zoukankan      html  css  js  c++  java
  • Pie(二分)

    Pie

    Time Limit : 5000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 190   Accepted Submission(s) : 72
    Problem 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
    题意:就是几块馅饼,高是一,每人分的最大体积;
    代码:
     1 #include<stdio.h>
     2 #include<math.h>
     3 #define PI acos(-1.0)
     4 double pie[10010];
     5 int N,F;
     6 int find(double mid){int sum=0;
     7     for(int i=0;i<N;i++){
     8         sum+=(int)(pie[i]/mid);
     9     }
    10     if(sum>=F+1)return 1;
    11     else return 0;
    12 }
    13 int main(){
    14     int temp,T;
    15     double sum,anser;
    16     scanf("%d",&T);
    17     while(T--){sum=0;
    18         scanf("%d%d",&N,&F);
    19         for(int i=0;i<N;i++){
    20             scanf("%d",&temp);
    21             pie[i]=PI*temp*temp;
    22             sum+=pie[i];
    23         }
    24         //printf("%lf
    ",sum);
    25         double l=0,r=sum/(F+1.0),mid;
    26     while(r-l>1e-7){
    27         mid=(l+r)/2;
    28         if(find(mid))l=mid;
    29         else r=mid;
    30     }
    31         printf("%.4f
    ",mid);
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    8天学会Hadoop基础(2)
    [Java]剑指offer:扑克牌顺子
    8天学会Hadoop基础(1):NameNode的工作机制
    [Java]剑指offer:构建乘积数组
    关于Hadoop启动之后jps没有namenode节点的解决方法
    剑指offer:对称的二叉树
    两只小熊队高级软件工程第七次作业敏捷冲刺5
    两只小熊队高级软件工程第七次作业敏捷冲刺4
    两只小熊队高级软件工程第七次作业敏捷冲刺3
    两只小熊队高级软件工程第七次作业敏捷冲刺2
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4700667.html
Copyright © 2011-2022 走看看