Untitled
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 947 Accepted Submission(s): 538
Problem Description
There is an integer a and n integers b1,…,bn. After selecting some numbers from b1,…,bn in any order, say c1,…,cr, we want to make sure that a mod c1 mod c2 mod… mod cr=0 (i.e., a will become the remainder divided by ci each time, and at the end, we want a to become 0). Please determine the minimum value of r. If the goal cannot be achieved, print −1 instead.
Input
The first line contains one integer T≤5, which represents the number of testcases.
For each testcase, there are two lines:
1. The first line contains two integers n and a (1≤n≤20,1≤a≤106).
2. The second line contains n integers b1,…,bn (∀1≤i≤n,1≤bi≤106).
Output
Print T answers in T lines.
Sample Input
2
2 9
2 7
2 9
6 7
Sample Output
2
-1
Source
本次题目就是个纯暴力搜索的一道题目，唯一一点要注意的就是序列应该先降序排列，为什么呢？(因为如果升序排列的话i<j，Ci小于Cj,Ci先进行取余运算，那么Cj就没有意义了)
除此之外就是简单的深搜，附上本人的代码
1 #include<stdio.h>
2 #include<stdlib.h>
3 #define maxn 30
4 int min, sum,n;
5 int a[maxn], b[maxn],c[maxn];
6 int Min(int a, int b)
7 {
8 return a> b ? b:a;
9 }
10 int dfs(int m)
11 {
12 if (m == n + 1)
13 {
14 int temp = sum;
15 int length = 0;
16 for (int j = 1; j <=n; j++)
17 {
18 if (temp == 0)
19 break;
20 if (c[j] == 0)
21 {
22 temp %= a[j];
23 length++;
24 }
25 }
26 if (temp==0)
27 min = Min(min, length);
28 return 0;
29 } //这里利用选择为0，未选为1，方便代码调试时按照二进制运算来看，比较直观
30 c[m] = 0;//用来标记哪些被选了
31 dfs(m + 1);
32 c[m] = 1;//标记未被选
33 dfs(m + 1);
34 }
35 int cmp(const void*a, const void*b)
36 {
37 return *(int *)b - *(int*)a;
38 }
39 int main()
40 {
41 int num;
42 scanf("%d", &num);
43 while (num--)
44 {
45 min= 1000000;
46 scanf("%d%d", &n, &sum);
47 for (int i = 1; i <= n; i++)
48 scanf("%d", &a[i]);
49 qsort(&a[1], n, sizeof(a[1]), cmp);
50 dfs(1);
51 if (min!=1000000)
52 printf("%d\n", min);
53 else printf("-1\n");
54 }
55 }