“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 1001000using 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 longusing 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;    }