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poj 3122 （二分查找）
链接：http://poj.org/problem?id=3122
Pie
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:
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 33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2Sample Output 25.13273.141650.2655Source Northwestern Europe 2006 | ||||||||||
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一开始没理解题意，粗心了，原来是（f+1）个人
然后直接二分就出来结果了
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <iostream> 6 7 using namespace std; 8 #define eps 1e-6 9 #define MAXX 1000510 #define PI 3.1415926535911 12 double area(double x)13 {14 return PI*x*x;15 }16 17 int n,m;18 double str[MAXX];19 20 int judge(double x)21 {22 int sum=0;23 for(int i=0; i<n; i++)24 sum+=(int)(str[i]/x);25 return sum;26 }27 28 int main()29 {30 int t,i,j;31 double tmp;32 scanf("%d",&t);33 while(t--)34 {35 double maxx=0.0,left=0.0,right;36 scanf("%d%d",&n,&m);37 for(i=0; i<n; i++)38 {39 scanf("%lf",&tmp);40 str[i]=area(tmp);41 maxx=maxx<str[i]?str[i]:maxx;42 }43 right=maxx;44 while(right - left > eps)45 {46 double mid=(right+left)/2.0;47 if( judge(mid) >= m + 1)48 left=mid+eps;49 else right=mid-eps;50 }51 printf("%.4lf\n",left);52 }53 return 0;54 }
poj 3122 （二分查找）
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