Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14489   Accepted: 6735 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14391   Accepted: 6685 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…

                                                              Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 18608   Accepted: 8621 Description Suppose that the fourth generation mobile phone base stations in the Tampere a…

Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…

Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contain…

树状数组支持两种操作: Add(x, d)操作:   让a[x]增加d. Query(L,R): 计算 a[L]+a[L+1]……a[R]. 当要频繁的对数组元素进行修改,同时又要频繁的查询数组内任一区间元素之和的时候,可以考虑使用树状数组. 通常对一维数组最直接的算法可以在O(1)时间内完成一次修改,但是需要O(n)时间来进行一次查询.而树状数组的修改和查询均可在O(log(n))的时间内完成. 在二维情况下:数组A[][]的树状数组定义为: C[x][y] = ∑ a[i][j], 其中, …

题链: http://poj.org/problem?id=1195 题解: 二维树状数组 #include<cstdio> #include<cstring> #include<iostream> #define MAXN 1500 using namespace std; struct BIT{ int val[MAXN][MAXN],N; void Reset(int n){N=n; memset(val,0,sizeof(val));} int Lowbit(i…

题目链接:http://poj.org/problem?id=1195 题目意思:有一部 mobie phone 基站,它的面积被分成一个个小正方形(1 * 1 的大小),所有的小正方形的面积构成了一个 S * S 大小的矩阵(下标都是从 0 ~ S-1 变化的). 有四种指令: 第 一 行的指令默认输入是 0, 空格之后是矩阵的大小: S 最后一行的指令是 3, 代表 整个输入结束 注意:这两行的指令只会出现一次! 夹在它们中间的指令有可能是指令1,假设为X Y A,代表向第 X 行 第 Y…

二维的树状数组,,, 记得矩阵的求和运算要想好在写.... 代码如下: #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <cstdlib> #include <stack> #include <queue> #include <vector> #include <algorithm>…