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

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;
}