1 #include<cstdio>
 2 #include<cstring>
 3 #include<string>
 4 using namespace std;
 5 char str[17];
 6 int val[17][17];
 7 int len;
 8 int ans;
 9 
10 void cal()
11 {
12     memset(val,0,sizeof(val));
13     for(int i = 0 ; i < len ; i++)
14         for(int j = i ; j < len ; j++)
15             for(int k = i ; k <= j ; k++)
16                 val[i][j] = val[i][j]*10+(str[k]-‘0‘);
17 }
18 void dfsr(int lsum,int pos, int rsum)
19 {
20     if(pos>=len)
21     {
22         if(rsum==lsum)
23             ans++;
24             return ;
25     }
26     for(int k = pos+1 ; k <= len ; k++)
27         dfsr(lsum,k,rsum+val[pos][k-1]);
28 }
29 void dfsl(int equ , int pos , int lsum)
30 {
31     if(pos>=equ)
32         dfsr(lsum,equ,0);
33         for(int k = pos+1 ; k <= equ ; k++)
34             dfsl(equ,k,lsum+val[pos][k-1]);
35 }
36 
37 int main()
38 {
39     while(~scanf("%s",str)&&str[0]!=‘E‘)
40     {
41         ans = 0 ;
42         len = strlen(str);
43         cal();
44         int equ;
45         for(equ = 1 ; equ < len ; equ++)
46             dfsl(equ,0,0);
47         printf("%d\n",ans);
48     }
49     return 0;
50 }

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 using namespace std;
 5 #define N 17
 6 char a[N];
 7 int equ;
 8 int len;
 9 bool ck(int sum)
10 {
11     int l= 0 , r=0;
12     int cur = 0;
13     for(int i = 0  ; i <= equ ;i++)
14     {
15         if(sum&(1<<i)) {
16             cur = cur * 10 + a[i]-‘0‘;
17             l+=cur;
18             cur = 0;
19         }
20         else cur = cur*10+a[i]-‘0‘;
21     }
22     if(cur != 0) l+=cur; 
23     cur = 0 ;
24     for(int i = equ+1 ; i< len ; i++)
25     {
26         if(sum&(1<<i)){
27             cur = cur * 10 + a[i]-‘0‘;
28             r+=cur;
29             cur = 0;
30         }
31         else cur = cur*10+a[i]-‘0‘;
32     }
33     if(cur!=0) r += cur;
34     if(l==r) return true;
35     else return false;
36 }
37 int main()
38 {
39     while(~scanf("%s",a)&&a[0]!=‘E‘)
40     {
41         int ans = 0 ;
42         len = strlen(a);
43         for( equ = 0 ; equ < len-1 ; equ++)
44         {
45             int tm = 1<<(len-1);
46             for(int j = 0 ; j < tm ;j++)
47             {
48                 if(ck(j)){
49                     ans++;
50             //        printf("%d %d\n", equ, j);
51                 }
52             }
53         }
54         printf("%d\n",ans>>1);
55     }
56     return 0;
57 }