Pie
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 11178 | Accepted: 3899 | Special Judge | ||
Description
My birthday is coming up and traditionally I‘m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various
sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.
My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is
better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.
What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.
Input
One line with a positive integer: the number of test cases. Then for each test case:
- One line with two integers N and F with 1 ≤ N, F ≤ 10 000: the number of pies and the number of friends.
- One line with N integers ri with 1 ≤ ri ≤ 10 000: the radii of the pies.
Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10?3.
Sample Input
3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2
Sample Output
25.1327 3.1416 50.2655
Source
Northwestern Europe 2006
题目大意：给出n个馅饼，m+1个人平均分这些馅饼，每人一块，问每个人最多分多少。
二分可以得到的馅饼大小。double二分
while( (high-low) > eqs )
{
mid = (low+high)/2.0 ;
if( solve(mid) )
{
low = mid ;
last = mid ;
}
else
high = mid ;
}
注意点：1 double 减法 （high - low）> eqs ，eqs用来卡精度，eqs太大精度降低，eqs太小时间变高。
2 二分的时候不加0.0001， 即low = mid high = mid；这样分可能会慢一点，但精度会高
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std ;
#define PI 3.1415926535898
#define eqs 1e-5
double s[11000] ;
int n , m ;
double f(double x)
{
int k = (x+eqs) * 10000 ;
x = k * 1.0 / 10000 ;
return x ;
}
int solve(double x)
{
int i , j , num = 0 ;
for(i = n-1 ; i >= 0 && (s[i]-x) > eqs ; i--)
{
j = s[i] / x ;
num += j ;
if( num >= m+1 )return 1 ;
}
if( num >= m+1 ) return 1 ;
return 0 ;
}
int main()
{
int t , i , k ;
double low , mid , high , last ;
while( scanf("%d", &t) != EOF )
{
while(t--)
{
scanf("%d %d", &n, &m) ;
for(i = 0 ; i < n ; i++)
scanf("%lf", &s[i]) ;
sort(s,s+n) ;
for(i = 0 ; i < n ; i++)
s[i] = s[i]*s[i]*PI ;
low = 0 ;
high = s[n-1] ;
while( (high-low) > eqs )
{
mid = (low+high)/2.0 ;
if( solve(mid) )
{
low = mid ;
last = mid ;
}
else
high = mid ;
}
printf("%.4lf\n", last) ;
}
}
return 0;
}