• TOJ 4383: n % ( pow( p , 2) ) ===0


    4383: n % ( pow( p , 2) ) ===0 分享至QQ空间

    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;
    }
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  • 原文地址:https://www.cnblogs.com/BobHuang/p/7805731.html
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