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 C: Divisibility

Consider an arbitrary sequence of integers. One can place + or - operators between integers in the sequence, thus deriving different arithmetical expressions that evaluate to different values. Let us, for example, take the sequence: 17, 5, -21, 15. There are eight possible expressions:

17	+	5	+	-21	+	15	=	16
17	+	5	+	-21	-	15	=	-14
17	+	5	-	-21	+	15	=	58
17	+	5	-	-21	-	15	=	28
17	-	5	+	-21	+	15	=	6
17	-	5	+	-21	-	15	=	-24
17	-	5	-	-21	+	15	=	48
17	-	5	-	-21	-	15	=	18

We call the sequence of integers divisible by K if + or - operators can be placed between integers in the sequence in such way that resulting value is divisible by K. In the above example, the sequence is divisible by 7 (17+5+-21-15=-14) but is not divisible by 5.

You are to write a program that will determine divisibility of sequence of integers.

Input

The first line of the input file contains a integer M indicating the number of cases to be analyzed. Then M couples of lines follow.
For each one of this couples, the first line contains two integers, N and K (1 <= N <= 10000, 2 <= K <= 100) separated by a space. The second line contains a sequence of N integers separated by spaces. Each integer is not greater than 10000 by it's absolute value.

Output

For each case in the input file, write to the output file the word "Divisible" if given sequence of integers is divisible by K or "Not divisible" if it's not.

Sample input

2
4 7
17 5 -21 15
4 5
17 5 -21 15

Sample Output

Divisible
Not divisible

---

第一次用 set 就跑很慢, 結果最後還是用老方法

//ANSI C 0.032 s
#include <stdio.h>
#include <string.h>

int main() {
    int t, n, m, x, y, i, j;
    scanf("%d", &t);
    char ch[2][100];
    while(t--) {
        scanf("%d %d", &n, &m);
        memset(ch, 0, sizeof(ch));
        scanf("%d", &x);
        x = x%m;
        if(x < 0)   x += m;
        ch[0][x] = 1;
        char flag;
        for(i = 1, flag = 0; i < n; i++, flag ^= 1) {
            scanf("%d", &x);
            x = x%m;
            if(x < 0)   x += m;
            for(j = 0; j < m; j++) {
                if(ch[flag][j]) {
                    y = j+x;
                    if(y >= m)  y -= m;
                    ch[flag^1][y] = 1;
                    y = j-x;
                    if(y < 0)  y += m;
                    ch[flag^1][y] = 1;
                }
            }
            memset(ch[flag], 0, sizeof(ch[flag]));
        }
        int ans = ch[flag][0];
        puts(ans ? "Divisible" : "Not divisible");
    }
    return 0;
}