首页 > 代码库 > HDU 3988 Harry Potter and the Hide Story（数论-整数和素数）

HDU 3988 Harry Potter and the Hide Story（数论-整数和素数）

2024-07-15 02:16:59   223人阅读

Harry Potter and the Hide Story



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
 

Source
2011 Multi-University Training Contest 15 - Host by WHU
 

Recommend
lcy   |   We have carefully selected several similar problems for you:  3987 3986 3983 3984 3985 
 

题目大意：

给定 n和k ， 求 n! % k^i 等于0时，i 的最大取值是多少？


解题思路：

将 k分解质因素，n也根据k的质因素求出关系限制i，最后算出最大的i即可。


解题代码：

#include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <cstring>
using namespace std;

typedef unsigned long long ll;
ll n,k;

const int maxn=10000010;
bool isPrime[maxn];
vector <ll> v;
ll tol;

void get_prime(){
    tol=0;
    memset(isPrime,true,sizeof(isPrime));
    for(ll i=2;i<maxn;i++){
        if(isPrime[i]){
            tol++;
            v.push_back(i);
        }
        for(ll j=0;j<tol && i*v[j]<maxn;j++){
            isPrime[i*v[j]]=false;
            if(i%v[j]==0) break;
        }
    }
    //for(ll i=v.size()-1;i>=v.size()-100;i--) cout<<v[i]<<endl;
}

map <ll,ll> getPrime(ll x){
    map <ll,ll> mp;
    for(ll i=0;i<tol && x>=v[i];i++){
        while(x>0 && x%v[i]==0){
            x/=v[i];
            mp[v[i]]++;
        }
    }
    if(x>1) mp[x]++;
    return mp;
}

void solve(){
    if(k==1){
        printf("inf\n");
        return;
    }
    map <ll,ll> mp=getPrime(k);
    ll ans=1e19;
    for(map <ll,ll>::iterator it=mp.begin();it!=mp.end();it++){
        ll tmp=n,sum=0;
        while(tmp>0){
            sum+=tmp/(it->first);
            tmp/=(it->first);
        }
        if(sum/(it->second)<ans) ans=sum/(it->second);
    }
    cout<<ans<<endl;
}

int main(){
    get_prime();
    int t;
    scanf("%d",&t);
    for(int i=0;i<t;i++){
        cin>>n>>k;
        printf("Case %d: ",i+1);
        solve();
    }
    return 0;
}



联系
我们