1 solutions
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0
#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
- 2
- Uploaded By