1 solutions

  • 0
    @ 2024-12-4 19:37:02

    C :

    #include<stdio.h>
    #include<math.h>
    int yueshu(int a,int b)
    {
    	int t;
    	while(b)
    	{
    		t=a%b;
    		a=b;
    		b=t;
    	}
    	return a;
    }
    int beishu(int a,int b)
    {
    	int t=yueshu(a,b);
    	return a*b/t;
    }
    int main()
    {
    	int i,j,a,b;
    	scanf("%d%d",&a,&b);
    	printf("%d %d",yueshu(a,b),beishu(a,b));
    } 
    

    C++ :

    #include<iostream>
    using namespace std;
    int gcd(int m,int n)
    {
    	int r=m%n;
    	while (r)
    	{
    		m=n;
    		n=r;
    		r=m%n;
    	}
    	return n;
    }
    int lcd(int m,int n)
    {
    	return m*n/gcd(m,n);
    }
    int main()
    {
    	int m,n;
    	cin>>m>>n;
    	cout<<gcd(m,n)<<" "<<lcd(m,n)<<endl;
    	return 0;
    }
    
    • 1

    C语言程序设计教程(第三版)课后习题8.1

    Information

    ID
    2203
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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