POJ 1195 Mobile phones

//二维数状数组 需要注意的是下标从1开始 lowbit(0) = 0.. 超时陷阱，一定要谨记啊！ 还有是c++的输入输出2600ms scanf 547ms
#include <iostream>
#include <stdio.h>
using namespace std;
const int MAXN = 1025;
int tree[MAXN][MAXN];
int p[MAXN];
int n;            //坐标范围 1 ~ 1
inline int lowbit(int t){
    return t & (-t);
}
void add(int x, int y, int value){
    for(;x <= n; x += lowbit(x)){
        for(int tempy = y; tempy <= n; tempy += p[tempy])//lowbit(tempy))
            tree[x][tempy] += value;
    }
}
int sum(int x, int y){
    int ans = 0;;
    for(; x > 0; x -= lowbit(x)){
        for(int tempy = y; tempy > 0; tempy -= p[tempy])//lowbit(tempy))
            ans += tree[x][tempy];
    }
    return ans;
}
int main(){
    for (int i = 1;i <= 1024; i++)     
        p[i] = i & (-i);
    int cmd;
    //cin >> cmd >> n;
    scanf("%d%d",&cmd, &n);
    while(scanf("%d",&cmd) == 1){//cin >> cmd){
        switch(cmd)
        {
        case 1:
            int x, y, value;
            //cin >> x >> y >> value;
            scanf("%d%d%d",&x,&y,&value);
            add(x + 1, y + 1, value);
            break;
        case 2:
            int x1, y1, x2, y2;
            //cin >> x1 >> y1 >> x2 >> y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            /*cout << sum(x2 + 1, y2 + 1) + sum(x1, y1) - sum(x1, y2 + 1) - sum(x2 + 1, y1) << endl;    //下标要加一 这题目做的一波三折 我怎么写成sum(x2 + 1, y2 + 1) - sum(x1, y1)了呢 ？ 我靠..脑袋抽筋啊*/
            printf("%d\n",sum(x2 + 1, y2 + 1) + sum(x1, y1) - sum(x1, y2 + 1) - sum(x2 + 1, y1));
            break;
        default:
            return 0;
        }
    }
    return 0;
}