Problem Description

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

 

Author

iSea@WHU

 

题意：给你n和k,让你求出最大的i 满足n的阶乘能被k的i次方整除。

思路:对k进行质因数分解。求出每一个质因数在阶乘中的幂的大小。答案即为最小的那个幂。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <vector>

#include <string>

#include <algorithm>

#include <queue>

using namespace std;

#define LL unsigned long long

const int maxn = 10000005;

bool isPrime[maxn];

vector<LL> prime,digit,cnt;

void getPrime(){

    prime.clear();

    memset(isPrime,0,sizeof isPrime);

    for(LL i = 2; i < maxn; i++){

        if(!isPrime[i]){

            prime.push_back(i);

            for(LL j = i*i; j < maxn; j += i)

                isPrime[j] = 1;

        }

    }

}

void getDigit(LL k){

    for(int i = 0; i < prime.size() && k >= prime[i]; i++){

        if(k%prime[i]==0){

            int tt = 0;

            digit.push_back(prime[i]);

            while(k%prime[i]==0){

                tt++;

                k /= prime[i];

            }

            cnt.push_back(tt);

        }

    }

    if(k!=1){

        digit.push_back(k);

        cnt.push_back(1);

    }

}

LL getSum(LL n,LL p){

    LL res = 0;

    while(n){

        n /= p;

        res += n;

    }

    return res;

}

int main(){

    int ncase,T=1;

    LL k,n;

    getPrime();

    cin >> ncase;

    while(ncase--){

        cin >> n >> k;

        if(k==1){

            printf("Case %d: inf\n",T++);

            continue;

        }

        LL ans = -1;

        digit.clear();

        cnt.clear();

        getDigit(k);

        for(int i = 0; i < digit.size(); i++){

            LL tk = getSum(n,digit[i])/cnt[i];

            if(ans == -1) ans = tk;

            else ans = min(ans,tk);

        }

        printf("Case %d: %I64u\n",T++,ans);

    }

    return 0;

}

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