zoj 1337 pi

Professor Robert A. J. Matthews of the Applied Mathematics and Computer Science Department at the University of Aston in Birmingham, England has recently described how the positions of stars across the night sky may be used to deduce a surprisingly accurate value of Pi. This result followed from the application of certain theorems in number theory.

Here, we don't have the night sky, but can use the same theoretical basis to form an estimate for Pi:

Given any pair of whole numbers chosen from a large, random collection of numbers, the probability that the two numbers have no common factor other than one (1) is 6/Pi^2

For example, using the small collection of numbers: 2, 3, 4, 5, 6; there are 10 pairs that can be formed: (2,3), (2,4), etc. Six of the 10 pairs: (2,3), (2,5), (3,4), (3,5), (4,5) and (5,6) have no common factor other than one. Using the ratio of the counts as the probability we have:

6/Pi^2 = 6/10

Pi = 3.162

In this problem, you'll receive a series of data sets. Each data set contains a set of pseudo-random positive integers. For each data set, find the portion of the pairs which may be formed that have no common factor other than one (1), and use the method illustrated above to obtain an estimate for Pi. Report this estimate for each data set.

Input

The input consists of a series of data sets.

The first line of each data set contains a positive integer value, N, greater than one (1) and less than 50.

There is one positive integer per line for the next N lines that constitute the set for which the pairs are to be examined. These integers are each greater than 0 and less than 32768.

Each integer of the input stream has its first digit as the first character on the input line.

The set size designator, N, will be zero to indicate the end of data.

Output

A line with a single real value is to be emitted for each input data set encountered. This value is the estimate for Pi for the data set. An output format like the sample below should be used. Answers must be rounded to six digits after the decimal point.

For some data sets, it may be impossible to estimate a value for Pi. This occurs when there are no pairs without common factors. In these cases, emit the single-line message:

 

No estimate for this data set.

 

exactly, starting with the first character, "N", as the first character on the line.

Sample Input

5
2
3
4
5
6
2
13
39
0

Sample Output

3.162278
No estimate for this data set

 

简单题。

问题：那个公式必须清楚的。6/Pi^2 = 6/10，后面的6是固定的。

代码里有明确注释

#include<stdio.h>
#include<math.h>
int BJ(int a,int b)//函数功能：判断是否存在公约数
{
 int i;
 for(i=2;i<=a&&i<=b;i++)//i是小于较大数还是较小数？看是小于较大数！？？//解决：||或&&加上=号，就ok  
 {
  if(a%i==0&&b%i==0)
  return 1;
 }
 return 0;//全部判断完，说明这对数不存在公约数，就返回0，作为后面计数的条件。
}
int main()
{
 int N,a[50],i,j,k,M;//k:计数器累计不存在公约数的对数；M:计数器累计总对数。
 while(scanf("%d",&N)&&N)
 {  
  double t,x;
  k=0;
  M=0;
  for(i=0;i<N;i++)
   scanf("%d",&a[i]); //s输入测试数据
  for(i=0;i<N-1;i++)//对数据进行处理，注意i<N-1;j<N;
   for(j=i+1;j<N;j++) 
   {
     M++;
    if(BJ(a[i],a[j])==0) //
     k++;
   }
  // printf("%d/n",k);//k是计数器，累计不存在公约数的对数！
   if(k)
   {
    x=double(M)/double(k);//注意类型
    t=double(sqrt(6*x));
    printf("%.6lf/n",t);
   } 
    else
   {
    printf("No estimate for this data set./n");
   }
  
 }
 return 0;
}