HDU 4344 大数分解

题意：给出一条长n的绳子，则有很多长为L的均分方法，找出一个L的集合，集合内元素两两不互素，并且要求元素最多，然后求和。

很容易看出来，有多少个素因子就有多少个元素，再把求出次幂求和就行，我的模板超时。。网上找了模板。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
using namespace std;
typedef __int64 ll;
ll gcd(ll a,ll b)
{
    return (b==0)?a:gcd(b,a%b);
}
ll Mulmod(ll a,ll b,ll n)
{
    ll  exp = a%n,res = 0;
    while(b)
    {
        if(b&1)
        {
            res += exp;
            if(res>n) res -= n;
        }
        exp <<= 1;
        if(exp>n)
            exp -= n;

        b>>=1;
    }
    return res;
}

ll exp_mod(ll a,ll c)
{
    ll k = 1;
    while(b)
    {
        if(b&1)
            k = Mulmod(k,a,c);
        a = Mulmod(a,c);
        b>>=1;
    }
    return k;
}
bool Miller_Rabbin(ll n,ll times)
{
    if(n==2)return 1;
    if(n<2||!(n&1))return 0;

    ll a,u=n-1,x,y;
    int t=0;
    while(u%2==0)
    {
        t++;
        u/=2;
    }
    srand(100);
    for(int i=0; i<times; i++)
    {
        a = rand() % (n-1) + 1;
        x = exp_mod(a,u,n);
        for(int j=0; j<t; j++)
        {
            y = Mulmod(x,n);
            if ( y == 1 && x != 1 && x != n-1 )
                return false; //must not
            x = y;
        }
        if( y!=1) return false;
    }
    return true;
}
ll Pollard_Rho(ll n,ll c)
{
    ll x,y,d,i=1,k=2;
    y = x = rand() % (n-1) + 1;
    while(1)
    {
        i++;
        x = (Mulmod(x,n) + c)%n;
        d = gcd((x-y+n)%n,n);
        if(d>1&&d<n)
            return d;
        if(x==y)
            return n;
        if(i==k)
        {
            k<<=1;
            y = x;
        }
    }
}
ll factor[200],cnt;
void Find_factor(ll n,ll c)
{
    if(n==1)
        return;
    if(Miller_Rabbin(n,6))
    {
        factor[cnt++] = n;
        return;
    }
    ll p = n;
    ll k = c;
    while(p>=n)
        p = Pollard_Rho(p,c--);
    Find_factor(p,k);
    Find_factor(n/p,k);
}
ll a_b(ll a,ll b)
{
    ll ans = 1;
    for(ll i=0; i<b; i++)
        ans*=a;
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    ll n;
    while(t--)
    {
        scanf("%I64d",&n);
        cnt = 0;
        Find_factor(n,120);
        map<ll,int>m0;
        for(int i=0; i<cnt; i++)
        {
            m0[factor[i]]++;
        }
        map<ll,int>::iterator iter;
        int size = m0.size();
        if(size==1)
        {
            printf("%d %I64dn",size,n/factor[0]);
            continue;
        }
        ll sum = 0;
        for(iter=m0.begin(); iter!=m0.end(); iter++)
            sum+=a_b(iter->first,iter->second);
        printf("%d %I64dn",sum);
    }
    return 0;
}