UVa 294-Divisors

找到L和U之间其中一个数，这个数有最多的因子数

/*************************************************************************    > File Name: 294.cpp    > Author: Toy    > Mail: [e-mail]     > Created Time: 2013年06月06日 星期四 21时02分58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;void prime_sieve ( ) {for ( int i = 3; i <= SQRP; i += 2 ) if ( !IsComp ( i ) ) for ( int j = i * i; j <= MAXP; SetComp ( j ), j += i + i ) ;primes.push_back ( 2 );for ( int i = 3; i <= MAXP; i += 2 ) if ( !IsComp ( i ) ) primes.push_back ( i );}int count_divisors ( int n ) {int ret = 1;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 ) ;ret *= e + 1;sn = sqrt ( n );}if ( n > 1 ) ret *= 2;return ret;}int Case, n, m;int main ( ) {prime_sieve();scanf ( "%d", &Case );while ( Case-- ) {int Max = 0, ii;scanf ( "%d%d", &n, &m );for ( int i = n; i <= m; ++i ) if ( Max < count_divisors ( i ) ) Max = count_divisors ( i ), ii = i;printf ( "Between %d and %d, %d has a maximum of %d divisors.\n", n, m, ii, Max );}    return 0;}