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：

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 ;

 }