#include<cmath>#include<cstdio>#include<iostream>#include<algorithm>#define N 50010#define M 100010#define RG register#define inf 0x
https://www.u72.net/daima/nwa0u.html - 2024-11-03 12:38:39 - 代码库公约数和<em>公倍数</em>描述小明被一个问题给难住了,现在需要你帮帮忙。问题是:给出两个正整数,求出它们的最大公约数和<em>最小公倍数</em>。
https://www.u72.net/daima/nax9r.html - 2024-07-30 18:48:51 - 代码库<em>最小公倍数</em>#include<stdio.h>int main(){ int i,a,b; scanf("%d%d"
https://www.u72.net/daima/vk3z.html - 2024-08-23 09:52:29 - 代码库P1029 最大公约数和<em>最小公倍数</em>问题题目描述输入二个正整数x0,y0(2<=x0<100000,2<=y0<=1000000),求出满足下列条件的
https://www.u72.net/daima/nbbdm.html - 2024-10-03 00:15:01 - 代码库Lowest Common Multiple PlusTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31453
https://www.u72.net/daima/nzz9.html - 2024-07-03 10:51:38 - 代码库Least Common MultipleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30285 Accep
https://www.u72.net/daima/06du.html - 2024-07-18 11:22:06 - 代码库水题#include <iostream>using namespace std;int fun(int x,int y){ int i; for(i=(x<y?x:y);!(x%i==0&&y%i==0);i--);
https://www.u72.net/daima/6r8k.html - 2024-09-08 06:34:03 - 代码库http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1419题意:思路:要想最大,肯定去找尽量大的互质的数,如果不是互质的数,去掉约数后也就变成
https://www.u72.net/daima/ebkx.html - 2024-09-14 21:17:11 - 代码库题目描述 http://acm.nyist.net/JudgeOnline/problem.php?pid=972小张是软件项目经理,他带领3个开发组。工期紧,今天都在加班呢。为鼓舞士气,小张打算给
https://www.u72.net/daima/nva53.html - 2024-10-27 17:38:39 - 代码库求最大公约数和最大<em>公倍数</em>一、求三个数的最大公约数和最大<em>公倍数</em>/* 时间:2017年6月30日20:14:33 功能:求三个数的最大公约数与<em>最小公倍数</em>*
https://www.u72.net/daima/nr047.html - 2024-10-14 14:46:02 - 代码库本文出自:http://blog.csdn.net/svitter原题:http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1632题意:给你一个数字N,让你
https://www.u72.net/daima/zmse.html - 2024-07-05 10:55:38 - 代码库1、最大公约数所谓“最大公约数”是指两个数(A和B)都能够被C整除,求这个C的最大值问题。辗转相除的方法假设存在A和B两个正整数(且A>B),那么令R= A
https://www.u72.net/daima/f7sx.html - 2024-07-10 10:39:45 - 代码库http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1355 斐波那契数列定义如下: F(0) = 0 F(1) = 1F(n) = F(n-1) + F(n-2) 给
https://www.u72.net/daima/uawm.html - 2024-08-21 10:34:29 - 代码库最大公约数:代码: 1 #include <stdio.h> 2 #include <math.h> 3 long long gcd(int x,int y ) 4 { 5 return (y==0)?x:gcd(y,x%y); 6 } 7 8 int mai
https://www.u72.net/daima/uh4z.html - 2024-07-13 20:24:54 - 代码库#include <stdio.h>int gcd(int m,int n){if(n==0) return m;else return gcd(n,m%n); }*************变形一int gcd(int m,int n) {if(m%n==0)
https://www.u72.net/daima/v5s0.html - 2024-07-15 11:40:53 - 代码库Description 给定一张N个顶点M条边的无向图(顶点编号为1,2,&hellip;,n),每条边上带有权值。所有权值都可以分解成2^a*3^b的形式。现在有q个询问,每次
https://www.u72.net/daima/7x78.html - 2024-09-10 06:54:13 - 代码库写了一段相关的代码,好久不写了#include <iostream>#include <exception>#include <stack>using namespace std;int func(int a, int b) {
https://www.u72.net/daima/4741.html - 2024-09-05 15:55:51 - 代码库Lowest Common Multiple PlusTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34980
https://www.u72.net/daima/nhecs.html - 2024-08-03 12:03:57 - 代码库【算法】欧几里德算法#include<cstdio>int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){ int a,b; scanf("%d%d",&a,&b);
https://www.u72.net/daima/nkf67.html - 2024-09-26 04:12:01 - 代码库题目链接:点击打开链接题目大意:有n个行星,给出每个行星的旋转的周期。问最少多少时间后n个行星会在一条直线上,初始点在一起,不存在全部的行星都有同一
https://www.u72.net/daima/nfv38.html - 2024-10-07 08:23:39 - 代码库