1 条题解
-
0
#include <bits/stdc++.h> #define int long long using namespace std; const int MAXN = 4096; int n, m; int lowbit(int x) { return x & (-x); } // t[x][y] 存 t[x-lowbit(x)+1 ~ x][y-lowbit(y)+1 ~ y] 的元素之和 int t[MAXN + 5][MAXN + 5]; void update(int x, int y, int z) { for (int i = x; i <= n; i += lowbit(i)) for (int j = y; j <= m; j += lowbit(j)) t[i][j] += z; } int query(int x, int y) { int res = 0; for (int i = x; i >= 1; i -= lowbit(i)) for (int j = y; j >= 1; j -= lowbit(j)) res += t[i][j]; return res; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cin >> n >> m; int op; while (cin >> op) { if (op == 1) { int x, y, k; cin >> x >> y >> k; update(x, y, k); } if (op == 2) { int a, b, c, d, delta; cin >> a >> b >> c >> d; cout << query(c, d) - query(c, b - 1) - query(a - 1, d) + query(a - 1, b - 1) << "\n"; } } return 0; }
- 1
信息
- ID
- 14246
- 时间
- 1000ms
- 内存
- 256MiB
- 难度
- 6
- 标签
- (无)
- 递交数
- 45
- 已通过
- 15
- 上传者