HDU3988-Harry Potter and the Hide Story（数论-质因数分解）

Harry Potter and the Hide Story

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2193    Accepted Submission(s): 530

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

22 210 10

 

Sample Output

Case 1: 1Case 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 longconst 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;}