1 solutions
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0
#include <bits/stdc++.h> using namespace std; int main() { int k,g,m; int x=0,y=0,z; cin>>k>>g>>m; for(int i=0;i<k;i++) { if (x==g)x=0; //第一种操作 else if(y==0)y=m; //第二种操作 else //第三种操作 { z=min(g-x,y); //取决于第一个杯子可以装的和第二个杯子有的最小值 x+=z,y-=z; } } cout<<x<<" "<<y<<endl; return 0; }
- 1
Information
- ID
- 2341
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 1
- Tags
- (None)
- # Submissions
- 8
- Accepted
- 3
- Uploaded By