B – How Many Equations Can You Find dfs

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.

InputEach 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.OutputThe 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题目大意就是 在一个在来两个数字之间添加+ — 或者不做处理 使他可以得到后一个数，并记录次数就好了
思路：DFS遍历每一种情况

#include<iostream>
#include<string >
using namespace std;
string a;
typedef long long ll;
ll n,ans;
void dfs(int row,int sum){
    if(row==a.size())
    {
        if(sum==n)
            ans++;
        return ;
    }    ll t=;
    for(int i=row;i<a.size();i++)
    {
        t=t*+a[i]-'';
        dfs(i+,sum+t);
        if(row==) continue ;  //因为当row==0时 sum=0，如过加减号的话相当于在第一个数前边加负号，所以要跳过
　　　　　dfs(i+,sum-t);
    }
}int main()
{
    while(cin>>a>>n){
        ans=;
        dfs(,);
        cout<<ans<<endl;
    }
    return ;
}