN = 1e6 + 5;
  18. const double eps = 1e-8;
  20. int n, x;
  21. std::vector<int> mp[MXN];
  22. LL ar[MXN], mi[40], dp[MXN];
  23. void dfs(int u,int ba) {
  24.     dp[u] = ar[u];
  25.     for(auto v: mp[u]) {
  26.         if(v == ba) continue;
  27.         dfs(v, u);
  28.         dp[u] = max(dp[u], dp[v] + ar[u]);
  29.     }
  30. }
  31. bool cmp(LL a, LL b) {
  32.     return a > b;
  33. }
  34. int main(){
  35.     mi[0] = 1;
  36.     for(int i = 1; i <= 32; ++i) mi[i] = mi[i-1] * 2;
  37.     scanf("%d", &n);
  38.     for(int i = 1; i <= n; ++i) {
  39.         scanf("%lld", &ar[i]);
  40.         if(ar[i] >= 0) ar[i] = mi[ar[i]-1];
  41.         else ar[i] = -mi[-ar[i]-1];
  42.         //printf("%lld\n", ar[i]);
  43.     }
  44.     for(int i = 1, a, b; i < n; ++i) {
  45.         scanf("%d%d", &a, &b);
  46.         mp[a].push_back(b);
  47.         mp[b].push_back(a);
  48.     }
  49.     int Q; scanf("%d", &Q);
  50.     while(Q --) {
  51.         scanf("%d", &x);
  52.         for(int i = 1; i <= n; ++i) dp[i] = 0;
  53.         dfs(x, x);
  54.         std::vector<LL> vs;
  55.         for(auto v: mp[x]) {
  56.             if(dp[v] >= 0) vs.push_back(dp[v]);
  57.         }
  58.         sort(vs.begin(), vs.end(), cmp);
  59.         LL ans = ar[x];
  60.         for(int i = 0; i < vs.size() && i < 4; ++i) ans += vs[i];
  61.         //printf("**%lld\n", ans);
  62.         if(ans < 0) printf("-"), ans = -ans;
  63.         std::vector<int> v;
  64.         while(ans) {
  65.             if(ans & 1) v.push_back(1);
  66.             else v.push_back(0);
  67.             ans /= 2;
  68.         }
  69.         reverse(v.begin(), v.end());
  70.         for(auto s: v) {
  71.             printf("%d", s);
  72.         }
  73.         printf("\n");
  74.     }
  75.     return 0;
  76. }

D div1&2 欧拉回路 (性质)

  1. //D欧拉回路
  2. //保证度数为偶数即可，注意特判2的情况
  3. #include<bits/stdc++.h>
  4. #define fi first
  5. #define se second
  6. #define iis std::ios::sync_with_stdio(false)
  7. #define eb emplace_back
  8. #define o2(x) (x)*(x)
  9. using namespace std;
  10. typedef long long LL;
  11. typedef pair<int, int> pii;
  12. inline LL read(){
  13.   LL x=0;int f=0;char ch=getchar();
  14.   while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
  15.   while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
  16.   return x=f?-x:x;
  17. }
  18. inline void write(LL x) {
  19.     if(x==0){putchar('0'),putchar('\n');return;}
  20.     if(x < 0) {putchar('-');x=-x;}
  21.     static char s[23];int l = 0;
  22.     while(x!=0)s[l++]=x%10+48,x/=10;
  23.     while(l)putchar(s[--l]);
  24.     putchar('\n');
  25. }
  26. const int INF = 0x3f3f3f3f;
  27. const LL mod = 998244353;
  28. const int MXN = 1e3 + 7;
  30. int n, m;
  32. int main(int argc, char const *argv[]) {
  33.     scanf("%d%d", &n, &m);
  34.     if(m % 2 == 0 && n % 2 == 1) swap(n, m);
  35.     if(m == 2) swap(n, m);
  36.     LL ans = (n-1)*m + n*(m-1);
  37.     if(n == 2) {
  38.         ans = ans + ((m-2)/2)*2+(m%2==1);
  39.     }else if(n % 2 == 1) {
  40.         ans = (ans+((m-2)/2)*2+4+((n-2)/2)*2);
  41.     }else if(m % 2 == 1) {
  42.         ans = (ans+((m-2)/2)*2+4+(n-2)/2+(n-4)/2);
  43.     }else {
  44.         ans = (ans+m+n-4);
  45.     }
  46.     printf("%lld\n", ans);
  47.     return 0;
  48. }

