Play with GCD 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/play-with-gcd Description Minka is very smart kid who recently started learning computer programming. He learned how to calculate the Greatest Common Divisor (GC…

Xtreme8.0 - Kabloom 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/kabloom Description The card game Kabloom is played with multiple decks of playing cards. Players are dealt 2 n cards, face up and arranged in two rows of…

Xtreme8.0 - Kabloom 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/kabloom Description The card game Kabloom is played with multiple decks of playing cards. Players are dealt 2 n cards, face up and arranged in two rows of…

Xtreme8.0 - Magic Square 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/magic-square Description Johnny designed a magic square (square of numbers with the same sum for all rows, columns and diagonals i.e. both the main di…

注意以下几点: 搜索维度非约束条件的都要记录,否则大概率出错,比如_0 st参数传递和_0的互相影响要分辨清楚 num==-1就要返回0而不是1 #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<string> #include<vec…

题目: 聪明的猴子 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1123 Accepted Submission(s): 294 Problem Description 森林中有一排香蕉树(无限长),一只猴子站在其中一棵树上,猴子在跳跃前要先抽取一张卡片,卡片上写有A+1个自然数,其中最后一个是B,前A个数只能小于等于B,卡片上的数字可以…

D. Array GCD 题目连接: http://codeforces.com/contest/624/problem/D Description You are given array ai of length n. You may consecutively apply two operations to this array: remove some subsegment (continuous subsequence) of length m < n and pay for it m·…

CA Loves GCD 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5656 Description CA is a fine comrade who loves the party and people; inevitably she loves GCD (greatest common divisor) too. Now, there are N different numbers. Each time, CA will select s…

http://acm.uestc.edu.cn/#/problem/show/923 给定一堆数字,求其所有数字的gcd. 现在要删除最多的数字,使得剩下的数字的gcd和原来的一样. 设dp[i][val]表示在前i个数中,得到val这个数字所需的最小数字,怎么得到val这个数字?就是gcd得到. dp[i][gcd] 然后转移就是 dp[i][a[i]] = 1是必然的,自己一个 枚举新数字得到新的gcd  val dp[i][val] = min(dp[i][val], dp[i - 1][…

定义:dp[i][j] 表示 在前i个数中,使整个gcd值为j时最少取的数个数. 则有方程: gg = gcd(a[i],j) gg == j : 添加这个数gcd不变,不添加,  dp[i][j] = dp[i-1][j] gg != j: t添加,更新答案,                dp[i][gg] = dp[i-1][j] + 1 最后答案为dp[n][g] (g为原始的所有数的gcd) 时间复杂度: O(n*max(a[i])) 代码: #include <iostream>…