UVa 10780-Again Prime? No Time.

求N！中M的最高次幂。

思路：先将M质数分解，然后对M的每个素数因子求N！的最高次幂。例如M = 45 N = 67. 45 = 3 ^ 2 * 5 ^ 1. 45的素数因子为3 和 5，             而67！中3的最高次幂为31，5的最高次幂为15.这样就可以确定67！中45的最高幂为15，因为67！中每2个3，1个5就包含                 一个45，换句话说也就是求67！中包含了多少（3^2*5^1）这样的组合，也就是求min （ 31 / 2， 15 / 1 ）。

/*************************************************************************    > File Name: 10780.cpp    > Author: Toy    > Mail: [e-mail]     > Created Time: 2013年06月03日 星期一 18时29分03秒 ************************************************************************/#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <cstdlib>#include <climits>#include <sstream>#include <fstream>#include <cstdio>#include <string>#include <vector>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>using namespace std;const int INF = 0x7fffffff;typedef pair<int,int> II;typedef vector<int> IV;typedef vector<II> IIV;typedef vector<bool> BV;typedef long long i64;typedef unsigned long long u64;typedef unsigned int u32;#define For(t,v,c) for(t::const_iterator v=c.begin(); v!=c.end(); ++v)#define IsComp(n) (_c[n>>6]&(1<<((n>>1)&31)))#define SetComp(n) _c[n>>6]|=(1<<((n>>1)&31))const int MAXP = 46341; //sqrt(2^31)const int SQRP = 216; //sqrt(MAX)int _c[(MAXP>>6)+1];IV primes;IIV opt;int Case, m, n;void prime_sieve ( ) {for ( int i = 3; i <= SQRP; i += 2 ) if ( !IsComp ( i ) ) for ( int j = i * i; j <= MAXP; j += i + i ) SetComp ( j );primes.push_back ( 2 );for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp ( i ) ) primes.push_back ( i );}void prime_factorize ( int n, IIV &f ) {f.clear();int sn = sqrt ( n );For ( IV, it, primes ) {int prime = *it;if ( prime > sn ) break;if ( n % prime ) continue;int e = 0; for ( ; n % prime == 0; e++, n /= prime ) ;f.push_back ( II ( prime, e ) );sn = sqrt ( n );}if ( n > 1 ) f.push_back ( II ( n, 1 ) );}int get_powers ( int n, int p ) {int res = 0;for ( int power = p; power <= n; power *= p ) res += n / power;return res;}int main ( ) {prime_sieve ();scanf ( "%d", &Case );for ( int cnt = 1; cnt <= Case; ++cnt ) {int ans = INF;scanf ( "%d%d", &m, &n );prime_factorize ( m, opt );For ( IIV, it, opt ) {int tmp = get_powers ( n, it -> first );tmp /= it -> second;if ( tmp < ans ) ans = tmp;}printf ( "Case %d:\n", cnt );if ( ans == 0 ) printf ( "Impossible to divide\n" );else printf ( "%d\n", ans );}    return 0;}