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[][];
 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 = ;
     for(int i = x; i > ; i -= lowbit(i))
     {
         for(int j = y; j > ; j -= lowbit(j))
             ans += num[i][j];
     }
     return ans;
 }
 int main()
 {
     int t, x1, y1, x2, y2, d;
     while(cin >> t >> n)
     {
         memset(num, , sizeof(num));
         while(cin >> t)
         {
             if(t == )break;
             else if(t == )
             {
                 scanf("%d%d%d", &x1, &y1, &d);
                 x1++, y1++;
                 add(x1, y1, d);
             }
             else if(t == )
             {
                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                 x1++, y1++, x2++, y2++;
                 cout<<(sum(x2, y2) - sum(x2, y1 - ) - sum(x1 - , y2) + sum(x1 - , y1 - ))<<endl;
             }
         }
     }
     return ;
 }