山东省第六届“浪潮杯”ACM程序设计大赛：D:Square Number

Description：

In mathematics, a square number is an integer that is the square of an integer. In other words, it is the product of some integer with itself. For example, 9 is a square number, since it can be written as 3 * 3.

Given an array of distinct integers (a1, a2, ..., an), you need to find the number of pairs (ai, aj) that satisfy (ai * aj) is a square number.

Input：

The first line of the input contains an integer T (1 ≤ T ≤ 20) which means the number of test cases.

Then T lines follow, each line starts with a number N (1 ≤ N ≤ 100000), then N integers followed (all the integers are between 1 and 1000000).

Output：

For each test case ,you should output the answer of each case

Sample Input:

1

5

1 2 3 4 12

Sample Output:

2

参考代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <string>
 6 #include <algorithm>
 7 #include <cstdlib>
 8 #define M 100000 + 1000
 9 using namespace std;
10 int a[M] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};
11 int b[M];
12 int fun(int n)
13 {
14     for(int i = 0; i < 168; i ++)           //这里i一定不能大于等于168，不然可能除0；可见边界并不是越大越好：
15     {
16         while(n % (a[i] * a[i]) == 0)
17             n = n / (a[i] * a[i]);
18         if(n < (a[i] * a[i]))
19             break;
20     }
21     return n;
22 }
23 int main()
24 {
25     //freopen("in.txt","r",stdin);
26     //freopen("out.txt","w",stdout);
27     int t;
28     scanf("%d",&t);
29     while(t --)
30     {
31         int n;
32         scanf("%d",&n);
33         for(int i = 0 ; i < n; i ++)
34         {
35             scanf("%d",&b[i]);
36             b[i] = fun(b[i]);
37         }
38         sort(b,b + n);
39         long long int cnt = 0;
40         for(int i = 0; i < n; i ++)
41         {
42             long long int temp = b[i];
43             long long int cnttemp = 0;
44             while(temp == b[i])                 //第一次把i ++写进了while里面，错了，查了好久检查不出错误。唉，i ++ 不是都能写进去的；
45             {
46                 cnttemp ++;
47                 i ++;                           //改到这里就对了
48             }
49             cnt = cnt + cnttemp * (cnttemp - 1) / 2;
50             i --;
51         }
52         printf("%lld\n",cnt);
53     }
54     return 0;
55 }