HDU/HDOJ 3988 2011 WHU多校联合 Harry Potter and the Hide Story

Harry Potter and the Hide Story

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1Accepted Submission(s): 1

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).

SampleInput

2
2 2
10 10

SampleOutput

Case 1: 1
Case 2: 2

Author

iSea@WHU

解题方法当然是把k质因数分解之后，分别判断每一个素因子在阶乘里面出现了多少次，然后取所有素因子出现次数的最小值即可

由于数据量比较大，我用的米勒罗宾素数测试进行的素因子分解。

我的代码：

#include<iostream> #include<string.h> #include<stdlib.h> #include<stdio.h> using namespace std; typedef unsigned __int64 ll; #define INF (__int64)9223372036854775807 const int MAX=100; ll myrandom() { ll a; a=rand(); a=a*rand(); a=a*rand(); a=a*rand(); return a; } ll mulmod(ll a,ll b,ll c) { ll ret=0; while(b) { if(b&1) { ret=ret+a; if(ret>c) ret=ret-c; } a<<=1; if(a>c) a-=c; b>>=1; } return ret; } ll powmod(ll a,ll b,ll c) { ll ret=1; while(b) { if(b&1) ret=mulmod(ret,a,c); a=mulmod(a,a,c); b>>=1; } return ret; } bool miller(ll base,ll n) { ll m=n-1,k=0,i; while(m%2==0) { m>>=1; k++; } if(powmod(base,m,n)==1) return true; for(i=0;i<k;i++) { if(powmod(base,m<<i,n)==n-1) return true; } return false; } bool miller_rabin(ll n) { int i; for(i=2;i<100;i++) if(n%i==0) if(n==i) return true; else return false; for(i=0;i<MAX;i++) { ll base=myrandom()%(n-1)+1; if(!miller(base,n)) return false; } return true; } ll gcd(ll a,ll b) { if(b==0) return a; else return gcd(b,a%b); } ll f(ll a,ll b) { return (mulmod(a,a,b)+1)%b; } ll pollard_rho(ll n) { int i; if(n<=2) return 0; for(i=2;i<100;i++) if(n%i==0) return i; ll p,x,xx; for(i=1;i<MAX;i++) { x=myrandom()%n; xx=f(x,n); p=gcd((xx+n-x)%n,n)%n; while(p==1) { x=f(x,n); xx=f(f(xx,n),n); p=gcd((xx+n-x)%n,n)%n; } if(p) return p; } return 0; } ll prime(ll a) { if(miller_rabin(a)) return 0; ll t=pollard_rho(a); ll p=prime(t); if(p) return p; return t; } ll max(ll a,ll b) { if(a>b) return a; else return b; } ll getans(ll m,ll n) { ll ans=0,t=n; while(t) { t=t/m; ans=ans+t; } return ans; } int main() { int t,T; ll n,k,maxfac,tmp,ans,bingshen,acm; scanf("%d",&T); for(t=1;t<=T;t++) { scanf("%I64d%I64d",&n,&k); if(k==1) { printf("Case %d: inf\n",t); continue; } maxfac=0,ans=INF; while(k>1) { bingshen=0; if(miller_rabin(k)) { acm=getans(k,n); if(acm<ans) ans=acm; break; } tmp=prime(k); k=k/tmp; bingshen++; while(k%tmp==0) { k=k/tmp; bingshen++; } acm=getans(tmp,n); acm=acm/bingshen; if(acm<ans) ans=acm; } if(ans>=INF) printf("Case %d: inf\n",t); else printf("Case %d: %I64d\n",t,ans); } return 0; }