UVa 10487 Closest Sums

UVa 10487 Closest Sums

题目大意：给出n个数字，给出一个数字x，求出n个数字中哪两个数字的和与x最接近。
我的做法是这样的，把n个数字两两求和的结果存在数组里面，排序，去重（这一步无所谓）。每当给出一个x的时候，二分查找大于等于x的第一个数字，和小于x的最后一个数字。比较即可得出结果。
以下是我的代码：

#include < vector >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
using   namespace  std;

int  main()
{
    #ifndef ONLINE_JUDGE
    freopen( " data.in " , " r " ,stdin);
    freopen( " data.out " , " w " ,stdout);
     #endif

     int  case_num( 0 ),n;
     while (scanf( " %d " , & n) == 1   &&  n)
    {
        vector < int >  a,r;
         for ( int  i = 0 ;i < n;i ++ )
        {
             int  t;
            scanf( " %d " , & t);
            a.push_back(t);
             for ( int  j = 0 ;j < i;j ++ )
                r.push_back(a[j] + t);
        }

        sort(r.begin(),r.end());
        r.erase(unique(r.begin(),r.end()),r.end());

         /*
        for(int i=0;i<r.size();i++)
        {
            if(i!=0)
                printf(" ");
            printf("%d",r[i]);
        }
        printf("\n");
        // */
        case_num ++ ;
        printf( " Case %d:\n " ,case_num);

         int  m;
        scanf( " %d " , & m);
         for ( int  i = 1 ;i <= m;i ++ )
        {
             int  x;
            scanf( " %d " , & x);

             int  dist( 0x7f7f7f7f ),ans;
            vector < int > ::iterator lpos(lower_bound(r.begin(),r.end(),x));
             if (lpos != r.end())
            {
                 int  t =* lpos;
                 if (dist > abs(t - x))
                {
                    dist = abs(t - x);
                    ans = t;
                }
            }
             if (lpos != r.begin())
            {
                 int  t =* (lpos - 1 );
                 if (dist > abs(t - x))
                {
                    dist = abs(t - x);
                    ans = t;
                }
            }

            printf( " Closest sum to %d is %d.\n " ,x,ans);
        }
    }
}