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): 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：

 #include<stdio.h>
 #include<string.h>
 __int64 ans,N;
 int len;
 char s[];
 void dfs(int t,__int64 sum){
     if(t==len){
         //printf("%I64d\n",sum);
         if(sum==N)ans++;
         return;
     }
     __int64 k=;
     for(int i=t;i<len;i++){
         k=k*+s[i]-'';
         dfs(i+,sum+k);
         if(t!=)dfs(i+,sum-k);
     }
 }
 int main(){
     while(~scanf("%s%I64d",s,&N)){
         len=strlen(s);
         ans=;
         dfs(,);
         printf("%I64d\n",ans);
     }
     return ;
 }
 /*#include<stdio.h>
 #include<string.h>
 __int64 ans,N;
 int len;
 char s[15];
 void dfs(int top,__int64 sum){
     if(top==len){//写成N了错的心酸。。。
         //printf("%I64d %d\n",sum,len);
         if(sum==N)ans++;
         return ;
     }
     __int64 x=0;
     for(int i=top;i<len;i++){
         x=x*10+s[i]-'0';
         dfs(i+1,sum+x);
         if(top!=0)dfs(i+1,sum-x);
     }
 }
 int main(){
     while(~scanf("%s%I64d",s,&N)){
         len=strlen(s);
         //printf("%d\n",len);
         ans=0;
         dfs(0,0);
         printf("%I64d\n",ans);
     }
     return 0;
 }*/