杭电2266 How Many Equations Can You Find【DFS】

How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 892    Accepted Submission(s): 590

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

 

Author

dandelion

 

Source

HDU 8th Programming Contest Online

 

Recommend

lcy

 

 

AC代码：

#include<cstring>
#include<cstdio>
using namespace std;
#define LL long long
LL num,n,len;
char str[15];
void DFS(LL x, LL sum)
{//x是取的字符串的长度, sum是当前的值 
    if(x == len)//当取得字符串长度==字符串总串长度时 
    {
        if(sum==n)//当前结果== n时 
            num++;
        return ;  
    }
    LL k=0;
    for(int i=x;i<len;i++)/
    {
        k=k*10+str[i]-'0';
        
        //加减两种可能 
        DFS(i+1, sum+k);
        if(x != 0)//第一个数字前不能加“-”号
            DFS(i+1, sum-k);
    }
}
int main()
{
    while(scanf("%s %lld",str,&n)!=EOF)
    {
        len=strlen(str);
        num = 0;
        DFS(0,0);
        printf("%lld\n", num);
    }
    return 0;
}