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 b,ll c){    ll k = 1;    while(b)    {        if(b&1)            k = Mulmod(k,a,c);        a = Mulmod(a,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, 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,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 %I64d\n",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 %I64d\n",size,sum);    }    return 0;}