A - Always Online

Unsolved.

B - Beautiful Now

Solved.

题意：

给出一个n, k  每次可以将n这个数字上的某两位交换，最多交换k次，求交换后的最大和最小值

思路：

很明显有一种思路，对于最小值，尽可能把小的放前面， 对于最大值，尽可能把打的放前面。但是如果有多个最小数字或者最大数字会无法得出放哪个好，因此BFS一下即可

  1.  #include<bits/stdc++.h>
  2.  
  3.  using namespace std;
  4.  
  5.  const int INF = 0x3f3f3f3f;
  6.  const int maxn = 1e5 + ;
  7.  
  8.  struct node{
  9.      int num, idx, step;
  10.      node(){}
  11.      node(int num, int idx, int step) :num(num), idx(idx), step(step){}
  12.  };
  13.  
  14.  int n, k;
  15.  int cnt;
  16.  int arr[maxn];
  17.  
  18.  int BFS1()
  19.  {
  20.      queue<node>q;
  21.      q.push(node(n, cnt, ));
  22.      int ans = INF;
  23.      while(!q.empty())
  24.      {
  25.          node now = q.front();
  26.          q.pop();
  27.          ans = min(ans, now.num);
  28.          if(now.step == k) continue;
  29.          if(now.idx == ) continue;
  30.          int tmp = now.num;
  31.          cnt = ;
  32.          while(tmp)
  33.          {
  34.              arr[++cnt] = tmp % ;
  35.              tmp /= ;
  36.          }
  37.          int Min = now.idx;
  38.          for(int i = now.idx - ; i >= ; --i)
  39.          {
  40.              if(arr[i] == arr[now.idx]) continue;
  41.              if(arr[i] ==  && now.idx == cnt) continue;
  42.              if(arr[i] < arr[Min]) Min = i;
  43.          }
  44.          if(Min == now.idx)
  45.          {
  46.              q.push(node(now.num, now.idx - , now.step));
  47.          }
  48.          else
  49.          {
  50.              for(int i = now.idx - ; i >= ; --i)
  51.              {
  52.                  if(arr[i] == arr[Min])
  53.                  {
  54.                      swap(arr[i], arr[now.idx]);
  55.                      tmp = ;
  56.                      for(int j = cnt; j >= ; --j)
  57.                      {
  58.                          tmp = tmp *  + arr[j];
  59.                      }
  60.                      q.push(node(tmp, now.idx - , now.step + ));
  61.                      swap(arr[i], arr[now.idx]);
  62.                  }
  63.              }
  64.          }
  65.      }
  66.      return ans;
  67.  }
  68.  
  69.  int BFS2()
  70.  {
  71.      queue<node>q;
  72.      q.push(node(n, cnt, ));
  73.      int ans = -INF;
  74.      while(!q.empty())
  75.      {
  76.          node now = q.front();
  77.          q.pop();
  78.          ans = max(ans, now.num);
  79.          if(now.step == k) continue;
  80.          if(now.idx == ) continue;
  81.          int tmp = now.num;
  82.          cnt = ;
  83.          while(tmp)
  84.          {
  85.              arr[++cnt] = tmp % ;
  86.              tmp /= ;
  87.          }
  88.          int Max = now.idx;
  89.          for(int i = now.idx - ; i >= ; --i)
  90.          {
  91.              if(arr[i] == arr[now.idx]) continue;
  92.              if(arr[i] > arr[Max]) Max = i;
  93.          }
  94.          if(Max == now.idx)
  95.          {
  96.              q.push(node(now.num, now.idx - , now.step));
  97.          }
  98.          else
  99.          {
  100.              for(int i = now.idx - ; i >= ; --i)
  101.              {
  102.                  if(arr[i] == arr[Max])
  103.                  {
  104.                      swap(arr[i], arr[now.idx]);
  105.                      tmp = ;
  106.                      for(int j = cnt; j >= ; --j)
  107.                      {
  108.                          tmp = tmp *  + arr[j];
  109.                      }
  110.                      q.push(node(tmp, now.idx - , now.step + ));
  111.                      swap(arr[i], arr[now.idx]);
  112.                  }
  113.              }
  114.          }
  115.      }
  116.      return ans;
  117.  }
  118.  
  119.  int main()
  120.  {
  121.      int t;
  122.      scanf("%d", &t);
  123.      while(t--)
  124.      {
  125.          scanf("%d %d", &n ,&k);
  126.          int tmp = n;
  127.          cnt = ;
  128.          while(tmp)
  129.          {
  130.              cnt++;
  131.              tmp /= ;
  132.          }
  133.          k = min(k, cnt - );
  134.          int ans1 = BFS1();
  135.          int ans2 = BFS2();
  136.          printf("%d %d\n", ans1, ans2);
  137.      }
  138.      return ;
  139.  }

C - Call It What You Want

Unsolved.

