1 solutions

  • 1
    @ 2025-7-22 10:52:32
    #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