[2018 Dahua Software Innovation Challenge] The fourth question of the simulation competition

topic	enter	output	limit	Example
There is a special positive integer (int type) that divides the number into two sides, each side has at least one digit, and each number on both sides adds up to be equal. Please write a judgment method to detect whether it is such a number. For example: 1236, can be split into 123 and 6.	The first line of input specifies the number of use cases T; the second line of use cases inputs a positive integer;	whether the output is a special positive integer		bool is_magicnum(int number) intput: 2 1232145 4543622 output: 1 1

 

[implementation code]:

 1 #include <stdio.h>
 2 
 3 int result[100000];
 4 int count;
 5 
 6 void combination(int* arr,int i,int cnt)
 7 {
 8         if(i >= cnt)
 9         {
10                 int sum = 0;
11                 for(int j=0; j<cnt; j++)
12                 {
13                         sum += arr[j];
14                 }
15                 for(int j=0; j<count; j++)
16                 {
17                         if(sum == result[j]) return;
18                 }
19                 result[count++] = sum;
20                 return;
21         }
22 
23         int temp = arr[i];
24         arr[i] = 0;
25         combination(arr,i+1,cnt);
26         arr[i] = temp;
27         combination(arr,i+1,cnt);
28 }
29 
30 int main()
31 {
32         int t = 0;
33         scanf("%d",&t);
34         while(t--)
35         {
36                 int n = 0;
37                 scanf("%d",&n);
38 
39                 int val[n];
40                 for(int i=0; i<n; i++)
41                 {
42                         scanf("%d",&val[i]);
43                 }
44 
45                 int num[n];
46                 int cnt = 0;
47                 for(int i=0; i<n; i++)
48                 {
49                         scanf("%d",&num[i]);
50                         cnt += num[i];
51                 }
52 
53                 int arr[n*cnt];
54                 cnt = 0;
55                 for(int i=0; i<n; i++)
56                 {
57                         for(int j=0; j<num[i]; j++)
58                         {
59                                 arr[cnt++] = val[i];
60                         }
61                 }
62                 printf("\n");
63                 combination(arr,0,cnt);
64                 printf("%d \n",count);
65         }
66 }

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[Code effect]: