SDUT 3258 Square Number(2015年山东省第六届ACM大学生程序设计竞赛)

Square Number

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

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

分析：

判断两个数的乘积是否为平方数，需要先将该数字拆分多个素数的平方数的形式，例如：

12 = 2*2*3

3 = 3

这样2*2就可以忽略不看，因为它已经是平方数，只需要找3与12的质因数3配成平方数即可.

先打一个1000的素数表，因为最大数据是1e6，所以取根号大小就行。然后根据素数表打一个平方数表。

代码如下：

#include <stdio.h>#include <string.h>#define MAX 1000010int pri[1010];int dp[200];int vis[MAX];int amount=0;void init(){for(int i=2;i<=1000;i++){//素数打表 for(int j=i+i;j<=1000;j+=i){if(!pri[j])pri[j]=1;}}//平方数打表 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];//除去平方因数后的值 ，有多少个该平方因数的值就可以组成多少对 vis[t]++;}printf("%d\n",ans);}return 0;}