G div1&2 置置置换

\(dp[i] = \sum_{j=1,j+=2}^{j<=i}dp[j-1]*dp[i-j]*COMB(i-1, j-1)\)

  1. /*
  2. 题意：有多少个n的排列满足：偶数位置小于奇数位置，奇数位置大于偶数位置；减增减增。。。的序列
  3. */
  4. //上凸和下凸的答案是一样的
  5. //dp[i]表示1-i的排列的合法情况总数
  6. //枚举最大的数字i所在的位置：dp[i] = \sum_{j=1,j+=2}^{j<=i}dp[j-1]*dp[i-j]*COMB(i-1, j-1)
  7. #include<bits/stdc++.h>
  8. #define fi first
  9. #define se second
  10. #define iis std::ios::sync_with_stdio(false)
  11. #define eb emplace_back
  12. #define o2(x) (x)*(x)
  13. using namespace std;
  14. typedef long long LL;
  15. typedef pair<int, int> pii;
  16. inline LL read(){
  17.   LL x=0;int f=0;char ch=getchar();
  18.   while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
  19.   while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
  20.   return x=f?-x:x;
  21. }
  22. inline void write(LL x) {
  23.     if(x==0){putchar('0'),putchar('\n');return;}
  24.     if(x < 0) {putchar('-');x=-x;}
  25.     static char s[23];int l = 0;
  26.     while(x!=0)s[l++]=x%10+48,x/=10;
  27.     while(l)putchar(s[--l]);
  28.     putchar('\n');
  29. }
  30. const int INF = 0x3f3f3f3f;
  31. const LL mod = 1e9 + 7;
  32. const int MXN = 1e4 + 7;
  34. LL F[MXN], Inv[MXN], invF[MXN];
  35. LL ksm(LL a, int b) {
  36.     LL res = 1;
  37.     for(;b;b>>=1,a=a*a%mod) {
  38.         if(b&1) res=res*a%mod;
  39.     }
  40.     return res;
  41. }
  42. int n;
  43. LL dp[MXN];
  44. void init() {
  45.     Inv[1] = F[1] = invF[1]  = 1;
  46.     Inv[0] = F[0] = invF[0] = 1;
  47.     for(int i = 2; i < MXN; ++i) {
  48.         F[i] = F[i-1] * i % mod;
  49.         Inv[i] = (mod-mod/i)*Inv[mod%i] % mod;
  50.         invF[i] = invF[i-1] * Inv[i] % mod;
  51.     }
  52. }
  53. LL COMB(int n, int m) {
  54.     if(n == m || m == 0) return 1;
  55.     if(n < m) return 0;
  56.     return F[n]*invF[m]%mod*invF[n-m]%mod;
  57. }
  58. int main(){
  59.     n = read();
  60.     init();
  61.     dp[0] = dp[1] = dp[2] = 1;
  62.     for(int i = 3; i <= n; ++i) {
  63.         for(int j = 1; j <= i; j += 2) {
  64.             dp[i] = (dp[i] + dp[j-1] * dp[i-j]%mod * COMB(i-1, j-1)%mod)%mod;
  65.         }
  66.     }
  67.     printf("%lld\n", dp[n]);
  68.     return 0;
  69. }
  70. /*
  71. d4G:
  72. f[i][j]表示前i个位置，在剩下的第j小的方案数，枚举所有大于它的k。用前缀和或后缀和维护
  73. 1I/4G/7J
  74. */

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