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| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 37865 | Accepted: 8051 |
Description
Input
Output
Sample Input
4 11 8.02 7.43 4.57 5.39
Sample Output
2.00
Source
#include <cstdio>
#include <cmath>
double a[10005];
int n, k;
bool ok(double x)
{
int num = 0;
for(int i = 0; i < n; ++i){
num += (int)(a[i]/x);
}
return num >= k;
}
int main()
{
scanf("%d%d", &n, &k);
for(int i = 0; i < n; ++i)
scanf("%lf", &a[i]);
double l = 0, r = 1e5+5;
for(int i = 1; i < 100; ++i){
double mid = (l+r)/2;
if(ok(mid)) l = mid;
else r = mid;
}
printf("%.2f\n", floor(l*100)/100);
return 0;
}
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原文地址:http://www.cnblogs.com/inmoonlight/p/5746981.html