题意: for (int i = 0; ; ++i) { for (int j = 0; j <= i; ++j) { M[j][i - j] = A[cursor]; cursor = (cursor + 1) % L; }} 给定序列A[1..L],二维数组M的规律由以上代码给出.Q个查询,每次给x0,y0,x1,y1 (0≤x0≤x1≤1e8,0≤y0≤y1≤1e8)四个数,求以(x0,y0)和(x1,y1)两个点为端点的矩形中数的和. 分析:根据推导可得,M[i][j] = M[i+2L…

6336.Problem E. Matrix from Arrays 不想解释了,直接官方题解: 队友写了博客,我是水的他的代码 ------>HDU 6336 子矩阵求和 至于为什么是4倍的,因为这个矩阵是左上半边有数,所以开4倍才能保证求的矩阵区域里面有数,就是图上的红色阴影部分,蓝色为待求解矩阵. 其他的就是容斥原理用一下,其他的就没什么了. 代码: //1005-6336-矩阵求和-二维前缀和+容斥-预处理O(1)查询输出 #include<iostream> #include&…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6336 Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1711    Accepted Submission(s): 794 Problem Description Kazari…

Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1384    Accepted Submission(s): 630 Problem Description Kazari has an array A length of L, she plans to generate a…

Problem E. Matrix from Arrays Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Kazari has an array A length of L, she plans to generate an infinite matrix M using A. The p…

题目 给出长度为n 的A矩阵 , 按 int cursor = 0; for (int i = 0; ; ++i) { for (int j = 0; j <= i; ++j) { M[j][i - j] = A[cursor]; cursor = (cursor + 1) % L; }}构造出无限矩阵M , 然后给出l1 , r1 , l2, r2 ; 查询以(l1,r1)左上角 (l2,r2)右上角 的矩阵和 题意:用上面的转化规则十分容易的想到可能是有什么规律 , 所以我们打了个表出来发现…

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6336 题目: 题意:给你一个l个元素的数组a,用题目中的程序构造一个新的矩阵,询问q次,问以(x1,y1)为左上角,(x2,y2)为右下角的矩阵内的元素之和(原点在左上角). 思路:我们通过打表可以发现这个大矩阵都是以左上角2l*2l的小矩阵M循环出现的,所以对于每次查询我们只需统计他要查询的矩阵包含多少个完整的M,对于那些不构成完整的行列和,我们首先用前缀和统计出来,最后加起来即可.我的方法用下图…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - BZOJ4972 八月月赛Problem B 题目概括 一个矩阵,一坨询问,问矩阵中一个特定方向的等腰直角三角形范围的sum. 题解 一开始毫无头绪. 看完9题,一题也不会. 发现这题A的人多,于是我花了15分钟仔细思考. 发现可以了. 对于一个三角形区域,我们可以看下图: 我们把求右下黑色三角形区域转化成一个矩形和3个左上的三角形,然后就OK了. 矩形只要前缀和就可以了,O(nm) 求贴在上面和左边的,各…

Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around the abandoned Eikou Cram School building, Oshino's makeshift residence. The space is represented by a rectangular grid of n × m cells, arranged into n…

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom.…

#include<cstdio> #include<cstring> using namespace std; typedef long long ll; ; ; ll c[N][N]; ]; int lowbit(int x) { return x&-x; } 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)) c[i][j]=(…

Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 501    Accepted Submission(s): 210 Problem Description Kazari has an array $A$ length of $L$, she plans to generat…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 找个规律会发现 M[i][j] = M[i-2*L][j] = M[i][j-2*L] 先预处理出来(1,1)-(2L,2L)这个矩阵以及他的二维前缀和 那么对于要求的(x0,y0)-(x1,y1)这个矩阵. 可以用若干个(1,1)-(x,y)这样的前缀矩阵通过加加减减算出来. 对于(1,1)-(x,y)这样的矩阵. 显然是由若干个(1,1)-(2L,2L)矩阵合并而成的(x/L)*(y/L)个. 多余的部分(下边,右下角以及右…

Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 474    Accepted Submission(s): 202 Problem Description Kazari has an array A length of L, she plans to generate an…

Problem E. Matrix from Arrays Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1162    Accepted Submission(s): 522 Problem Description Kazari has an array A length of L, she plans to generate a…

11.6 Given an M x N matrix in which each row and each column is sorted in ascending order, write a method to find an element. LeetCode上的原题,请参见我之前的博客Search a 2D Matrix 搜索一个二维矩阵和Search a 2D Matrix II 搜索一个二维矩阵之二. class Solution { public: bool findElemen…

/*Arrays jdk中为了便于开发,给开发者提供了Arrays类, 其中包含了很多数组的常用操作.例如快速输出.排序.查找等.*/ import java.util.Arrays; public class ShuZun { public static void main(String[] args) { //数组的字符串形式 int[] arr={8,3,6,7,2,9}; //数组的[输出] String str=Arrays.toString(arr); System.out.prin…

Birthday Toy Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 644    Accepted Submission(s): 326 Problem Description AekdyCoin loves toys. It is AekdyCoin’s Birthday today and he gets a special “…

题目: Problem F. Matrix GameInput file: standard inputOutput file: standard inputTime limit: 1 secondMemory limit: 256 mebibytesAlice and Bob are playing the next game. Both have same matrix N × M filled with digits from 0 to 9.Alice cuts the matrix ve…

6343.Problem L. Graph Theory Homework 官方题解: 一篇写的很好的博客: HDU 6343 - Problem L. Graph Theory Homework - [(伪装成图论题的)简单数学题] 代码: //1012-6343-数学 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset&g…

FatMouse's Speed Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14980    Accepted Submission(s): 6618 Special Judge Problem Description FatMouse believes that the fatter a mouse is, the faster…

FATE Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11908    Accepted Submission(s): 5645 Problem Description 最近xhd正在玩一款叫做FATE的游戏,为了得到极品装备,xhd在不停的杀怪做任务.久而久之xhd开始对杀怪产生的厌恶感,但又不得不通过杀怪来升完这最后一级.现在的…

FATE http://acm.hdu.edu.cn/showproblem.php?pid=2159 Problem Description 最近xhd正在玩一款叫做FATE的游戏,为了得到极品装备,xhd在不停的杀怪做任务.久而久之xhd开始对杀怪产生的厌恶感,但又不得不通过杀怪来升完这最后一级.现在的问题是,xhd升掉最后一级还需n的经验值,xhd还留有m的忍耐度,每杀一个怪xhd会得到相应的经验,并减掉相应的忍耐度.当忍耐度降到0或者0以下时,xhd就不会玩这游戏.xhd还说了他最多只杀…

(Nim积相关资料来自论文曹钦翔<从"k倍动态减法游戏"出发探究一类组合游戏问题>) 关于Nim积计算的两个函数流程: 代码实现如下: ][]={,,,}; int Nim_Multi_Power(int x,int y) { ) return m[x][y]; ; for(;;a++) <<(<<a))&&x<(<<(<<(a+)))) break; <<(<<a); int p…

Triple Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 388    Accepted Submission(s): 148 Problem Description Given the finite multi-set A of n pairs of integers, an another finite multi-set B …

Check Corners Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2377    Accepted Submission(s): 859 Problem Description Paul draw a big m*n matrix A last month, whose entries Ai,j are all intege…

题目链接 Problem Description Given the finite multi-set A of n pairs of integers, an another finite multi-set B of m triples of integers, we define the product of A and B as a multi-set C=A∗B={⟨a,c,d⟩∣⟨a,b⟩∈A, ⟨c,d,e⟩∈B and b=e} For each ⟨a,b,c⟩∈C, its B…

题目链接:http://poj.org/problem?id=2155 Matrix Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 32950   Accepted: 11943 Description Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th col…