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2013
11-12

POJ 3122 Pie [解题报告] Java

Pie

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2

样例输出:

25.1327
3.1416
50.2655

解题代码:

/* @author: */
import java.util.Scanner;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
 
    int r[]=new int[10000], t, i, n, f, count;
    double a, b, c;
    t=sc.nextInt();
    while(( t--)!=0 ) {
      n=sc.nextInt();
      f=sc.nextInt();
      f++;
      b = 0;
      for( i=0; i< n; i++ ) {
	r[i]=sc.nextInt();
	r[i] *= r[i];
	b += r[i];
      }

      a = 0;
      while( b-a> 1e-5 ) {
	c = (b+a)/2;
	count = 0;
	for( i=0; i< n; i++ )
	   count += (int)(r[i]/c);
	if( count >= f )
	   a = c;
	else
	   b = c;
	}
	System.out.printf( "%.4f\n", a*3.14159265358979324 );
    }
   }
}

  1. 有两个重复的话结果是正确的,但解法不够严谨,后面重复的覆盖掉前面的,由于题目数据限制也比较严,所以能提交通过。已更新算法

  2. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count