hdu 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): 691    Accepted Submission(s): 450

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

知识点：DFS

题意：给你一个数字字符串，在字符之间加最多一个符号是之等于所给的数。

思路：状态：（当前位置，已得到的数值）

难点：像这样情况如何给出 4+34567-12334；这里要用for循环具体见代码

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 using namespace std;
 5 long long n,cnt,num;
 6 int vis[20];
 7 char s[20];
 8 int len;
 9 void dfs(int p,int num)
10 {
11     if(p==len)
12     {if(num==n)
13     {
14     cnt++;
15     return;
16     }}
17     long long temp=0;
18     for(int i=p;i<len;i++)//难点 19     {
20     temp=temp*10+(s[i]-'0');
21     dfs(i+1,num+temp);
22     if(p!=0)
23     dfs(i+1,num-temp);
24     }
25 }
26 int main()
27 {
28     while(~scanf("%s%lld",s,&n))
29     {
30          len=strlen(s);
31          cnt=0;
32          //temp=0;
33         dfs(0,0);
34         printf("%d\n",cnt);
35     }
36     return 0;
37 }