TOJ 4383: n % ( pow( p , 2) ) ===0

4383: n % ( pow( p , 2) ) ===0 

Time Limit(Common/Java):10000MS/30000MS     Memory Limit:65536KByte
Total Submit: 237            Accepted:54

Description

 

There is a number n , determine whether there is a p (p>1) that p^2 is a divisor of n.

Input

 

The first line contains an integer T , the number of test case.

The following T lines , each contains an integer n.

( 1<= T <=10^2 , 1<= n <=10^18 )

Output

 

A integer p , if there exist multiple answer ,output the minimum one.

Or print “oh,no.” .

Sample Input

 

3
8
16
17

Sample Output

 

2
2
oh,no.

Source

数信学院第5届新生程序设计竞赛

这个题是错了一个数据，最后我去修了这个数据并加强了，好开心

就是问一下一个数是不是一个素数的平方，这个题的思路比较简单，就是一个大数的素数分解，直接搞一个miller_rabin的素数检测，再来一个pollard_rho大数分解质因数就好的

那个判断次数一般选20

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<algorithm>
#include<cstring>
using namespace std;
typedef __int64 ll;
const int times=20;
const int N=100;
ll mult_mod(ll a,ll b,ll mod)
{
    a%=mod;
    b%=mod;
    ll res=0;
    while(b)
    {
        if(b&1)
        {
            res+=a;
            res%=mod;
        }
        a<<=1;
        if(a>=mod) a%=mod;
        b>>=1;
    }
    return res;
}
ll pow_mod(ll x,ll n,ll mod)
{
    if(n==1) return x%mod;
    x%=mod;
    ll t=x;
    ll res=1;
    while(n)
    {
        if(n&1) res=mult_mod(res,t,mod);
        t=mult_mod(t,t,mod);
        n>>=1;
    }
    return res;
}
bool test(ll a,ll n,ll x,ll t)
{
    ll res=pow_mod(a,x,n);
    ll last=res;
    for(int i=1; i<=t; i++)
    {
        res=mult_mod(res,res,n);
        if(res==1&&last!=1&&last!=n-1) return true;
        last=res;
    }
    if(res!=1) return true;
    return false;
}
bool miller_rabin(ll n)
{
    if(n<2) return false;
    if(n==2) return true;
    if((n&1)==0) return false;
    ll x=n-1,t=0;
    while((x&1)==0)
    {
        x>>=1;
        t++;
    }
    for(int i=0; i<times; i++)
    {
        ll a=rand()%(n-1)+1;
        if(test(a,n,x,t)) return false;
    }
    return true;
}
ll factor[N];
int tot;
ll gcd(ll a,ll b)
{
    if(a==0) return 1;
    if(a<0) return gcd(-a,b);
    while(b)
    {
        ll c=a%b;
        a=b;
        b=c;
    }
    return a;
}
ll pollard_rho(ll x,ll c)
{
    ll i=1,k=2;
    ll x0=rand()%x;
    ll y=x0;
    while(1)
    {
        i++;
        x0=(mult_mod(x0,x0,x)+c)%x;
        ll d=gcd(y-x0,x);
        if(d!=1&&d!=x) return d;
        if(y==x0) return x;
        if(i==k)
        {
            y=x0;
            k+=k;
        }
    }
}
void find_factor(ll n)
{
    if(miller_rabin(n))
    {
        factor[tot++]=n;
        return ;
    }
    ll p=n;
    while(p>=n) p=pollard_rho(p,rand()%(n-1)+1);
    find_factor(p);
    find_factor(n/p);
}
int main()
{
    srand(time(0));
    int t;
    scanf("%d",&t);
    ll n;
    while(t--)
    {
        scanf("%I64d",&n);
        if(miller_rabin(n)||n<=1)
        {
            printf("oh,no.
");
            continue;
        }
        tot=0;
        find_factor(n);
        sort(factor,factor+tot);
        int f=1;
        for(int i=1; i<tot; i++) if(factor[i]==factor[i-1])
            {
                printf("%I64d
",factor[i]);
                f=0;
                break;
            }
        if(f)printf("oh,no.
");
    }
    return 0;
}