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

最新推荐文章于 2018-12-13 23:17:17 发布

转载最新推荐文章于 2018-12-13 23:17:17 发布 · 119 阅读

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原文链接：http://www.cnblogs.com/toyking/p/3893157.html

本文探讨了如何通过分解质因数和利用模运算解决Harry Potter与隐藏故事中的数学难题，提供了详细的解题步骤和代码实现。

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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

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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;
}

转载于:https://www.cnblogs.com/toyking/p/3893157.html