Codeforces Round #527 (Div. 3) 总结 A B C D1 D2 F

传送门

A

贪心的取

每个字母n/k次

令r=n%k

让前r个字母各取一次

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 int t, n, k;

 int main() {
     cin >> t;

     while (t--) {
         cin >> n >> k;
         int x = n / k, r = n - x * k;
         rep(i, , k) rep(j, , x) {
             cout << (char)('a' + i - );
         }
         rep(i, , r) {
             cout << (char)('a' + i - );
         }
         cout << '\n';
     }
     return ;
 }

B

排序完连续两个比较

证明一下吧：max(a[2] - a[1]), a[4] - a[3]) <= max(a[3] - a[1]), a[4] - a[2]) (后者的间隙大)

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 const int N = 1e5 + ;

 int n, a[N];

 int main() {
     cin >> n;

     rep(i, , n) {
         cin >> a[i];
     }

     sort(a + , a + n + );

     int ans = ;
     rep(i, , n / ) {
         ans += a[i * ] - a[i *  - ];
     }

     cout << ans << '\n';

     return ;
 }

C

题意难理解

找出最长的两个串 判断哪个是最长前缀

f[len]代表着长度为len的前后缀是否被选 用来避免长度为len的两个子串都是前缀或都是后缀的情况

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (ll i = a; i <= b; ++i)

 const int N = ;

 ll n;
 bool f[N];
 string s[N];

 int main() {
     cin >> n; ll m =  * n - ;

     string s1 = "", s2 = "";
     rep(i, , m) {
         cin >> s[i];
         if (s[i].size() > s1.size()) s1 = s[i];
         else if (s[i].size() == s1.size()) s2 = s[i];
     } 

     ll cnt = ;
     rep(i, , m) if (s1.substr(, s[i].size()) == s[i]) {
         if (s[i] != s2) cnt++;
     }

     string pre;
     if (cnt >= n -  && s1.substr(, s1.size() - ) == s2.substr(, s1.size() - )) pre = s1; else pre = s2;

     rep(i, , m) {
         if (pre.substr(, s[i].size()) == s[i] && !f[s[i].size()]) {
             cout << 'P';
             f[s[i].size()] = ;
         }
         else cout << 'S';
     }

     return ;
 }

D1 D2

两题类似

用一个栈 若当前高度和栈顶一致就弹出 具体判断见代码

 //D1
 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 const int N = 2e5 + ;

 int n;
 bool a[N];
 stack <bool> st;

 int main() {
     scanf("%d", &n);

     int x;
     rep(i, , n) {
         scanf("%d", &x);
         a[i] = x & ;
     }

     rep(i, , n) {
         if(st.empty())
             st.push(a[i]);
         else if(a[i] == st.top())
             st.pop();
         else
             st.push(a[i]);
     }

     st.size() >  ? puts("NO") : puts("YES");

     return ;
 }

 //D2
 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 int n, mx;
 stack <int> s;

 int main() {
     scanf("%d", &n);

     int x;
     rep(i, , n) {
         scanf("%d",&x);
         mx=max(mx,x);
         if (s.empty())
             s.push(x);
         else if (s.top()==x)
             s.pop();
         else if (s.top()>=x)
             s.push(x);
         else {
             puts("NO");
             return ;
         }
     }

     if (!s.size())
         puts("YES");
     else if (s.size() ==  && s.top() == mx)
         puts("YES");
     else
         puts("NO");

     return ;
 }

E

blank

F

初始思路是：维护树上的一堆和乱搞

其实dfs2的操作十分巧妙

1.res -= sum[v] 代表 v子树下的每个点的距离都少了1

2.sum[u] -= sum[v]; 为下一步做准备

3.res += sum[u]; 代表 u子树下(除)的每个点的距离都多了1

4.sum[v]+=sum[u]; dfs2到下一层的时候 (假如边为（v,w）) res+= sum[v](即步骤3)的时候v点外u点子树下的一些点到w的距离也多了1

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;

 #define rep(i, a, b) for (int i = a; i <= b; ++i)

 const int N = 2e5 + ;

 int n;

 ll a[N], sum[N], res, ans;

 vector <int> e[N];

 void dfs1(int u, int p = , int d = ) {
     res += d * a[u];
     sum[u] = a[u];
     for (auto v : e[u]) if (v != p) {
         dfs1(v, u, d + );
         sum[u] += sum[v];
     }
 }

 void dfs2(int u, int p = ) {
     ans = max(ans, res);
     for (auto v : e[u]) if (v != p) {
         res -= sum[v];
         sum[u] -= sum[v];
         res += sum[u];
         sum[v] += sum[u];

         dfs2(v, u);

         sum[v] -= sum[u];
         res -= sum[u];
         sum[u] += sum[v];
         res += sum[v];
     }
 }

 int main() {
     scanf("%d", &n);

     rep(i, , n) {
         scanf("%lld", &a[i]);
     }

     int u, v;
     rep(i, , n - ) {
         scanf("%d%d", &u, &v);
         e[u].push_back(v);
         e[v].push_back(u);
     }

     dfs1();
     dfs2();

     printf("%lld\n", ans);

     return ;
 }