UVa 406-Prime Cuts

筛素数的问题，1~n的素数cnt为偶数个时，输出c*2个素数，当cnt为奇数时输出c*2-1个素数，都是从1~n的素数表的中间开始的。

当要求输出的素数个数大于cnt时，将1~n之间的素数全部输出。

/*************************************************************************
    > File Name: 406.cpp
    > Author: AcToy
    > Mail: ycsgldy@163.com
    > Created Time: 2013年05月31日 星期五 22时31分11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;
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 ( ) {
    primes.push_back(1);
    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 );
}
int n, c;

int main ( ) {
    prime_sieve();
    while ( scanf ( "%d%d", &n, &c ) == 2 ) {
        int A[n];
        int cnt = 0;
        for ( int i = 0; i < primes.size(); ++i )
            if ( primes[i] <= n ) A[i] = primes[i], cnt++;
            else break;

        printf ( "%d %d:", n, c );
        if ( cnt % 2 == 0 ) c *= 2;
        else c = c * 2 - 1;

        if ( c >= cnt ) {
            for ( int i = 0; i < cnt; ++i )
                printf ( " %d", A[i] );
            printf ( "\n\n" );
            continue;
        }
        else {
            int tmp = cnt / 2;
            int left = tmp, right = tmp;
            c--;
            while ( c > 0 ) {
                left--;
                c--;
                if ( c > 0 ) right++, c--;
            }

            for ( int i = left; i <= right; ++i )
                printf ( " %d", primes[i] );
            printf ( "\n\n" );
        }

    }
    return 0;
}