zoj 1337

 

Pi

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Time Limit: 1 Second      Memory Limit: 32768 KB

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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.

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Source: East Central North America 1995

 

最近一直都在做水题，毕竟是初学者嘛，其实我觉得不管做什么题，只要有收获就行了。

这道题真的没什么难度，题目的意思就是从n*(n-1)/2组数据中，算出互质数的组数，然后再根据公式算出Pi就行了

主要是考察最多公约数gcd()

刚开始提交，老是WA，都不知道哪里错了，最后才发现是sqrt(3*n*(n-1)/hz)中出错了，不应该用3，而应该用3.0，不然精度就降低了。。。。看来以后要在细节上多注意下

 

#include<stdio.h>#include<math.h>int gcd(int a,int b){int r=0;while(b!=0){r=a%b;a=b;b=r;}return a;}int main(){int n;while(scanf("%d",&n)!=EOF&&n!=0){int num[51];int i;int hz=0;for(i=0;i<n;i++)scanf("%d",&num[i]);for(i=0;i<n;i++)for(int t=i+1;t<n;t++)if(gcd(num[i],num[t])==1)hz++;if(hz==0)printf("No estimate for this data set./n");else{printf("%.6lf/n",sqrt((3.0*n*(n-1))/hz));}}return 0;}