/**copyright(c) 2014,烟台大学计算机学院*All rights reserved。*文件名称:16周(oj 4)*作者:王忠*完成日期:2014.12.15*版本号:v1.0**问题描述:输入
https://www.u72.net/daima/ncdun.html - 2024-08-08 00:42:40 - 代码库1 import java.util.Scanner; 2 3 public class ZuiDaGongYueShuClass { 4 5 public static void main(String[] args) { 6 Scan
https://www.u72.net/daima/nwhwb.html - 2024-11-04 06:23:39 - 代码库1.求最大<em>公约数</em>辗转相除求值例如:12,8求最大<em>公约数</em>12%8=4[不为零]8%4=0[为零,则4为最大<em>公约数</em>]【原理就是12,8的最大<em>公约数</em>和
https://www.u72.net/daima/nub7z.html - 2024-10-22 10:13:02 - 代码库a=int(raw_input(‘a‘))b=int(raw_input(‘b‘))su=[]if a>b: smaller=b else: smaller=a for i in range(1,smaller+1): if
https://www.u72.net/daima/nc5w8.html - 2024-10-11 19:00:39 - 代码库#include<stdio.h>int gcd(int a,int b){ if(b!=0) gcd(b,a%b); else return a;}int lcm(int a,int b){ return a*b/gcd(a,b); //a/gcd(a,b
https://www.u72.net/daima/nfkre.html - 2024-10-06 07:10:39 - 代码库include <cstdio> 2 #include <cstring> 3 using namespace std; 4 #define N 1010 5 6 //两个数的最大<em>公约数</em>和最小公倍数
https://www.u72.net/daima/nb23f.html - 2024-08-06 09:57:15 - 代码库1、求最大<em>公约数</em> 利用辗转相除法求最大<em>公约数</em>int gcd(int a,int b) {
https://www.u72.net/daima/7xxn.html - 2024-09-10 06:25:11 - 代码库思路:m和n如果有<em>公约数</em>,则安全洞存在,无<em>公约数</em>或<em>公约数</em>为1,则无 #include <stdio.h>int gcd(int a,int b){
https://www.u72.net/daima/9r2.html - 2024-07-03 05:54:14 - 代码库描述:计算两个非负整数p和q的最大<em>公约数</em>:若q是0,则最大<em>公约数</em>为p。否则,将p除以q得到余数r,p和q的最大<em>公约数</em>即为q和r的最大<em>公约数</em>。
https://www.u72.net/daima/z861.html - 2024-08-12 23:25:20 - 代码库链接: http://soj.me/1732ConstraintsTime Limit: 1 secs, Memory Limit: 32 MBDescription:Alice is a beautiful and clever girl. Bob would li
https://www.u72.net/daima/zacr.html - 2024-07-04 10:25:43 - 代码库# include <stdio.h>int main(void){ int x, y,temp; int r; printf("请输入两个正整数:\n"); scanf("%d %d", &num1, &num2); r = num1 % num2; temp
https://www.u72.net/daima/s93s.html - 2024-07-13 15:02:31 - 代码库# include<stdio.h>int main(){int a,b,c,x,y;printf("请输入两个正整数,用逗号间隔:");scanf("%d,%d",&a,&b);x=a;y=b;if (a<b){c=a;a=b;
https://www.u72.net/daima/3k3f.html - 2024-09-02 14:39:10 - 代码库Divided LandTime Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 123 Accepted Submiss
https://www.u72.net/daima/67nc.html - 2024-07-24 17:56:53 - 代码库1 #include<iostream> 2 using namespace std; 3 int main(){ 4 int a,b,m,n,x; 5 cin>>a>>b; 6 m=a>b?a:b; 7 n=a+b-m; 8 x=n;
https://www.u72.net/daima/ca1d.html - 2024-07-10 15:26:11 - 代码库题目链接:http://lightoj.com/volume_showproblem.php?problem=1024题意:给你n(2<=n<=1000)个数, 然后求n个数的最小公倍数,每个数的大小是1---10000;所以
https://www.u72.net/daima/cv0u.html - 2024-08-17 18:29:02 - 代码库function gcd(a,b){return b == 0 ? a : gcd(b,a%b);}function lcm(a,b){return a * b / gcd(a,b);}console.log(gcd(24,42));console.lo
https://www.u72.net/daima/na369.html - 2024-07-30 22:37:18 - 代码库<pre name="code" class="cpp">/**coyright(c)2014 龙城无泪*All rights reserved*文件名称 digui.c*作者:封尘之魂*完成日期:20141108*版本号V1.
https://www.u72.net/daima/nn148.html - 2024-07-31 22:26:05 - 代码库比欧几里得算法高效的用来计算gcd和lcm的stein算法,用来大数的计算:function gcd(a,b){if(a == b){return a;}var bigger;var smaller;if(a>b){
https://www.u72.net/daima/nnvzn.html - 2024-07-31 18:00:14 - 代码库题目描述 Description输入二个正整数x0,y0(2<=x0<100000,2<=y0<=1000000),求出满足下列条件的P,Q的个数条件: 1.P,Q是正整数2.要求P,Q以x0为最大
https://www.u72.net/daima/ncc9k.html - 2024-10-10 07:17:01 - 代码库<em>约数</em>总和
https://www.u72.net/daima/nva50.html - 2024-10-27 17:37:02 - 代码库