6 条题解

  • 0
    @ 2025-8-10 21:28:44
    #include<bits/stdc++.h>
    using namespace std;
    int n,m,a,b,c;
    int main(){
    	while(cin >> n >> m){
    		a=n;
    		b=m;
    		int ans=m;
    		while(a%b!=0) {
    			c=a%b;
    			a=b;
    			b=c;
    		}
    		cout << b << " " << n*m/(b) << "\n";
    	} 
    	return 0;
    }
    
    
    
    • 0
      @ 2025-3-9 21:26:01

      主函数才两行

      #include<bits/stdc++.h>
      using namespace std;
      const int N=1e5+5,INF=0x3f3f3f3f;
      typedef long long LL;
      int n,m; 
      int main()
      {
      	while(cin>>n>>m)cout<<__gcd(n,m)<<" "<<n/__gcd(n,m)*m<<endl;
      	return 0;
      }
      
      
    • 0
      @ 2022-7-6 21:01:03
      /*****************************************
      备注:
      ******************************************/
      #include <math.h>
      #include <stdio.h>
      #include <iostream>
      #include <string.h>
      using namespace std;
      int main()
      {
      	int a,b;
      	while(cin >> a >> b)
      	{
      		int sum = a* b;
      		while(b != 0)  // 辗转相除法
      		{
      			int r = a%b;
      			a = b;
      			b = r;
      		}
      		cout << a << " " << sum/a << endl;;
      	}
      
      
      }
      
      • -2
        @ 2024-11-16 11:35:40
        #include<bits/stdc++.h>
        using namespace std; 
        int main(){
        	int a,b,c,d;
        	while(cin>>a>>b){
        		c=a%b;
        		d=a*b;
        		while(c!=0){
        			a=b;
        			b=c;
        			c=a%b;
        		}
        		cout<<b<<" "<<d/b<<endl;
        	}
        }
        
        • -3
          @ 2023-5-27 13:50:05

          #include #include #include #include #include #include #include #include using namespace std; const int INF = 0x3f3f3f3f; int main() { int a , b; while(cin >> a >> b) { int sum = a * b; while(b != 0) { int r = a % b; a = b; b = r; } cout << a << " " << sum / a << endl; } return 0; }

          • -3
            @ 2023-5-1 15:28:57
            #include<algorithm>
            using namespace std;
            int main(){
            	int m,n;
            	while(cin >>m>>n){
            		int sum = m*n;
            		while (n!=0){
            			int y=m%n;
            			m=n;
            			n=y; 
            		}
            		cout << m << " " << sum/m << endl;
            		
            	}
            	
            	return 0;
            }
            
            • 1

            信息

            ID
            977
            时间
            1000ms
            内存
            128MiB
            难度
            6
            标签
            递交数
            817
            已通过
            223
            上传者