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

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lcy

AC代码：

 1 #include<cstring>
 2 #include<cstdio>
 3 using namespace std;
 4 #define LL long long
 5 LL num,n,len;
 6 char str[15];
 7 void DFS(LL x, LL sum)
 8 {//x是取的字符串的长度, sum是当前的值
 9     if(x == len)//当取得字符串长度==字符串总串长度时
10     {
11         if(sum==n)//当前结果== n时
12             num++;
13         return ;
14     }
15     LL k=0;//核心的东东  短短的几句  却是指数级的搜索量啊
16     for(int i=x;i<len;i++)//这里刚开始写成了从0开始，哎合适能搜到头啊
17     {
18         k=k*10+str[i]-‘0‘;
19
20         //加减两种可能
21         DFS(i+1, sum+k);
22         if(x != 0)//第一个数字前不能加“-”号
23             DFS(i+1, sum-k);
24     }
25 }
26 int main()
27 {
28     while(scanf("%s %lld",str,&n)!=EOF)
29     {
30         len=strlen(str);
31         num = 0;
32         DFS(0,0);
33         printf("%lld\n", num);
34     }
35     return 0;
36 }