Posted on 2010-08-11 14:37
MiYu 阅读(1073)
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题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1969
题目描述:
Pie
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 65
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
题目分析:
2分求解.
题目大意是要办生日Party，有n个馅饼，有f个朋友，接下来是n个馅饼的半径。然后是分馅饼了，
注意咯自己也要，大家都要一样大，形状没什么要求，但都要是一整块的那种，也就是说不能从两个饼中
各割一小块来凑一块，像面积为10的和6的两块饼（饼的厚度是1，所以面积和体积相等），
如果每人分到面积为5，则10分两块，6切成5，够分3个人，如果每人6，则只能分两个了！
题目要求我们分到的饼尽可能的大！
只要注意精度问题就可以了,一般WA 都是精度问题.
代码如下:
#include <iostream>
#include <cmath>
using namespace std;
double a[10005];
int N,f;
double pi = acos ( -1.0 );
bool test ( double x )
{
int count = 0;
for ( int i = 1; i <= N; ++ i )
{
count += int ( a[i] / x );
}
if ( count >= f + 1 )
{
return true;
}
else
{
return false;
}
}
int main()
{
int T;
scanf ( "%d", &T );
while ( T -- )
{
double sum = 0;
double rad;
scanf("%d%d",&N,&f);
for ( int i = 1; i <= N; ++ i )
{
scanf ( "%lf", &rad );
a[i] = rad * rad * pi;
sum += a[i];
}
double max = sum / ( f + 1 );
double l = 0.0;
double r = max;
double mid = 0.0;
while ( r - l > 1e-6 )
{
mid = ( l + r ) / 2;
if ( test ( mid ) )
{
l = mid;
}
else
{
r = mid;
}
}
printf("%.4lf\n",mid);
}
return 0;
}