poj3122--Pie（二分的精度问题）

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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

Output

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

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;
}

poj3122--Pie（二分的精度问题）

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原文地址：http://blog.csdn.net/winddreams/article/details/43053069