poj 3399 Product

#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
__int64 data[105],pos[105],neg[105],len1,len2,count0;
__int64 n,k,i,ans[2][105];
void printout(__int64 a,__int64 b,__int64 c,bool tag)
{
    for(i=len1-1;i>len1-1-a;--i)
        cout<<pos[i]<<" ";
    while(b--)
        cout<<"0 ";
    if(tag==0)
        for(i=0;i<c;++i)
            cout<<neg[i]<<" ";
    else
        for(i=len2-c;i<len2;++i)
            cout<<neg[i]<<" ";
    cout<<endl;
}
int main()
{
    cin>>n>>k;
    for(i=0;i<n;++i)
    {
        cin>>data[i];
        if(data[i]==0)
            count0++;
        else if(data[i]>0)
            pos[len1++]=data[i];
        else
            neg[len2++]=data[i];
    }
    sort(pos,pos+len1);
    sort(neg,neg+len2,greater<__int64>());
    if( len1+len2<k || ( len1+len2==k && len2%2==1 ) || ( len1+len2>k && len1==0 && k%2==1 ) )    //考虑最大乘积不大于0的情况
    {
        __int64 t=min(k-len1,count0);
        printout(len1,t,k-len1-t,0);
    }
    else 
    {
        __int64 a=0,c=0;
        if(k%2==1)
            a++,k--;
        while(k>0)
        {
            if( len2-c-2<0 || (len1-a-2>0 && pos[len1-a-1]*pos[len1-a-2] > neg[len2-c-1]*neg[len2-c-2] ) )
                a+=2;
            else 
                c+=2;
            k-=2;
        }
        printout(a,0,c,1);
    }
    return 0;
}
//注意到当len1+len2>k 时，只要正数的个数不为0，那么乘积一定可以为正