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…

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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…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 暴力遍历 日期 题目地址:https://leetcode.com/problems/greatest-common-divisor-of-strings/ 题目描述 For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenate…

题目如下: For strings S and T, we say "T divides S" if and only if S = T + ... + T  (T concatenated with itself 1 or more times) Return the largest string X such that X divides str1 and X divides str2. Example 1: Input: str1 = "ABCABC", st…

这是小川的第391次更新,第421篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第253题(顺位题号是1071).对于字符串S和T,当且仅当S = T + ... + T(T与自身连接1次或更多次)时,我们说"T除S". 返回最大的字符串X,使得X除以str1,X除以str2. 例如: 输入:str1 ="ABCABC",str2 ="ABC" 输出:"ABC" 输入:str1 ="AB…

题目 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". 返回最长字符串 X,要求满足 X 能除尽 str1 且 X 能除尽 str2. 来源:力扣(LeetCode) 著作权归领扣网络所有.商业转载请联系官方授权,非商业转载请注明出处. 题解 首先假设这两个字符串都是由一个(且最长)子串X生成的,那么很容易想到S串和T串都是X的重复,则S=k个X,T=n个X,则S+T == T+S == (k+n)个X…

题目 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". 返回最长字符串 X,要求满足 X 能除尽 str1 且 X 能除尽 str2. 示例 1: 输入:str1 = "ABCABC", str2 = "ABC" 输出:"ABC" 示例 2: 输入:str1 = "ABABAB", str2 = "ABAB"…

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≤…

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…

UPC备战省赛组队训练赛第十七场 with zyd,mxl G: Greatest Common Divisor 题目描述 There is an array of length n, containing only positive numbers. Now you can add all numbers by many times. Please find or state that it is impossible. You should notice that , you need to…

一 暴力枚举法 原理:试图寻找一个合适的整数i,看看这个整数能否被两个整形参数numberA和numberB同时整除.这个整数i从2开始循环累加,一直累加到numberA和numberB中较小参数的一半为止.循环结束后,上一次寻找到的能够被两整数整除的最大i值,就是两数的最大公约数. int getGreatestCommonDivisor(int numberA, int numberB) { ||numberB < ) { ; } || numberB <= ) { ; } int max…

欧几里得算法求最大公约数 If A = 0 then GCD(A,B)=B, since the GCD(0,B)=B, and we can stop. If B = 0 then GCD(A,B)=A, since the GCD(A,0)=A, and we can stop. Write A in quotient remainder form (A = B⋅Q + R) Find GCD(B,R) using the Euclidean Algorithm since GCD(A,B)…

两个数的最大公约数.一个典型的解决方案是欧几里德,叫欧几里德算法. 原理:(m,n)代表m和nGCD,和m>n.然后,(m,n)=(n,m%n)=.....直到余数为0. 码如下面: public class GCD { public static int gcd(int m, int n){ if(m*n<0){ return -1; } if(n==0){ return m; } if(m==0){ return n; } //辗转相除法 if(m<n){ int temp=m; m…

#include<stdio.h> #include<string.h> #include<math.h> ]; ]; int main() { int sb; scanf("%d", &sb); int u; ; u <= sb; u++) { int n, i; ; scanf("%d", &n); memset(flag, , sizeof(flag)); ; i <= n; i++) { sca…

function gcd(a, b) return a else return gcd(b, a mod b)…

题目链接: hdu:http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=153598 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=577&pid=1002 题解: 1.对于每个输入,枚举它的因子,并统计,结果存在mmp数组中,最后再倒过来扫一遍mmp数组,其中第一个mmp[i]>=2的,就是答案. 时间复杂度:O(…

B. Weakened Common Divisor time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output During the research on properties of the greatest common divisor (GCD) of a set of numbers, Ildar, a famous mat…

一.题目描述 A common divisor for two positive numbers is a number which both numbers are divisible by. It's easy to calculate the greatest common divisor between tow numbers. But your teacher wants to give you a harder task, in this task you have to find…

Weakened Common Divisor time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output During the research on properties of the greatest common divisor (GCD) of a set of numbers, Ildar, a famous mathem…

题目链接: 题目 Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 问题描述 This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence. 输入 Each sequen…

题目地址:http://poj.org/problem?id=2127 Description You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length. Sequence S1 , S2 , . . . , SN of length N is called an increa…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1423 Problem Description This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.   Input Each sequence is described with M - its length (1 <= M <= 500) and…

Greatest Common Increasing Subsequence 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1432 题目大意:给出两串数字,求他们的最长公共上升子序列(LCIS),并且打印出来. Sample Input 1 51 4 2 5 -124-12 1 2 4 Sample Output 21 4 分析:神奇就神奇在是LIS与LCS的组合 令dp[i][j]表示A串的前i个,与B串的前j…