c语言入门编程题 输入两个正整数m和n,求其最大公约数和最小公倍数
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//输入两个正整数m和n,求其最大公约数.//用辗转相除法求最大公约数//算法描述:m对n求余为a,若a不等于0则m0){m_cup=m;n_cup=n;res=m_cup%n_cup;while(r
输入两个正整数m和n,求其最大公约数和最小公倍数.用辗转相除法求最大公约数算法描述:m对n求余为a,若a不等于0则m0){m_cup=m;n_cup=n;res=m_cup%n_cup;while(r
#include<stdio.h>#include<math.h>int min(int x,int y)\x09\x09//求m和n的最小值{\
#includeintmain(){intm,n;intm_cup,n_cup,res;/*被除数,除数,余数*/printf("Entertwointeger:\n");scanf("%d%d",&
#include int main() { int m, n; int m_cup, n_cup,
main(){inta,b,num1,num2,temp;printf("请输入两个正整数:\n");scanf("%d,%d",&num1,&num2);if(num1
#includeintmain(){inta,sum=0;scanf("%d",&a);while(a){sum+=a%10;a/=10;}printf("sum=%d\n",sum);return0
#include<stdio.h>int main(){\x09int n,s=0;\x09scanf("%d",&n);\x09while
利用辗除法公约数,再算公倍数.#include <stdio.h>void main(){ int a,&nbs
#include#includeusingnamespacestd;intmain(){intm,n,m_cup,n_cup,res;cin>>m>>n;if(m>0&&n>0){m_cup=m;n_
定义unsignedintn,longlongintn1,计算过程用for循环每次*10,保存到n1,最后输出n1再答:算法思路大致就是这样了,具体代码应该不难写再问:再答:哦,原来是这个再答:那么把
#include#includeintmain(){printf("请输入一个正整数:");intn1,n2;boolflag1=false,flag2=false;scanf("%d",&n1);n
#include#includeintmain(void){intn,m,i,j,t;scanf("%d%d",&n,&m);i=m>n?m:n;j=m>n?n:m;while(j){t=i%j;i=
scanf("%d%d",&a,&b);/*这里能不能加个空格或者逗号?否则计算机怎么知道你从哪里分割?*/if(!(e-i+2))/*这里是什么意思,说实话没想明白*/再问:有必要加空格么,该有的逗
#include <stdio.h>#include <math.h>int main(void){  
最大公约数:intGcd(inta,intb){if(a%b==0)returna;return(b,a%b);}最小公倍数:intGbs(inta,intb){returna*b/Gcd(a,b);
#includevoidmain(){\x09inti,n;\x09inta[10];\x09ints,p;\x09printf("n:");\x09scanf("%d",&n);\x09for(i=
main(){intp,r,n,m,temp;printf("Pleaseenter2numbersn,m:");scanf("%d,%d",&n,&m);//输入两个正整数.if(n
if(m>=6){count=0;for(number=m;numberk是为了预防i==k且都是素数的情况){judge=false;//有一个不是素数就不行break;}}if(judge)//如
修改好了#include "stdio.h"#include "math.h"int prime(int n){&nbs