比赛链接：Here

835A. Key races

理解题意，然后根据情况输出

835B. The number on the board

根据题目意思贪心处理下，总和和k的关系，问你要改几位，改成9就好了啊

  1. const int N = 100000 + 10;
  2. int a[N];
  3. bool cmp(int a, int b) {return a > b;}
  4. int main() {
  5.     cin.tie(nullptr)->sync_with_stdio(false);
  6.     ll k; string s;
  7.     cin >> k >> s;
  8.     int n = int(s.size());
  9.     int cnt = 0;
  10.     for (int i = 0; i < n; i++) {
  11.         cnt += s[i] - '0';
  12.         a[i] = 9 - (s[i] - '0');
  13.     }
  14.     sort(a, a + n, cmp);
  15.     for (int i = 1; i < n; i++) a[i] += a[i - 1];
  16.     if (cnt >= k)  cout << "0";
  17.     else {
  18.         int ans = k - cnt;
  19.         for (int i = 0; i < n; ++i) {
  20.             if (a[i] >= ans) {
  21.                 cout << i + 1;
  22.                 break;
  23.             }
  24.         }
  25.     }
  26. }

835C. Star sky

尝试写转移方程，但发现没必要那么麻烦，考虑二/三维前缀和，由于坐标范围在 Codeforces Round #427 (Div. 2) (A~C) - 图1 之间那么写暴力也没事

  1. const int N = 110;
  2. int sum[N][N][12];
  3. int main() {
  4.     cin.tie(nullptr)->sync_with_stdio(false);
  5.     int n, q, c;
  6.     cin >> n >> q >> c;
  7.     for (int i = 1; i <= n; ++i) {
  8.         int x, y, s;
  9.         cin >> x >> y >> s;
  10.         sum[x][y][s] += 1;
  11.     }
  12.     for (int i = 1; i <= 100; ++i)
  13.         for (int j = 1; j <= 100; ++j)
  14.             for (int k = 0; k <= c; k++)
  15.                 sum[i][j][k] += sum[i - 1][j][k] + sum[i][j - 1][k] - sum[i - 1][j - 1][k];
  17.     while (q--) {
  18.         int t, sx, sy, ex, ey;
  19.         cin >> t >> sx >> sy >> ex >> ey;
  20.         int ans = 0;
  21.         for (int i = 0; i <= c; ++i) {
  22.             int x = (i + t) % (c + 1);
  23.             ans += x * (sum[ex][ey][i] - sum[sx - 1][ey][i] - sum[ex][sy - 1][i] + sum[sx - 1][sy - 1][i]);
  24.             // cout << "demo: " << ans << "\n";
  25.         }
  26.         cout << ans << "\n";
  27.     }
  28. }