poj 1195 Mobile phones

// 题意: n*n的矩阵 ,定义两种操作：
// (1) x y a : [x,y]的值加上a  (2) x1 y1 x2 y2 : 查询子矩阵[x1,y1] -[x2,y2]的和

#include <iostream>            //简单二维树状数组
using namespace std;
int C[1025][1025];    //树状数组初始化为0
int side;
int lowbit(int x)
{
    return x&(-x);
}
void modify(int x,int y,int a)
{
    x++;y++;        //让数组下标从 1 开始
    while(x<=side)
    {
        int j=y;
        while(j<=side)
        {
            C[x][j]+=a;
            j+=lowbit(j);
        }
        x+=lowbit(x);
    }
}
int sum(int x,int y)
{
    x++;y++;
    int s=0;
    while(x>0)
    {
        int j=y;
        while(j>0)
        {
            s+=C[x][j];
            j-=lowbit(j);
        }
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    int op,ans;
    while(scanf("%d",&op)&&op!=3)
    {
        switch (op)
        {
            case 0:
                scanf("%d",&side);
                break;
            case 1:
                int x,y,a;
                scanf("%d%d%d",&x,&y,&a);
                modify(x,y,a);
                break;
            case 2:
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                ans=sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1);
                printf("%d\n",ans);
                break;
            default:
                break;
        }
    }
    return 0;
}