Problem Introduction The greatest common divisor \(GCD(a, b)\) of two non-negative integers \(a\) and \(b\) (which are not both equal to 0) is the greatest integer \(d\) that divides both \(a\) and \(b\). Problem Description Task.Given two integer \(…

One efficient way to compute the GCD of two numbers is to use Euclid's algorithm, which states the following: GCD(A, B) = GCD(B, A % B) GCD(A, 0) = Absolute value of A In other words, if you repeatedly mod A by B and then swap the two values, eventua…

定义: 最大公约数(英语:greatest common divisor,gcd).是数学词汇,指能够整除多个整数的最大正整数.而多个整数不能都为零.例如8和12的最大公因数为4. 最小公倍数是数论中的一个概念.若有一个数\[X\],可以被另外两个数\[A\].\[B\]整除,且\[X\]大于(或等于)\[A\]和\[B\],则\[X\]X为\[A\]和\[B\]的公倍数.\[A\]和\[B\]的公倍数有无限个,而所有的公倍数中,最小的公倍数就叫做最小公倍数.两个整数公有的倍数称为它们的公倍数,…

描述 Given two numbers, number a and number b. Find the greatest common divisor of the given two numbers. In mathematics, the greatest common divisor (gcd) of two or more integers, which are not all zero, is the largest positive integer that divides ea…

Greatest Greatest Common Divisor Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5207 Description 在数组a中找出两个数ai,aj(i≠j),使得两者的最大公约数取到最大值. Input 多组测试数据.第一行一个数字T,表示数据组数.对于每组数据,第一行是一个数n,表示数组中元素个数,接下来一行有n个数,a1到an.1≤T≤…

1071. 字符串的最大公因子 1071. Greatest Common Divisor of Strings 题目描述 对于字符串 S 和 T,只有在 S = T + ... + T(T 与自身连接 1 次或多次)时,我们才认定 "T 能除尽 S". 返回字符串 X,要求满足 X 能除尽 str1 且 X 能除尽 str2. 每日一算法2019/6/17Day 45LeetCode1071. Greatest Common Divisor of Strings 示例 1: 输入:s…

problem 1071. Greatest Common Divisor of Strings solution class Solution { public: string gcdOfStrings(string str1, string str2) { return (str1+str2==str2+str1) ? (str1.substr(, gcd(str1.size(), str2.size()))) : ""; } }; 参考 1. Leetcode_easy_1071…

题目描述 There is an array of length n, containing only positive numbers.Now you can add all numbers by 1 many times. Please find out the minimum times you need to perform to obtain an array whose greatest common divisor(gcd) is larger than 1 or state th…

Greatest Common Divisor 题目链接 题目描述 There is an array of length n, containing only positive numbers. Now you can add all numbers by 1 many times. Please find out the minimum times you need to perform to obtain an array whose greatest common divisor(gcd…

lc1071 Greatest Common Divisor of Strings 找两个字符串的最长公共子串 假设:str1.length > str2.length 因为是公共子串,所以str2一定可以和str1前面一部分匹配上,否则不存在公共子串. 所以我们比较str2和str1的0~str2.length()-1部分, 若不同,则直接返回””,不存在公共子串. 若相同,继续比较str2和str1的剩下部分,这里就是递归了,调用原函数gcd(str2, str1.substring(str…