DESCRIPTION
Bob and his girl friend Alice like factors very much.
One day, Bob find a beautiful path: 4, 2, 1, Becauese 2 is the factor of 4 and 1 is the factor of 2.
Alice is excited, says “why can’t we find a longest path A1,A2,A3,…,Ak,Ai+1
is the factor of Ai and Ai+1
#include
#include
#include
#include
#include
#include
using namespace std;
const int S=20;
typedef long long LL;
const long long mod=1e9+7;
long long mult_mod(long long a,long long b,long long c)
{
a%=c;
b%=c;
long long ret=0;
while(b)
{
if(b&1){ret+=a;ret%=c;}
a<<=1;
if(a>=c)a%=c;
b>>=1;
}
return ret;
}
long long pow_mod(long long x,long long n,long long mod)//x^n%c
{
if(n==1)return x%mod;
x%=mod;
long long tmp=x;
long long ret=1;
while(n)
{
if(n&1) ret=mult_mod(ret,tmp,mod);
tmp=mult_mod(tmp,tmp,mod);
n>>=1;
}
return ret;
}
bool check(long long a,long long n,long long x,long long t)
{
long long ret=pow_mod(a,x,n);
long long last=ret;
for(int i=1;i<=t;i++)
{
ret=mult_mod(ret,ret,n);
if(ret==1&&last!=1&&last!=n-1) return true;//合数
last=ret;
}
if(ret!=1) return true;
return false;
}
bool Miller_Rabin(long long n)
{
if(n<2)return false;
if(n==2)return true;
if((n&1)==0) return false;//偶数
long long x=n-1;
long long t=0;
while((x&1)==0){x>>=1;t++;}
for(int i=0;ilong long a=rand()%(n-1)+1;//rand()需要stdlib.h头文件
if(check(a,n,x,t))
return false;//合数
}
return true;
}
map<long long,long long> factor;
map<long long,long long>::iterator it1;
int tol;
long long gcd(long long a,long long b)
{
if(a==0)return 1;//???????
if(a<0) return gcd(-a,b);
while(b)
{
long long t=a%b;
a=b;
b=t;
}
return a;
}
long long Pollard_rho(long long x,long long c)
{
long long i=1,k=2;
long long x0=rand()%x;
long long y=x0;
while(1)
{
i++;
x0=(mult_mod(x0,x0,x)+c)%x;
long long 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 findfac(long long n)
{
if(Miller_Rabin(n))
{
factor[n]++;
//factor[tol++]=n;
return;
}
long long p=n;
while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1);
findfac(p);
findfac(n/p);
}
void exgcd(LL a, LL b, LL &x, LL &y) //ax+by=1; x为a mod b 的逆元,y为 b mod a的逆元 a,b互质
{
if(b == 0)
{
x = 1;
y = 0;
return;
}
LL x1, y1;
exgcd(b, a%b, x1, y1);
x = y1;
y = x1-(a/b)*y1;
}
int main()
{
int t;
scanf("%d",&t);
long long n;
for(int i=1;i<=t;i++)
{
factor.clear();
scanf("%lld",&n);
printf("Case #%d: ",i);
if(n==1){
printf("1 1\n");
continue;
}
if(Miller_Rabin(n)){
printf("2 1\n");
continue;
}
findfac(n);
long long sum=0;
for(it1=factor.begin();it1!=factor.end();it1++){
sum+=it1->second;
}
printf("%lld ",(sum+1)%mod);
long long ans=1,x,y;
for(int i=2;i<=sum;i++) ans=ans*i%mod;
for(it1=factor.begin();it1!=factor.end();it1++){
long long gg=it1->second,an=1;
for(int i=2;i<=gg;i++) an=an*i%mod;
exgcd(an,mod,x,y);
ans=ans*x%mod;
}
printf("%lld\n",ans%mod);
}
return 0;
}