HDU1551&&HDU1064 Cable master 2017-05-11 17:50 38人阅读 评论(0) 收藏
Cable master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit:
65536/32768 K (Java/Others)
Total Submission(s): 4673 Accepted Submission(s): 1769
- i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it.
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants
as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter, and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not
known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.
is followed by N lines with one number per line, that specify the length of each cable in the stock in meters. All cables are at least 1 centimeter and at most 100 kilometers in length. All lengths in the input are written with a centimeter precision, with
exactly two digits after a decimal point.
The input is ended by line containing two 0's.
after a decimal point.
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output must contain the single number "0.00" (without quotes).
4 11
8.02
7.43
4.57
5.39
0 0
2.00
#include <iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<cstring>
using namespace std;
#define LL long long int n,k;
double a[100005];
int b[100005];
bool ok(int x)
{
int ans=0;
for(int i=0; i<n; i++)
{
ans+=b[i]/x;
}
if(ans>=k)
return 1;
else
return 0;
} int main()
{
while(~scanf("%d%d",&n,&k))
{
for(int i=0; i<n; i++)
{
scanf("%lf",&a[i]);
b[i]=a[i]*100;
}
int l=1,r=10000000;
int ans;
while(l<=r)
{
int mid=(l+r)/2;
if(ok(mid))
{
l=mid+1;
ans=mid;
}
else
{
r=mid-1;
}
}
double an=ans*1.0/100;
printf("%.2f\n",an);
}
return 0;
}
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