编程输入两正整数m.n,计算两数的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
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_
#includeintGcd(intm,intn)/*最大公约数*/{intt;if(m
#include#includevoidmain(){intm,n,i,j,mn,a,x,y;printf("请输入m,n\n");scanf("%d%d",&m,&n);mn=m*n;a=(mn)?
//输入范例假设n赋值为4/*1234(回车)2345(回车)3456(回车)5678(回车)注意输入时数字间要有空格间隔*/#includevoidmain(){inti,j;intsum=0;//
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
importjava.util.*;publicclassTest40014{publicstaticvoidmain(String[]args){Scannerin=newScanner(Syst
importjava.awt.*;importjava.awt.event.*;classTestimplementsActionListener{TextFieldintext,outtext;Bu
#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i
#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=
饭要一口一口吃,作业要自己做
#includeintmain(){intm,n;intsum1=0,sum2=0;scanf("%d%d",&m,&n);for(inti=m;i
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);
#includeintFun(intn){inti,c=1;for(i=1;i
#includeintfact(intn){intt=1;for(inti=1;i