D - Daylight

Unsolved.

E - Everything Has Changed

Solved.

题意：

求多个圆的周长并

思路：

对于不想交和内含的直接continue，相切的直接相加。对于相交的可以减去大圆上的弧长，加上小圆的弧长

  1.  #include<bits/stdc++.h>
  2.  
  3.  using namespace std;
  4.  
  5.  const double PI = acos(-1.0);
  6.  const double eps = 1e-;
  7.  
  8.  int sgn(double x)
  9.  {
  10.      if(fabs(x) < eps) return ;
  11.      else return x >  ?  : -;
  12.  }
  13.  
  14.  struct Point{
  15.      double x, y;
  16.      Point(){}
  17.      Point(double _x, double _y)
  18.      {
  19.          x = _x;
  20.          y = _y;
  21.      }
  22.  
  23.      double distance(Point p)
  24.      {
  25.          return hypot(x - p.x, y - p.y);
  26.      }
  27.  }P;
  28.  
  29.  int n;
  30.  double R, r;
  31.  
  32.  int main()
  33.  {
  34.      int t;
  35.      scanf("%d", &t);
  36.      while(t--)
  37.      {
  38.          scanf("%d %lf", &n, &R);
  39.          double ans =  * R * PI;
  40.          for(int i = ; i <= n; ++i)
  41.          {
  42.              scanf("%lf %lf %lf", &P.x, &P.y, &r);
  43.              double dis = P.distance(Point(0.0, 0.0));
  44.              if(sgn(dis - (r + R)) >= ) continue;
  45.              else if(sgn(dis - (R - r)) < ) continue;
  46.              else if(sgn(dis - (R - r)) == )
  47.              {
  48.                  ans +=  * PI * r;
  49.                  continue;
  50.              }
  51.              double arc1 = (R * R + dis * dis - r * r) / (2.0 * R * dis);
  52.              arc1 =  * acos(arc1);
  53.              double arc2 = (r * r + dis * dis - R * R) / (2.0 * r * dis);
  54.              arc2 =  * acos(arc2);
  55.              ans -= R * arc1;
  56.              ans += r * arc2;
  57.          }
  58.          printf("%.10f\n", ans);
  59.      }
  60.      return ;
  61.  }

F - Fireflies

Unsolved.

G - Glad You Came

Upsolved.

题意：

m个区间操作，每次给$[L, R]区间内小于v 的数变为v$

思路：

线段树，维护最大最小值+剪枝，因为数据随机才可以这样做。

  1.  #include <bits/stdc++.h>
  2.  using namespace std;
  3.  
  4.  #define N 5000010
  5.  #define M 100010
  6.  #define ui unsigned int
  7.  #define ll long long
  8.  int t, n, m;
  9.  ui x, y, z, w;
  10.  ui f[N << ];
  11.  ui MOD = (ui) << ;
  12.  
  13.  ui rng61()
  14.  {
  15.      x = x ^ (x << );
  16.      x = x ^ (x >> );
  17.      x = x ^ (x << );
  18.      x = x ^ (x >> );
  19.      w = x ^ y ^ z;
  20.      x = y;
  21.      y = z;
  22.      z = w;
  23.      return z;
  24.  }
  25.  
  26.  struct SEG
  27.  {
  28.      ui lazy[M << ], Max[M << ], Min[M << ];
  29.      void Init()
  30.      {
  31.          memset(lazy, , sizeof lazy);
  32.          memset(Max, , sizeof Max);
  33.          memset(Min, , sizeof Min);
  34.      }
  35.      void pushup(int id)
  36.      {
  37.          Max[id] = max(Max[id << ], Max[id <<  | ]);
  38.          Min[id] = min(Min[id << ], Min[id <<  | ]);
  39.      }
  40.      void pushdown(int id)
  41.      {
  42.          if (!lazy[id]) return;
  43.          lazy[id << ] = lazy[id];
  44.          Max[id << ] = lazy[id];
  45.          Min[id << ] = lazy[id];
  46.          lazy[id <<  | ] = lazy[id];
  47.          Max[id <<  | ] = lazy[id];
  48.          Min[id <<  | ] = lazy[id];
  49.          lazy[id] = ;
  50.      }
  51.      void update(int id, int l, int r, int ql, int qr, ui val)
  52.      {
  53.          if (Min[id] >= val) return;
  54.          if (l >= ql && r <= qr && Max[id] < val)
  55.          {
  56.              lazy[id] = Max[id] = val;
  57.              return;
  58.          }
  59.          pushdown(id);
  60.          int mid = (l + r) >> ;
  61.          if (ql <= mid) update(id << , l, mid, ql, qr, val);
  62.          if (qr > mid) update(id <<  | , mid + , r, ql, qr, val);
  63.          pushup(id);
  64.      }
  65.      int query(int id, int l, int r, int pos)
  66.      {
  67.          if (l == r) return Max[id];
  68.          pushdown(id);
  69.          int mid = (l + r) >> ;
  70.          if (pos <= mid) return query(id << , l, mid, pos);
  71.          else return query(id <<  | , mid + , r, pos);
  72.      }
  73.  }seg;
  74.  
  75.  int main()
  76.  {
  77.      scanf("%d", &t);
  78.      while (t--)
  79.      {
  80.          scanf("%d%d%u%u%u", &n, &m, &x, &y, &z);
  81.          for (int i = ; i <=  * m; ++i) f[i] = rng61();
  82.          seg.Init();
  83.          for (int i = , l, r, v; i <= m; ++i)
  84.          {
  85.              l = f[ * i - ] % n + ;
  86.              r = f[ * i - ] % n + ;
  87.              v = f[ * i] % MOD;
  88.              if (l > r) swap(l, r);
  89.              seg.update(, , n, l, r, v);
  90.          }
  91.          ll res = ;
  92.          for (int i = ; i <= n; ++i)
  93.              res ^= (ll)seg.query(, , n, i) * (ll)i;
  94.          printf("%lld\n", res);
  95.      }
  96.      return ;
  97.  }

