HDU 3988 Harry Potter and the Hide Story(数论-整数和素数)
2015-07-06 09:14
507 查看
Harry Potter and the Hide Story
Problem DescriptioniSea 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
field=problem&key=2011%20Multi-University%20Training%20Contest%2015%20-%20Host%20by%20WHU&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">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也依据k的质因素求出关系限制i,最后算出最大的i就可以。
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; }
相关文章推荐
- C#编写简单的聊天程序
- 用qemu搭建linux环境的最简单步骤(硬盘启动)
- Oracle中job的使用详解
- 最常用的两种C++序列化方案的使用心得(protobuf和boost serialization)
- android开发中屏幕适配问题
- javaee jsp中servlet发生java.lang.ClassNotFoundException可能原因
- Android WebView
- [Leetcode]Word Ladder
- 使用尾递归实现String To Int
- 如何自定义 UIPickerView 的行?
- thinkphp验证码在服务器显示不了
- Android 增强版百分比布局库 为了适配而扩展
- Android 增强版百分比布局库 为了适配而扩展
- Qt写入txt文件
- actionSheet报错的问题及解决方法
- 方法(六)
- Java数组List换算方法
- 一步一步搭建LVS-DR模型LB集群(三)
- UiAutomator学习笔记
- 小样本时的概率估算