POJ-1195 Mobile phones---裸的二维树状数组（注意下标从1，1开始）

题目链接：

https://vjudge.net/problem/POJ-1195

题目大意：

直接维护二维树状数组

注意横纵坐标全部需要加1，因为树状数组从(1,1)开始

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#include<list>
#include<deque>
#include<sstream>
#include<cctype>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, s, t) for(int i = (s); i < (t); i++)
using namespace std;
typedef long long ll;
int T, n, m, cases;
int num[1048][1048];
int lowbit(int x)
{
    return x & (-x);
}
///修改num[x][y] = d
void add(int x, int y, int d)
{
    for(int i = x; i <= n; i += lowbit(i))
    {
        for(int j = y; j <= n; j += lowbit(j))
            num[i][j] += d;
    }
}
///计算顶点为（1, 1）（x, y）的数总和
int sum(int x, int y)
{
    int ans = 0;
    for(int i = x; i > 0; i -= lowbit(i))
    {
        for(int j = y; j > 0; j -= lowbit(j))
            ans += num[i][j];
    }
    return ans;
}
int main()
{
    int t, x1, y1, x2, y2, d;
    while(cin >> t >> n)
    {
        memset(num, 0, sizeof(num));
        while(cin >> t)
        {
            if(t == 3)break;
            else if(t == 1)
            {
                scanf("%d%d%d", &x1, &y1, &d);
                x1++, y1++;
                add(x1, y1, d);
            }
            else if(t == 2)
            {
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                x1++, y1++, x2++, y2++;
                cout<<(sum(x2, y2) - sum(x2, y1 - 1) - sum(x1 - 1, y2) + sum(x1 - 1, y1 - 1))<<endl;
            }
        }
    }
    return 0;
}