Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

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;
}