1 solutions
-
0
C++ :
#include <iostream> using namespace std; int main() { int n,m,p; cin>>n>>m>>p; int i; int count=0; // double x1,x2, x; int x1,x2, x; for(i=1;i<m;i++) { // x1=(double)n*i/m; // x2=p+(double)(n-p)*i/m; // x=x2-x1; x1=n*i/m;x2=p+(n-p)*i/m; x=x2-x1; count+=x; if((int)x2==p+(n-p)*i/m) count--; //cout<<x1<<" "<<x2<<" "<<x<<" "<<count<<endl; } cout<<count<<endl; return 0; }
- 1
Information
- ID
- 19411
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By