Problem D
Closest Sums
Input: standard input
Output: standard output
Time Limit: 3 seconds

Given is a set of integers andthen a sequence of queries. A query gives you a number and asks to find a sum oftwo distinct numbers from the set, which is closest to the query number.

Input

Input contains multiple cases.

Each case starts with an integer n(1<n<=1000), which indicates, how many numbers are in the set of integer.Next n lines contain n numbers. Of course there is only one number in a singleline. The next line contains a positive integer m giving the number ofqueries, 0 < m < 25. The next m lines contain aninteger of the query, one per line.

Input is terminated by a case whose n=0. Surely,this case needs no processing.

Output

Output should be organized as in the samplebelow. For each query output one line giving the query value and the closestsum in the format as in the sample. Inputs will be such that no ties will occur.

Sample input

5

3

12

17

33

34

3

1

51

30

3

1

2

3

3

1

2

3

3

1

2

3

3

4

5

6

0

Sample output

Case 1:

Closest sum to 1 is 15.

Closest sum to 51 is 51.

Closest sum to 30 is 29.

Case 2:

Closest sum to 1 is 3.

Closest sum to 2 is 3.

Closest sum to 3 is 3.

Case 3:

Closest sum to 4 is 4.

Closest sum to 5 is 5.

Closest sum to 6 is 5.

//C 0.060 s
#include <stdio.h>
#include <stdlib.h>
#define max(x, y) ((x) > (y) ? (x) : (y))

int cmp(const void *i, const void *j) {
    return *(int *)i - *(int *)j;
}
int main() {
    int n, m, Case = 0, A[1000], B[25], C[25];
    int i, j, k;
    while(scanf("%d", &n) == 1 && n) {
        for(i = 0; i < n; i++)
            scanf("%d", &A[i]);
        scanf("%d", &m);
        int tmp;
        for(i = 0; i < m; i++) {
            scanf("%d", &B[i]);
            C[i] = 0xffffff;
        }
        qsort(A, n, sizeof(int), cmp);
        for(k = 0; k < m; k++) {
            for(i = 0; i < n; i++) {
                for(j = i+1; j < n; j++) {
                    tmp = A[i]+A[j];
                    if(abs(tmp-B[k]) <= abs(C[k]))
                        C[k] = tmp-B[k];
                    if(tmp-B[k] >= abs(C[k]))
                        break;
                }
            }
        }
        printf("Case %d:\n", ++Case);
        for(i = 0; i < m; i++) {
            printf("Closest sum to %d is %d.\n", B[i], B[i]+C[i]);
        }
    }
    return 0;
}