山东省第四届ACM省赛题——Square Number（平方数的性质）

题目描述
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.

输入
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).

输出
For each test case, you should output the answer of each case.
示例输入
1
5
1 2 3 4 12
示例输出
2

题意不难理解，主要是数据范围很大，暴力打表肯定不行。对于平方数n，它有一个性质是n一定等于奇数个质数相乘。比如36，它可以表示为2x18,3x12,4x9，其本质都是2x2x3x3。
所以对于题目中输入的每个数，要找出它的奇数的质数，比如3和12，两个数都有一个奇数的3，所以他们能相乘得到平方数，如果有3个以上奇个数的质数，取出任意两个都能组成平方数，因此这里要用到组合公式

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <cstdio>#include <set>#include <vector>#include <iomanip>#include <stack>#include <map>#define MAXN 1000005#define mod 9973#define inf 0x3f3f3f3fusing namespace std;int p[MAXN],a[MAXN],sq[MAXN],cnt;  //p保存素数，a判断当前数是否为素数，sq保存质数的平方数int vis[MAXN]; //vis保存奇个数的质数void Primer(){    int i,j, k;    for(i=2;i<MAXN;++i)   //i从2开始遍历    {        if(a[i]==0)    //a[i]==0说明该i是素数            p[cnt++]=i;   //cnt作为地址标志        for(j=0;j<i&&(k=i*p[j])<MAXN;++j)   //j从0开始遍历，k作为有质数p[j]的合数        {            a[k]=1;    //合数设1            if(i%p[j]==0)                break;        }    }}int sum(int x){    return x*(x-1)/2; //组合公式的化简}int main(){    ios::sync_with_stdio(false);    cnt=0;    Primer();    int t,n,top=0,x;    for(int i=0;i<cnt;++i)    {        sq[top++]=p[i]*p[i];    }    cin>>t;    while(t--)    {        memset(vis,0,sizeof(vis));        cin>>n;        for(int i=0;i<n;++i)        {            cin>>x;            for(int j=0;sq[j]<=x;++j)            {                while(x%sq[j]==0)                    x/=sq[j];            }            vis[x]++;        }        int ans=0;        for(int i=0;i<MAXN;++i)        {            ans+=sum(vis[i]);        }        cout<<ans<<endl;    }    return 0;}