学步园

Input:  Each problem is described on a single line. The line begins with an integer
N, such that 2 ≤ N ≤ 14, designating the number of numerals included in the problem. Following that are those
N numerals. There will always be at least 2 numerals that are nonzero. The end of the input is designated by a line containing only the value 0.

Output:  For each case, output a line with the minimum sum
S that can be achieved. Please keep in mind that by standard convention, the numeral
0 cannot appear as the first digit of either summand.

5 1 2 7 8 9
6 3 4 2 2 2 2
9 0 1 2 3 4 0 1 2 3
0

207
447
11257

#include<stdio.h>
#include<algorithm>
using namespace std;
#define mulit(i) (1<<(i))
__int64 a[15],sta[mulit(14)];
void init(int n)
{
    __int64 sum;
    int h,i;
    sort(a,a+n);
    for(int s=1;s<mulit(n);s++)
    {
        h=-1;
        for( i=0;mulit(i)<=s;i++)
        if((mulit(i)&s)&&a[i]>0)
        {
            h=i; sum=a[i]; break;
        }
        if(mulit(i-1)==s)
        {
            sta[s]=a[i-1]; continue;
        }
        if(mulit(i)>s)
        {
            sta[s]=-1; continue;
        }
        for( i=0;mulit(i)<=s;i++)
        if((mulit(i)&s)&&h!=i)
        sum=sum*10+a[i];
        sta[s]=sum;
    }
}
int main()
{
    int n;
    __int64 minsum;
    while(scanf("%d",&n)>0&&n)
    {
        for(int i=0;i<n;i++)
        scanf("%I64d",&a[i]);
        init(n);
        minsum=-1;
        for(int s=1;s<mulit(n);s++)
        if(sta[s]!=-1&&sta[s^(mulit(n)-1)]!=-1)
        if(minsum==-1)
            minsum=sta[s]+sta[s^(mulit(n)-1)];
        else if(minsum>sta[s]+sta[s^(mulit(n)-1)])
            minsum=sta[s]+sta[s^(mulit(n)-1)];
        printf("%I64d\n",minsum);
    }
}