山东理工3258 Square Number(质因数分解+乱搞)

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=3258

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

题意：输入一组数，查找其中任意两个（不重复）乘积为平方数的个数。（例题答案：1*4；3*12）.

理解：预处理素数，但是需要注意的是题目范围是1e6，意味着素数只需要1000以内的（否则会超时），原因很简单，如果一个数的最小质因数都大于1000，那么这个数就超过题目的范围了。所以预处理只要处理1~1000的素数就可以了。

    #include <stdio.h>
    #include <string.h>
    #include <math.h>
    #include <iostream>
    #include <map>
    #include <algorithm>
    using namespace std;
    int  p[200];
    bool prime[1005];
    int k=0;
    void isprime()
    {
        int i,j;
        memset(prime,true,sizeof(prime));
        for(i=2;i<=1000;i++)
        {
            if(prime[i])
            {
                p[k++]=i;
                for(j=i*i;j<=1000;j+=i)
                {
                    prime[j]=false;
                }
            }
        }
    }
    int test(int n)
    {
        int  flag=1;
        for(int i=0;i<k;i++)
        {
            while(!(n%(p[i]*p[i])))
            {
                n/=p[i]*p[i];
            }
            
        }
        return n;
    }
    int main()
    {

        int t,shu;
        long long ans;
        isprime();
        scanf("%d",&t);
        while(t--)
        {
           int n;
           ans=0;
           map<int,int> m1;
           map<int,int>::iterator i_it;
           scanf("%d",&n);
           for(int i=1;i<=n;i++)
           {
               scanf("%d",&shu);
               int x=test(shu);
              // ans-=1LL*m1[x]*(m1[x]-1)/2;
             m1[x]++;

          //     ans+=1LL*m1[x]*(m1[x]-1)/2;
           }
           for(i_it=m1.begin();i_it!=m1.end();i_it++)
                ans+=1LL*i_it->second*(i_it->second-1)/2;
           printf("%lld\n",ans);
        }
     return 0;
    }