H - Square Number
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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Status
Description
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.
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.
Sample Input
1
5
1 2 3 4 12
Sample Output
2
【分析】
题意:给定一串数字,求有多少对i,j满足a[i]*a[j]是一个平方数
经典题平方数....
首先,先对每个数进行质因数分解,质因数分解需要预处理一下素数,1000以内的素数所以直接暴力筛就行了
如果a[i]和a[j]满足条件,那么他们中某个质因数的个数和加起来一定是偶数,为了方便计算所以对每个数去掉它因子中已经成对的因子,因为自身的平方因子是无用的。
然后对剩下的那个数,在1-i中查找有没有能够组成偶数因子的数,其实就是找自己..在1-i中找自己就行了
【代码】
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int len=0;
int vis[1001000]={0};
int prime[10000]={0};
void init()
{
for(int i=2;i<1010;i++)
if(!vis[i])
{
prime[len++]=i;
for(int j=i+i;j<1010;j+=i) vis[j]=1;
}
}
int main()
{
init();
int pp,x;scanf("%d",&pp);
while(pp--)
{
memset(vis,0,sizeof(vis));
int n;scanf("%d",&n);
int ans=0;
while(n--)
{
scanf("%d",&x);
for(int i=0;i<len;i++)
{
int t=prime[i]*prime[i];
if (t>x) break;
while(x%t==0) x/=t;
}
ans+=vis[x];vis[x]++;
}
printf("%d\n",ans);
}
return 0;
}