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 0x3f3f3f3f
using 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;
}