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 12Sample Output
2【分析】
题意：给定一串数字，求有多少对i,j满足a[i]*a[j]是一个平方数
经典题平方数....
首先，先对每个数进行质因数分解，质因数分解需要预处理一下素数，1000以内的素数所以直接暴力筛就行了
如果a[i]和a[j]满足条件，那么他们中某个质因数的个数和加起来一定是偶数，为了方便计算所以对每个数去掉它因子中已经成对的因子，因为自身的平方因子是无用的。 然后对剩下的那个数，在1-i中查找有没有能够组成偶数因子的数，其实就是找自己..在1-i中找自己就行了
【代码】
#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; int len=0; int vis[1001000]={0}; int prime[10000]={0}; void init() { for(int i=2;i<1010;i++) if(!vis[i]) { prime[len++]=i; for(int j=i+i;j<1010;j+=i) vis[j]=1; } } int main() { init(); int pp,x;scanf("%d",&pp); while(pp--) { memset(vis,0,sizeof(vis)); int n;scanf("%d",&n); int ans=0; while(n--) { scanf("%d",&x); for(int i=0;i<len;i++) { int t=prime[i]*prime[i]; if (t>x) break; while(x%t==0) x/=t; } ans+=vis[x];vis[x]++; } printf("%d\n",ans); } return 0; }
