Square Number

Time Limit: 1000ms   Memory limit: 65536K  有疑問？點這裡^_^

題目描述

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.

輸入

 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).

輸出

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

示例輸入

1   
5   
1 2 3 4 12

示例輸出

2

分析：

判斷兩個數的乘積是否為平方數，需要先將該數字拆分多個素數的平方數的形式，例如：

12 = 2*2*3

3 = 3

這樣2*2就可以忽略不看，因為它已經是平方數，只需要找3與12的質因數3配成平方數即可.

先打一個1000的素數表，因為最大資料是1e6，所以取根號大小就行。然後根據素數表打一個平方數表。

程式碼如下：

#include <stdio.h>
#include <string.h>
#define MAX 1000010
int pri[1010];
int dp[200];
int vis[MAX];
int amount=0;
void init()
{
	for(int i=2;i<=1000;i++)
	{//素數打表 
		for(int j=i+i;j<=1000;j+=i)
		{
			if(!pri[j])
				pri[j]=1;
		}
	}
	//平方數打表 
	for(int i=2;i<=1000;i++)
		if(!pri[i])
			dp[amount++]=i*i;
}

int main()
{
	int T,n;
	int t;
	int i,j;
	init();
	scanf("%d",&T);
	while(T--)
	{
		memset(vis,0,sizeof(vis));
		int ans=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d",&t);
			for(j=0;j<amount;j++)
			{
				if(t%dp[j]==0)
				{
					while(t%dp[j]==0)
						t/=dp[j];
				}
			}
			ans+=vis[t];
			//除去平方因數後的值 ，有多少個該平方因數的值就可以組成多少對 
			vis[t]++;
		}
		printf("%d\n",ans);
	}
	return 0;
}