UVa 10780-Again Prime? No Time.
2013-06-03 22:40
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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 )。
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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