1 solutions
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1
#include<bits/stdc++.h> using namespace std; int n,k,g=0; double a[10005]; int suan(double x){ g=0; for(int i=1;i<=n;i++) { g+=floor(a[i]/x); } return g; } int main(){ cin>>n>>k; for(int i=1;i<=n;i++) { cin>>a[i]; } double l=0,r=100000,mid; while(r-l>1e-3) { double mid=(l+r)/2; if(suan(mid)<k) { r=mid; }else{ l=mid; } } printf("%.2lf",l); return 0; }
Information
- ID
- 5635
- Time
- 1000ms
- Memory
- 512MiB
- Difficulty
- 5
- Tags
- # Submissions
- 14
- Accepted
- 11
- Uploaded By