4 条题解

  • 2
    @ 2025-10-30 14:00:20
    #include <iostream>
    #include <algorithm>
    using namespace std;
    int n, a[100], l;
    int main()
    {
    	cin >> n;
    	for(int i = 0; i < n; i ++)
    		cin >> a[i];
    	l = a[0];
        for(int i = 1; i < n; i ++)
            l *= a[i] / __gcd(l, a[i]);
        for(int i = 0; i < n; i ++)
            cout << l / a[i] << endl;
    	return 0;
    }
    
    • 2
      @ 2023-4-4 22:17:13
      #include<iostream>
      using namespace std;
      int n,a[101],num;
      int lcm(int a,int b){//求最小公倍数
      	int r=a%b,x=a,y=b;
      	while(r){
      		a=b;
      		b=r;
      		r=a%b;
      	}
      	return x*y/b;
      }
      int main(){
      	cin>>n;
      	for(int i=1;i<=n;i++)cin>>a[i];
      	int num=a[1];
      	for(int i=2;i<=n;i++)num=lcm(num,a[i]);//求所有数的最小公倍数
      	for(int i=1;i<=n;i++)cout<<num/a[i]<<endl;//输出
      	return 0;
      }
      
      • 1
        @ 2026-3-22 19:31:04
        
        #include <stdio.h>
        #include <iostream>
        using namespace std;
        int a[1000];
        int main()
        {
        	int n;
        	cin >>n;
        	int num =1;
        	for(int i = 0 ; i < n ; i++)
        	{
        		cin >> a[i];
        		int p = num *a[i];
        		int x = a[i];
        		while(x)
        		{
        			int t = num % x;
        			num = x;
        			x=t;
        		}
        		num = p/num;
        	}
        	for(int i = 0 ; i < n ; i++)
                cout << num / a[i]	<< endl;
        }
        • -2
          @ 2022-3-27 19:49:48
          #include <iostream>
          #include <stdio.h>
          #include <string.h>
          using namespace std;
          #define LL long long
          const int N = 1e5 + 10;
          const int INF = 0x3f3f3f3f;
          int a[N];
          int main()
          {
          	int	n;
          	cin >> n;
          	int sum = 1;
          	for(int i = 1 , x , y ; i <= n ; i++)
          	{
          		cin >> a[i];
          		y = sum;
          		x = a[i];
          		sum *= x;
          		while(y)
          		{
          			int t = x%y;
          			x = y;
          			y = t;
          		}
          		sum /= x;
          	}
          	for(int i = 1 ; i <= n ; i++)
          	{
          		cout << sum / a[i] << endl;
          	}
          }
          • 1

          信息

          ID
          997
          时间
          1000ms
          内存
          128MiB
          难度
          5
          标签
          递交数
          429
          已通过
          179
          上传者