二维树状数组 区间求和模板（#1336 : Matrix Sum）

描述

You are given an N × N matrix. At the beginning every element is 0. Write a program supporting 2 operations: 1. Add x y value: Add value to the element Axy. (Subscripts starts from 02. Sum x1 y1 x2 y2: Return the sum of every element Axy for x1 ≤ x ≤ x2, y1 ≤ y ≤ y2.

输入

The first line contains 2 integers N and M, the size of the matrix and the number of operations.Each of the following M line contains an operation.1 ≤ N ≤ 1000, 1 ≤ M ≤ 100000For each Add operation: 0 ≤ x < N, 0 ≤ y < N, -1000000 ≤ value ≤ 1000000For each Sum operation: 0 ≤ x1 ≤ x2 < N, 0 ≤ y1 ≤ y2 < N 

输出

For each Sum operation output a non-negative number denoting the sum modulo 109+7.样例输入

5 8
Add 0 0 1
Sum 0 0 1 1
Add 1 1 1
Sum 0 0 1 1
Add 2 2 1
Add 3 3 1
Add 4 4 -1
Sum 0 0 4 4

样例输出

1
2
3

模板模板
代码：#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
const int ma=1100;
int mod=1e9+7;
int n,m;
int c[ma][ma];
int a[ma][ma];

int lowbit(int x)
{
return x&(-x);
}
void add(int x,int y,int v)
{
for(int i=x;i<=1010;i+=lowbit(i))
{
for(int j=y;j<=1010;j+=lowbit(j))
c[i][j]+=v;

}
}

long long sum(int x,int y)
{
long long su=0;
for(int i=x;i>0;i-=lowbit(i))
{
for(int j=y;j>0;j-=lowbit(j))
su=(c[i][j]+su)%mod;

}
return su;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(c,0,sizeof(c));
memset(a,0,sizeof(a));
while(m--)
{
char p[4];
scanf("%s",p);
if(p[0]=='A')
{
int w1,w2,e1;
scanf("%d%d%d",&w1,&w2,&e1);
w1++;w2++;
add(w1,w2,e1);
a[w1][w2]=(e1+a[w1][w2])%mod;

}
else if(p[0]=='S')
{
int r1,r2,w1,w2;
scanf("%d%d%d%d",&r1,&r2,&w1,&w2);
r1++;r2++;w1++;w2++;
if(r1>w1)swap(r1,w1);
if(r2>w2)swap(r2,w2);
int sun=0;
sun=sum(w1,w2)-sum(w1,r2-1)-sum(r1-1,w2)+sum(r1-1,r2-1);
while(sun<0)sun+=mod;
printf("%d\n",sun);
}
}

}
return 0;
}