How Many Equations Can You Find

点击打开链接

Now give you an string which only contains 0, 1 ,2 ,3 ,4 ,5 ,6 ,7 ,8 ,9.You are asked to add the sign ‘+’ or ’-’ between the characters. Just like give you a string “12345”, you can work out a string “123+4-5”. Now give you an integer N, please tell me how many ways can you find to make the result of the string equal to N .You can only choose at most one sign between two adjacent characters.

Input

Each case contains a string s and a number N . You may be sure the length of the string will not exceed 12 and the absolute value of N will not exceed 999999999999.

Output

The output contains one line for each data set : the number of ways you can find to make the equation.

Sample Input

123456789 321 1

Sample Output

181

算法：

#include <cstdio>  #include <stack>  #include <queue>  #include <cmath>  #include <vector>  #include <cstring>  #include <algorithm>  using namespace std;  #define CLR(a,b) memset(a,b,sizeof(a))  #define INF 0x3f3f3f3f  #define LL long long  char s[20];  LL n;  int l;  int ans;  void dfs(LL sum,LL num,int pos,int op)      //当前和，已积累数字，位置，积累数字正负   {      if (pos == l)       //递归终止条件       {          if (n == sum + num * op)              ans++;          return;      }      dfs(sum,num*10+s[pos]-'0',pos+1,op);        //无符号       dfs(sum+num*op,s[pos]-'0',pos+1,1);     //加号      dfs(sum+num*op,s[pos]-'0',pos+1,-1);        //减号   }  int main()  {      while (~scanf ("%s %lld",s,&n))      {          ans = 0;          l = strlen(s);          dfs(0,s[0]-'0',1,1);          printf ("%d\n",ans);      }      return 0;  }