版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/sxy201658506207/article/details/84676996
iSea is tired of writing the story of Harry Potter, so, lucky you, solving the following problem is enough.
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case contains two integers, N and K.
Technical Specification
1. 1 <= T <= 500
2. 1 <= K <= 1 000 000 000 000 00
3. 1 <= N <= 1 000 000 000 000 000 000
Output
For each test case, output the case number first, then the answer, if the answer is bigger than 9 223 372 036 854 775 807, output “inf” (without quote).
Sample Input
2
2 2
10 10
Sample Output
Case 1: 1
Case 2: 2
#include <bits/stdc++.h>
#define X 10005
#define inF 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
const int N = 1e4;
const int Times=10;
const ll inf= 9223372036854775807;
ll ct,cnt;
ll fac[N],num[N];
ll gcd(ll a,ll b)
{
if(!b)
return a;
else
gcd(b,a%b);
}
ll mul_mod(ll a,ll b,ll m)
{
ll ans=0;
while(b)
{
if(b&1)
{
ans=(ans+a)%m;
}
a=(a+a)%m;
b>>=1;
}
return ans;
}
ll Pow_mod(ll a, ll b,ll m)
{
ll ans=1;
a%=m;
while(b)
{
if(b&1)
{
ans=mul_mod(ans,a,m);//ans=ans*a%m;
}
a=mul_mod(a,a,m);//a=a*a%m;
b>>=1;
}
return ans;
}
bool Miller_Rabin(ll n)
{
if(n==2)
return true;
if(n<2||!(n&1))
return false;
ll a,m=n-1,x,y;
int k=0;
// while((m&1)==0)//判奇数
// {
// k++;
// m>>=1;
// }
while (~m & 1) //找到奇数
{
m >>= 1;
k++;
}
for(int i=0;i<Times;++i)
{
a=rand()%(n-1)+1;
x=Pow_mod(a,m,n);
for(int j=0;j<k;++j)
{
y=mul_mod(x,x,n);
if(y==1&&x!=1&&x!=n-1)
return false;
x=y;
}
if(y!=1) //a^(2*k*r)%m=1
return false;
}
return true;
}
ll f(ll x,ll n)
{
return (mul_mod(x,x,n)+2)%n;
}
ll Pollard_rho(ll n,ll c)
{
ll x,y,d,i=1,k=2;
y=x=rand()%(n-1)+1;
while(true)
{
i++;
x=f(x,n);//x=(mul_mod(x,x,n)+c)%n;
d=gcd((y-x+n)%n,n);
if(1<d&&d<n)
return d;
if(y==x)
return n;//没有随机查找素因子失败，返回
if(i==k)
{
y=x;
k<<=1;
}
}
// do
// {
// i++;
// x=f(x,n);//(mul_mod(x,x,n)+c)%n;
// d=gcd((y-x+n)%n,n);
// if(1<d&&d<n)
// return d;
//// if(y==x)
//// return n;//没有随机查找素因子失败，返回
// if(i==k)
// {
// y=x;
// k<<=1;
// }
// }while(y!=x);
return n;
}
ll get(ll p,ll n)
{
ll ans=0;
while(n)
{
ans+=n/p;/// no
n/=p;
}
return ans;
}
void Find(ll n,int c)
{
if(n==1)
return ;
if(Miller_Rabin(n))
{
fac[ct++]=n;
return ;
}
///不是素数：
ll p = n, k = c;
while(p>=n)
p = Pollard_rho(p,c);///找到n的一个因子，但是现在还不能保证是一个素因子，要进行Miller_rabin素数检测
Find(p,k);
Find(n/p,k);//因为n是偶数
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
int t=1,tt=1;
ll n,k,minx;
cin>>t;
for(int Case=1;Case<=t;++Case)
{
cin>>n>>k;
cout<<"Case "<<Case<<": ";
if(k==1)
{
cout<<"inf"<<endl;
continue;
}
ct=0;
Find(k,2);
sort(fac,fac+ct);
num[0]=1;
int k=1;
for(int i=1;i<ct;++i)
{
if(fac[i]==fac[i-1])
{
num[k-1]++;
}
else
{
num[k]=1;
fac[k++]=fac[i];
}
}
cnt=k;
minx=inf;
for(int i=0;i<cnt;++i)
{
ll tmp=get(fac[i],n)/num[i];
minx=min(minx,tmp);
}
if(minx==inf)
cout<<"inf"<<endl;
else
cout<<minx<<endl;
}
return 0;
}
/*
对大整数进行质因数分解 采用 Miller_Rabin 进行素数判定 + Pollard_rho 对整数进行因数分解(Mill_Rabin 判定是否是素数因子)
*/