SDUT3257Cube Number

Cube Number

Time Limit: 2000MS  Memory Limit: 65536KB

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Problem Description

In mathematics, a cube number is an integer that is the cube of an integer. In other words, it is the product of some integer with itself twice. For example, 27 is a cube number, since it can be written as 3 * 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 cube number.

Input

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).

Output

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

Example Input

1   
5   
1 2 3 4 9

Example Output

2

题意：

求乘积为立方数的对数。

分析：

先筛掉每个原数中的立方因子得到新数，对于每一个新数，若存在平方因子，则需要找该平方因子为1的对应的数与其匹配；若不存在平方因子（即只含单个因子），则找含有两个该单因子的数与其匹配。

注意，找到的匹配的数可能会超过1e^6，此时需要判断一下，并且用long long来存储，否则会RE。

具体代码实现的时候，使用一个变量tem，记录当前的数所含素因子的情况，即每次把当前的数所含的平方因子、单个因子都用tem累乘起来。该tem的个数加1。这样下一个数找对应匹配的数的时候，只要累加之前已经出现过的对应匹配的数的个数就可以了。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn = 1005;
bool vis[maxn];
int prim[maxn], cnt[1000005], tol;
void init()//1000以内的素数打表
{
    for(int i = 2; i <= 1000; i++)
    {
        if(!vis[i])
            prim[tol++] = i;
        for(int j = 0; j < tol && prim[j]*i <= 1000; j++)
        {
            vis[i*prim[j]] = true;
            if(i % prim[j] == 0)
                break;
        }
    }
}
int main()
{
    init();
    int n, ans, T;
    cin >> T;
    while(T--)
    {
        cin >> n;
        ans = 0;
        memset(cnt, 0, sizeof(cnt));
        for(int i = 1; i <= n; i++)
        {
            int x;
            bool flag = true;
            int self = 1;ll need = 1;
            cin >> x;
            for(int j = 0; j < tol && prim[j] <= x; j++)
            {
                int y = prim[j]*prim[j]*prim[j];
                while(x%y == 0)
                    x /= y;
                if(x%(prim[j]*prim[j]) == 0)
                {
                    x /= (prim[j]*prim[j]);
                    self = self * prim[j] * prim[j];
                    if(flag)
                        need *= prim[j];
                }
                else if(x%prim[j] == 0)
                {
                    x /= prim[j];
                    self *= prim[j];
                    if(flag)
                        need *= (prim[j]*prim[j]);
                }
                if(need > 1e6)
                    flag = false;
            }
            if(flag)
            {
                if(x != 1)
                    need = need*x*x;//这里不能用*=,因为x会爆int
                if(need <= 1e6)
                    ans += cnt[need];
            }
            if(x != 1)
                self *= x;
            cnt[self]++;
        }
        cout << ans << endl;
    }
    return 0;
}