题目大意:四种操作
- 0 s:清空(1,1)~(s,s)
- 1 x y a:在(x,y)点加上a
- 2 x1 y1 x2 y2:询问矩形(x1,y1)~(x2,y2)的和。
- 3:结束
解题思路:纯二维树状数组。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1030;
#define lowbit(x) ((x)&(-x))
int fenw[maxn+5][maxn+5];
void add (int x, int y, int a) {
for (int i = x; i <= maxn; i += lowbit(i)) {
for (int j = y; j <= maxn; j += lowbit(j))
fenw[i][j] += a;
}
}
int sum(int x, int y) {
int ret = 0;
for (int i = x; i; i -= lowbit(i)) {
for (int j = y; j; j -= lowbit(j))
ret += fenw[i][j];
}
return ret;
}
int main () {
int type;
int s, x, y, a, l, r;
while (scanf("%d", &type) == 1 && type != 3) {
if (type == 0) {
scanf("%d", &s);
for (int i = 1; i <= s; i++)
for (int j = 1; j <= s; j++)
fenw[i][j] = 0;
} else if (type == 1) {
scanf("%d%d%d", &x, &y, &a);
x++; y++;
add(x, y, a);
} else {
scanf("%d%d%d%d", &x, &y, &l, &r);
x++; y++; l++; r++;
printf("%d\n", sum(l, r) - sum(x-1, r) - sum(l, y-1) + sum(x-1, y-1));
}
}
return 0;
}