**【POJ - 3122】 Pie（二分寻值）

题干：

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

解题报告：

馅饼的面积排序然后从0到最大的面积二分寻找满足的值

ac代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>using namespace std;
const double PI = acos(-1.0);
const double eps = 1e-6;
double a[10000 + 5];
int n,f,t;
bool cmp(const double & a,const double & b) {return a>b;
}
double mid;
bool ok() {int sum=0;for(int i = 0; i<n; i++) {sum+=(int)(a[i] / mid);}return sum >= f ;
}int main()
{
//  freopen("in.txt","r",stdin);scanf("%d",&t);while(t--) {scanf("%d %d",&n,&f);f++;for(int i = 0; i<n; i++) {scanf("%lf",&a[i]);a[i]=a[i]*a[i]*PI;}sort(a,a+n,cmp);if(n>f) n=f;//如果人数没有蛋糕数多，那就取最大的那几个蛋糕。
//      sort(a,a+n);
//      double l=a[0];
//      double r=a[n-1];double l = 0;double r = a[0];while(r-l>=eps) {mid = (l+r )/2;if(ok() ) l=mid;else r=mid;}printf("%.4f\n",l);}return 0 ;
}

总结：

1. ok函数中for循环内的那一层判断，完全可以写成（吗？）（好像不太行但是为什么？）

    bool ok() {int sum=0;for(int i = 0; i<n; i++) {int tmp = a[i];while(tmp>=mid) {sum++;tmp-=mid;} }return sum >= f ;}

即：依次递减总是可以优化成除法！因为是线性结构，所以可以跳跃

2.还是ok函数中，加括号！！不然出来的全是整数！优先级问题！

3.最后的输出 mid 也能ac，但是不太符合原题的意思，所以最好还是输出l。

4.这件事情告诉我们二分不仅能查找量（可用stl），还能寻值

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