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>…

树状数组,开始的时候wa了,后来看看,原来是概率论没学好,以为求(L,B) - (R,T) 矩阵内的和只要用sum(R+1,T+1) - sum(L,B) 就行了,.傻x了.. 必须 sum(R,T) - sum(L,T) - sum(R,B) + sum(L,B) ;  (R,T 已经自加1)   诫之. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…

偶然发现这题还没A掉............速速解决了............. 树状数组和线段树比较下,线段树是在是太冗余了,以后能用树状数组还是尽量用......... #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <strin…

<题目链接> 题目大意: 一个由数字构成的大矩阵,开始是全0,能进行两种操作1) 对矩阵里的某个数加上一个整数(可正可负)2) 查询某个子矩阵里所有数字的和要求对每次查询,输出结果 解题分析: 二维树状数组模板题,需要注意的是,由于题目给的x,y坐标可以为0,所以我们应该将这些点的坐标全部+1,然后就是查询指定子矩阵中所有元素之和,很直观的就能得到式子 ans=sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1). #include <c…

题意:基础的二维数组,注意 0 + lowbit(0)会陷入无限循环----- 之前做一道一维的一直tle,就是因为这个-------------------------- #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #include…

题目连接: 题意:要求设计这样一个数据结构,支持下列操作 1.add(x,y,a).对二维数组的第x行,第y列加上a. 2.sum(l,b,r,t).求所有满足l<=x<=r,b<=y<=t,的数组元素的和 二位树状数组~ #include <iostream> #include <string.h> #include <stdio.h> #include <stdlib.h> #define loop(s,i,n) for(i =…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 13774   Accepted: 6393 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

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 contains a base sta…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 16893   Accepted: 7789 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: 18489   Accepted: 8558 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

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 contains a base sta…

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…

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Mobile phones Time Limit: 5000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate a…

Mobile phones Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 14291   Accepted: 6644 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: 19786   Accepted: 9133 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: 14288   Accepted: 6642 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: 18453   Accepted: 8542 Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The…

版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u013912596/article/details/33802561 题目链接:id=1195" rel="nofollow">http://poj.org/problem?id=1195 纯纯的二维树状数组,不解释.仅仅须要注意一点,由于题目中的数组从0開始计算.所以维护的时候须要加1.由于树状数组的下标是不能为1的 代码: #include <iostream&…

题目链接:http://poj.org/problem?id=1195 [题意] 给出一个全0的矩阵,然后一些操作 0 S:初始化矩阵,维数是S*S,值全为0,这个操作只有最开始出现一次 1 X Y A:对于矩阵的X,Y坐标增加A 2 L B R T:询问(L,B)到(R,T)区间内值的总和 3:结束对这个矩阵的操作   [思路] 二维树状数组单点更新+区域查询,可作为模板题. 注意坐标是从0开始,所以要+1   [代码] #include<cstdio> #include<cstrin…

1: #include <stdio.h> 2: #include <string.h> 3: #include <stdlib.h> 4: #include <algorithm> 5: #include <iostream> 6: using namespace std; 7:   8: #define LL(a) a<<1 9: #define RR(a) a<<1|1 10: const int MaxL = 10…