How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1477    Accepted Submission(s): 990


Problem Description


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 3
21 1

 


Sample Output

18
1 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
ll l,len;
ll cnt;
char str[15];

void DFS(int l,ll sum)// l记录当前所用的字符个数，sum传递的是前 [0,l-1] 个字符经过加减得到的值 
{
    if(l==len && sum==n)//当字符串全部用完  并且得到的值等于n 时才满足条件 
    {
        cnt++  ;  
        return ;//一个递归结束 
    }
    ll ans=0;// ans 保存的是以后的值 
    for(int i=l;i<len;i++)
    {
        ans=ans*10+str[i]-'0';
        DFS(i+1,sum+ans);
        if(l)
            DFS(i+1,sum-ans);//第一个数字前面不能是减号 
     } 
}

int main()
{
    while(~scanf("%s %lld",str,&n))
    {
          len=strlen(str);
          cnt=0;//记录有多少种方式   
          DFS(0,0);
          printf("%d\n",cnt);
    }
    return 0;
}

/*

对于任意一串数字，我们可以在其中：
    添加 +
    添加 -
    不添加

这一串数字的长度不超过11

复杂度应该是  O(3^11)





DFS(l,sum)

1234 10

DFS(0,0)DFS(1,1)//    l=0  sum=0;
            ->DFS(1,12)DFS(1,123)DFS(1,1234)//对于所有的数字全部用完的情况，都可以直接判断结果是否等于N,然后结束   

DFS(1,1)DFS(2,1+2)DFS(2,1-2)DFS(2,1+23)DFS(2,1-23)DFS(2,1+234)//对于所有的数字全部用完的情况，都可以直接判断结果是否等于N,然后结束
            ->DFS(2,1-234)//对于所有的数字全部用完的情况，都可以直接判断结果是否等于N,然后结束

DFS(1,12)DFS(2,12+3)DFS(2,12-3)DFS(2,12+34)//对于所有的数字全部用完的情况，都可以直接判断结果是否等于N,然后结束
            ->DFS(2,12-34)//对于所有的数字全部用完的情况，都可以直接判断结果是否等于N,然后结束

DFS(1,123)DFS(2,123+4)DFS(2,123-4)

DFS(2,1+2)DFS(3,1+2+3)DFS(3,1+2-3)DFS(3,1+2+34)DFS(3,1+2-34)

DFS(2,1-2)DFS(3,1-2+3)DFS(3,1-2-3)DFS(3,1-2+34)DFS(3,1-2-34)

DFS(2,1+23)..... 
DFS(2,1-23).....
DFS(2,1+234)->.....
DFS(2,1-234)->.....

DFS(3,...)->......
DFS(3,...)->......
DFS(3,...)->......
DFS(3,...)->......

DFS(4,...)->......
DFS(4,...)->......
DFS(4,...)->......
DFS(4,...)->......

*/