TJU Problem 1100 Pi

注：

　　1. 对于double计算，一定要小心，必要时把与double计算相关的所有都变成double型。

　　2. for (int i = 0; i < N; i++)         //N 不可写为N - 1，否则当N为1时无法进行；

原题：

1100.   Pi

Here, we don‘t have the night sky, but can use the same theoretical basis to
form an estimate for π:

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

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:

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 π. 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 π 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 π. 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.

Source: East Central
North America 1995

源代码：

 1 #include <iostream>
 2 #include <iomanip>
 3 #include <cmath>
 4 #include <stdio.h>
 5 using namespace std;
 6
 7 int num[55];
 8
 9 int gcd(int a, int b)    {
10     return b == 0 ? a : gcd(b, a % b);
11 }
12
13 int main()
14 {
15     int N;
16     while (cin >> N && N != 0)    {
17         int count = 0; double sum = 0;
18         for (int i = 0; i < N; i++)    cin >> num[i];
19         for (int i = 0; i < N; i++)         //N 不可写为N - 1，否则当N为1时无法进行；
20             for (int j = i + 1; j < N; j++)    {
21                 int m = num[i], n = num[j];
22                 if (gcd(m, n) == 1)
23                 {
24                     count++;
25                 }
26             }
27         sum = N * (N - 1) / 2;
28         if (count != 0)    {
29             double res = sqrt(6 * sum / count);
30             printf("%.6f\n",res);
31             //cout << fixed << setprecision(6) << res << endl;
32         }
33         else cout << "No estimate for this data set." << endl;
34     }
35     return 0;
36 }