How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 714    Accepted Submission(s): 467

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

题解，在一串数字内加+或-，使其等于N的方法数，深搜；

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 __int64 ans,N;
 4 int len;
 5 char s[15];
 6 void dfs(int t,__int64 sum){
 7     if(t==len){
 8         //printf("%I64d\n",sum);
 9         if(sum==N)ans++;
10         return;
11     }
12     __int64 k=0;
13     for(int i=t;i<len;i++){
14         k=k*10+s[i]-‘0‘;
15         dfs(i+1,sum+k);
16         if(t!=0)dfs(i+1,sum-k);
17     }
18 }
19 int main(){
20     while(~scanf("%s%I64d",s,&N)){
21         len=strlen(s);
22         ans=0;
23         dfs(0,0);
24         printf("%I64d\n",ans);
25     }
26     return 0;
27 }
28 /*#include<stdio.h>
29 #include<string.h>
30 __int64 ans,N;
31 int len;
32 char s[15];
33 void dfs(int top,__int64 sum){
34     if(top==len){//写成N了错的心酸。。。
35         //printf("%I64d %d\n",sum,len);
36         if(sum==N)ans++;
37         return ;
38     }
39     __int64 x=0;
40     for(int i=top;i<len;i++){
41         x=x*10+s[i]-‘0‘;
42         dfs(i+1,sum+x);
43         if(top!=0)dfs(i+1,sum-x);
44     }
45 }
46 int main(){
47     while(~scanf("%s%I64d",s,&N)){
48         len=strlen(s);
49         //printf("%d\n",len);
50         ans=0;
51         dfs(0,0);
52         printf("%I64d\n",ans);
53     }
54     return 0;
55 }*/