“浪潮杯”山东省第6届acm省赛 sdut3258 Square Number sdut 3257 Cube Number

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

平方数就一定能表示为若干（素数的偶次幂）的乘积。

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#define LL long long
#define MAX 1001000
using namespace std;
int a[MAX],flag[MAX];
bool visited[MAX];
int main()
{    
    int T,n,b;
    int num= 0;//计数器
    memset(visited,0,sizeof(visited));
    int m=sqrt(MAX+0.5);//素数筛选法将MAX之内素数的平方存入数组a[]
    for(int i=2;i<=m;i++)
        if(!visited[i])
    {
        a[num++]=i*i;
        for(int j=i*i;j<=MAX;j+=i)
            visited[j]=1;
    }
    
    
    cin>>T;
    while(T--)
    {
        memset(flag,0,sizeof(flag));
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>b;
            for(int j=0;a[j]<=b&&j<=num-1;j++)//将b分解为若干素数的2次幂的乘积。最后剩余部分记录入flag等待匹配
            {
                while(b % a[j] == 0)
                    b /= a[j];
            }
            flag[b]++;
        }
        
        
        LL ant = 0;
        for(int i=1;i<MAX;i++)
            ant += flag[i]*(flag[i]-1)/2;//flag[i]表示有flag[i]个数剩余部分为i，这些数可以进行两两匹配
        cout<<ant<<endl;
    }
    return 0;
}

题目描述
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.
 
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 9
示例输出
2

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define MAX 1000000
#define LL long long
using namespace std;
bool visited[1000050];
int prime[100050];
int flag[1000050];
int num[100050];
int main()
{
    int num=0;
    int m=sqrt(MAX+0.5);
    memset(visited,0,sizeof(visited));
    for(int i=2;i<=m;i++)
    {
        if(!visited[i])
        {
            prime[num++]=i;
        for(int j=i*i;j<=MAX;j+=i)
            visited[j]=1;
        }
    }
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        memset(flag,0,sizeof(flag));
        LL sum=0;
        for(int i=0;i<=n-1;i++)
        {
            int m;
            scanf("%d",&m);
            LL a=1,b=1;
            bool flagx=0;
            for(int j=0;j<=num-1&&prime[j]<=m;j++)
            {
                LL t=prime[j]*prime[j]*prime[j];
                while(m%t==0)
                    m=m/t;
                if(m%(prime[j]*prime[j])==0)
                {
                    m=m/(prime[j]*prime[j]);
                    a=a*prime[j]*prime[j];
                    if(!flagx)
                        b=b*prime[j];
                }
                else if(m%prime[j]==0)
                {
                    m=m/prime[j];
                    a=a*prime[j];
                    if(!flagx)
                        b=b*prime[j]*prime[j];
                }
                if(b>MAX)
                    flagx=1;
            }
            if(!flagx)
            {
                b=b*m*m;
                if(b<MAX)
                sum+=flag[b];
            }
            a=a*m;
            flag[a]++;
        }
        cout<<sum<<endl;
    }
    return 0;
}
 

素数筛选模板：

筛选出<=n的素数
 int num[]存筛选的素数值 ,bool visited[n];
    int top = 0;
    memset(visited,0,sizeof(visited));
    int m=sqrt(MAX+0.5);
    for(int i=2;i<=m;i++)
        if(!visited[i])
    {
        num[top++]=i;
        for(int j=i*i;j<=MAX;j+=i)
            visited[j]=1;
    }