sdut 3258 Square Number 打表

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

提示

 

来源

 

示例程序

 

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• 讨论
• 问题是要求两两相乘为平方数
  考虑到一个数可以分为若干个质数相乘的形式
  如果A*B为平方数的话那么A*B可以转化为偶对个质数相乘的形式
  那么我们可以把数A中偶数个质数相乘的因子去掉 如果经过相同操作后的B与之相等 那么他们就可以配对
  例如 3 和 12
  3 分解为 3 那么 就是 3
  12 分解为 3*2*2 除去2*2 为3
  那么 3和12这一对就满足条件
  对于45 分解为 3*3*5 化为 5 就不能和3配对了
  ACcode：

  #pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)#define rdl(x) scanf("%I64d,&x);#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define ull unsigned long long#define maxn 1000100#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)#define eps 1e-8ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}ll mul(ll a,ll b,ll p){ll sum=0;for(;b;a=(a+a)%p,b>>=1)if(b&1)sum=(sum+a)%p;return sum;}inline void Scan(int &x) {      char c;while((c=getchar())<'0' || c>'9');x=c-'0';      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';}using namespace std;int prime[maxn];int dp[222];int vis[maxn];void init(){    memset(prime,0,sizeof(prime));    for(int i=2;i<=maxn;++i){        if(!prime[i])prime[++prime[0]]=i;        for(int j=1;j<=prime[0]&&prime[j]<=maxn/i;j++){            prime[prime[j]*i]=1;            if(i%prime[j]==0)break;        }    }    for(int i=1;i<=200;++i)        dp[i]=prime[i]*prime[i];}int main(){    int n,loop,cnt=1;    init();    scanf("%d",&loop);    while(loop--){        scanf("%d",&n);        memset(vis,0,sizeof(vis));        int ans=0;        for(int i=0;i<n;++i){            int t;            scanf("%d",&t);            for(int j=1;j<=199;++j)                if(t%dp[j]!=0)continue;                else while(t%dp[j]==0)t/=dp[j];            ans+=vis[t];            vis[t]++;        }        printf("%d\n",ans);    }    return 0;}/*731 2 471 2 3 4 5 6 782 34123 344 35 33 23 434232 123 1234 3361 4 9 16 25 36*/