1 solutions

  • 0
    @ 2025-5-26 16:48:51
    #include <iostream>
    using namespace std;
    
    int main() {
      	int n,x;
      	cin>>n>>x;
      	for(int i=1;;i++)
      	{
      		int st=true;
      		int t=i; //枚举最后一值猫的雨的数量 
      		for(int j=2;j<=n;j++)
      		{
      			if((t*n+x)%(n-1)!=0) //前面的猫拿完以后不能够分成n-1份 
      			{
      				st=false;	
    			}
      			t=(t*n+x)/(n-1);//上一只猫得到的鱼的数量 
    		}
    		if(st)
    		{
    			cout<<t*n+x<<endl;
    			break;
    		}
    	}
        return 0;
    }
    
    • 1

    Information

    ID
    2534
    Time
    2000ms
    Memory
    256MiB
    Difficulty
    10
    Tags
    (None)
    # Submissions
    7
    Accepted
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