谁能做出来这道题（通过率0%）

全部评论 3

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  1

  

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  白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白白rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrgggggggggggggggggggggggggggggggggggggggggggggggg

  2025-03-13 来自 浙江

  0
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  复仇者_༺ཌༀ皴·泰ༀད༻

  

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  现在有人做了：

  #include <bits/stdc++.h>
  #define x first
  #define y second
  using namespace std;
  typedef pair<int, int> PII;
  const int N = 100010;
  int n, m, T;
  int score[N], last[N];
  bool st[N];
  PII q[N];
  int main()
  {
      scanf("%d%d%d", &n, &m, &T);
      for (int i = 0; i < m; i ++ ) 
          scanf("%d%d", &q[i].x, &q[i].y);
      sort(q, q + m);
  
      for (int i = 0; i < m;)
      {
          int j = i;
          while (j < m && q[j] == q[i]) j ++ ;
          int t = q[i].x, id = q[i].y, cnt = j - i;
          i = j;
  
          score[id] -= t - last[id] - 1;
          if (score[id] < 0) 
              score[id] = 0;
          if (score[id] <= 3) 
              st[id] = false;
  
          score[id] += cnt * 2;
          if (score[id] > 5) 
              st[id] = true;
  
          last[id] = t;
      }
      for (int i = 1; i <= n; i ++ )
          if (last[i] < T)
          {
              score[i] -= T - last[i];
              if (score[i] <= 3) st[i] = false;
          }
      int res = 0;
      for (int i = 1; i <= n; i ++ ) 
          res += st[i];
      printf("%d\n", res);
      return 0;
  }
  

  2025-03-12 来自 浙江

  0
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  NULL

  

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  很轻松啊

  2023-10-06 来自 广东

  0
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    只会WA的姜禹卓

    回复NULL

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    是的，并不是很难，但没人做，我暂时不想做这道题

    2023-10-06 来自 浙江

    0