链接：https://ac.nowcoder.com/acm/contest/7501/J
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
Given a matrix MM_{}M of size n×mn\times mn×m and two integers a,ba, b_{}a,b, determine weither it is possible to make all entrys of MM_{}M zero by repeatedly choosing a×ba\times ba×b submatrices and reduce the values in the chosen matrices by 1. If possible, print "^_^" in one line, or print "QAQ" in one line.
输入描述:
The first line contains one integer T (1≤T≤100), denoting the number of test cases.
For each test case:
The first line contains four integers n,m,a,b (1≤n,m≤1000,1≤a≤n,1≤b≤m)n,m,a,b~(1\le n,m \le 1000, 1\le a\le n, 1\le b\le m)n,m,a,b (1≤n,m≤1000,1≤a≤n,1≤b≤m), denoting the size of given matrix and the size of chosen submatrices respectively.
The next nn_{}n lines each contains mm_{}m integers Mi,j (0≤Mi,j≤109)M_{i,j}~(0\le M_{i,j} \le 10^9)Mi,j (0≤Mi,j≤109), denoting the entrys of matrix MM_{}M.
It's guaranteed that ∑nm≤106\sum nm \le 10^6∑nm≤106.
输出描述:
Print TT_{}T lines each containing a string "^_^" or "QAQ", denoting the answer to each test case.
示例1
输入
2 2 2 1 2 1 2 1 2 2 3 1 2 1 2 1 1 2 1
输出
QAQ ^_^
说明
For the second case, one possible scheme is to choose (1,1)−(1,2),(1,2)−(1,3),(2,1)−(2,2),(2,2)−(2,3)respectively.
题意：
有一个矩阵，每次可以选择一个大小为a * b的子矩阵，把子矩阵中的每个数减1，问若干操作后原矩阵是否可以都减为0
从左上角开始减，使用树状数组的区间修改和单点查询。
（今天下午为什么没AC：数组开大了导致memset超时
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1e3 + 20;
ll mp[N][N];
ll aa[N][N];
ll n, m, a, b, nn;
ll lowbit(ll k) {
return k & (-k);
}
void add(ll x, ll y, ll v) {
while(x <= nn) { ///nn需要修改
ll ty = y;
while(ty <= nn) ///nn需要修改
mp[x][ty] += v, ty += lowbit(ty);
x += lowbit(x);
}
}
void real_add(ll x1, ll y1, ll x2, ll y2, ll v) {
add(x1, y1, v);
add(x1, y2 + 1, -v);
add(x2 + 1, y1, -v);
add(x2 + 1, y2 + 1, v);
}
ll sum(ll x, ll y) {
ll res = 0;
while(x) {
ll ty = y;
while(ty)
res += mp[x][ty], ty -= lowbit(ty);
x -= lowbit(x);
}
return res;
}
int main() {
ll t;
scanf("%lld", &t);
while(t--) {
scanf("%lld%lld%lld%lld", &n, &m, &a, &b);
memset(mp, 0, sizeof(mp));
nn = max(n, m);
for(ll i = 1; i <= n; i++) {
for(ll j = 1; j <= m; j++) {
scanf("%lld", &aa[i][j]);
add(i, j, aa[i][j] - aa[i][j - 1] - aa[i - 1][j] + aa[i - 1][j - 1]);
}
}
bool flag = 1;
for(ll k = 2; k <= n + m; ++k) {
for(ll i = 1; i < k; ++i) {
ll x1 = i, y1 = k - i;
ll x2 = x1 + a - 1, y2 = y1 + b - 1;
if(x1 < 1 || x1 > n || y1 < 1 || y1 > m) continue;
if(x2 < 1 || x2 > n || y2 < 1 || y2 > m) continue;
ll tmp = sum(x1, y1);
if(tmp < 0) {
flag = 0;
break;
}
else if(tmp > 0) real_add(x1, y1, x2, y2, -tmp);
}
if(!flag) break;
}
if(flag) {
for(ll i = 1; i <= n; ++i) {
for(ll j = 1; j <= m; ++j) {
if(sum(i, j) != 0) {
flag = 0;
break;
}
}
if(!flag) break;
}
}
if(flag) printf("^_^\n");
else printf("QAQ\n");
}
return 0;
}