H - Hills And Valleys

Upsolved,

题意：

给出一个长为n的数字串，每一位范围是$[0, 9]$，可以翻转其中一段，使得最长非下降子序列最长

思路：

也就是说 可以存在这样一段

$0, 1, 2....., x, (x - 1) , y, (y - 1).... x, y + 1, y + 2..$

我们知道，如果不可以翻转，求最长上升子序列的话，我们可以将原串 和模式串$0123456789$ 求最长公共子序列

那么翻转的话，我们通过枚举翻转的区间$C(2, 10)$ 构造出上述的模式串，再求最长公共子序列

  1.  #include <bits/stdc++.h>
  2.  using namespace std;
  3.  
  4.  #define N 100010
  5.  #define INF 0x3f3f3f3f
  6.  int t, n, m, a[N], b[];
  7.  int dp[N][], tl[N][], tr[N][], res, l, r, ql, qr;
  8.  
  9.  void solve()
  10.  {
  11.      for (int i = ; i <= m; ++i) dp[][i] = , tl[][i] = -, tr[][i] = -;
  12.      for (int i = ; i <= n; ++i) for (int j = ; j <= m; ++j)
  13.      {
  14.          dp[i][j] = dp[i - ][j];
  15.          tl[i][j] = tl[i - ][j];
  16.          tr[i][j] = tr[i - ][j];
  17.          if (a[i] == b[j])
  18.          {
  19.              ++dp[i][j];
  20.              if (j == ql && tl[i][j] == -) tl[i][j] = i;
  21.              if (j == qr) tr[i][j] = i;
  22.          }
  23.          if (j >  && dp[i][j - ] > dp[i][j])
  24.          {
  25.              dp[i][j] = dp[i][j - ];
  26.              tl[i][j] = tl[i][j - ];
  27.              tr[i][j] = tr[i][j - ];
  28.          }
  29.      }
  30.      if (ql ==  && qr == )
  31.      {
  32.          res = dp[n][m];
  33.          l = ;
  34.          r = ;
  35.      }
  36.      else if (dp[n][m] > res && tl[n][m] != - && tr[n][m] != -)
  37.      {
  38.          res = dp[n][m];
  39.          l = tl[n][m];
  40.          r = tr[n][m];
  41.      }
  42.  }
  43.  
  44.  int main()
  45.  {
  46.      scanf("%d", &t);
  47.      while (t--)
  48.      {
  49.          scanf("%d", &n);
  50.          res = , l = , r = ;
  51.          for (int i = ; i <= n; ++i) scanf("%1d", a + i);
  52.          for (int i = ; i <= ; ++i) b[i] = i - ; m = ;    ql = , qr = ;
  53.          solve();
  54.          for (int i = ; i < m; ++i) for (int j = i + ; j <= ; ++j)
  55.          {
  56.              m = ;
  57.              for (int pos = ; pos <= i; ++pos) b[++m] = pos - ;
  58.              ql = m + ;
  59.              for (int pos = j; pos >= i; --pos) b[++m] = pos - ;
  60.              qr = m;
  61.              for (int pos = j; pos <= ; ++pos) b[++m] = pos - ;
  62.              solve();
  63.              //for (int i = 1; i <= m; ++i) printf("%d%c", b[i], " \n"[i == m]);
  64.          }
  65.          printf("%d %d %d\n", res, l, r);
  66.      }
  67.      return ;
  68.  }

I - Innocence

Unsolved.

J - Just So You Know

Unsolved.

K - Kaleidoscope

Unsolved.

L - Lost In The Echo

Unsolved.

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