UVa 10140- Prime Distance

求L，U区间之间的相邻素数的最小距离和最大距离。如果没有相邻的素数，输出

There are no adjacent primes.

/*************************************************************************
    > File Name: 10140.cpp
    > Author: Toy
    > Mail: ycsgldy@163.com 
    > Created Time: 2013年06月01日 星期六 09时52分32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; 
const int SQRP = 216; 
int _c[(MAXP>>6)+1];
IV primes;
IV prime;

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_seg_sieve ( i64 a, i64 b, IV &seg_primes ) {
	BV pmap ( b - a + 1, true );
	i64 sqr_b = sqrt ( b );
	For ( IV, it, primes ) {
		int p = *it;
		if ( p > sqr_b ) break;
		for ( i64 j = ( a + p - 1 ) / p, v = ( j == 1 ? p + p : j * p ); v <= b; v += p ) pmap[v - a] = false;
	}
	seg_primes.clear();
	if ( a == 1 ) pmap[0] = false;
	for ( int i = 0, I = b - a + 1; i < I; ++i ) if ( pmap[i] ) seg_primes.push_back ( a + i );
}

long long L, U;
int C1, C2, D1, D2, cnt;

int main ( ) {
    prime_sieve ( );
	while ( scanf ( "%lld%lld", &L, &U ) == 2 ) {
	    prime_seg_sieve ( L, U, prime );
		cnt = prime.size();
		if ( cnt < 2 ) printf ( "There are no adjacent primes.\n" );
		else {
			int min = INF, max = 0;
			for ( int i = 1; i < cnt; ++i ) {
				int tmp = prime[i] - prime[i - 1];
				if ( tmp < min ) min = tmp, C1 = prime[i - 1], C2 = prime[i];
				if ( tmp > max ) max = tmp, D1 = prime[i - 1], D2 = prime[i];
			}
			printf ( "%d,%d are closest, %d,%d are most distant.\n", C1, C2, D1, D2 );
		}
	}
    return 0;
}