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UVa 10780-Again Prime? No Time.

2013-06-03 22:40 253 查看
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: ycsgldy@163.com
> 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;
}
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