HihoCoder - 1336 Matrix Sum

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 A_xy. (Subscripts starts from 0

2. Sum x₁ y₁ x₂ y₁: Return the sum of every element A_xy for x₁ ≤ x ≤ x₂, y₁ ≤ y ≤ y₂.

Input

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 ≤ 100000

For each Add operation: 0 ≤ x < N, 0 ≤ y < N, -1000000 ≤ value ≤ 1000000

For each Sum operation: 0 ≤ x₁ ≤ x₂ < N, 0 ≤ y₁ ≤ y₂ < N

Output

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

 

Sample Input

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 

Sample Output

1
2
3 
二维的树状数组，做法与一维类似，注意求区间时加减范围。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=1005;
int g[maxn][maxn],n;

void update(int x,int y,int v)
{
    int j=y;
    for(;x<=n;x+=x&(-x))
        for(y=j;y<=n;y+=y&(-y))
        {
            g[x][y]+=v;
            g[x][y]%=1000000007;
        }
}

long long getsum(int x,int y)
{
    long long sum=0;
    int j=y;
    for(;x>0;x-=x&(-x))
        for(y=j;y>0;y-=y&(-y))
        {
            sum+=g[x][y];
            sum%=1000000007;    
        }
    return sum;
}

int main()
{
    int m,x1,y1,x2,y2,v;
    while(~scanf("%d%d",&n,&m))
    {
        memset(g,0,sizeof(g));
        char s[5];
        for(int i=0;i<m;i++)
        {
            cin>>s;
            if(s[0]=='A')
            {
                scanf("%d%d%d",&x1,&y1,&v);
                update(x1+1,y1+1,v);
            }
            else
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                long long ans=0;
                ans+=getsum(x2+1,y2+1)%(1000000007);
                ans-=getsum(x1,y2+1)%(1000000007);
                ans-=getsum(x2+1,y1)%(1000000007);
                ans+=getsum(x1,y1)%(1000000007);
                ans%=1000000007;
                if(ans<0)
                    ans+=1000000007;
                printf("%lld\n",ans);
            }
        }    
    }
    
    
    return 0;    
}