How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 985    Accepted Submission(s):
659

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

大意：

给出一个字符串，往任意二个字符间添加‘+’‘-‘，使得其值等于给定的aim，求方案个数，字符串前后均无符号；

自己第一次使用dfs保存一个加工过的字符串，再用一个pd函数判定此字符串的和；

其实可以直接暴力dfs水果，更省时：                      //ps：55555555~~~感觉自己好傻

#include<bits/stdc++.h>
using namespace std;
char num[15],b[105];
int aim,ans,n;
void dfs(int cur,int sum)
{
if(cur==n){
if(sum==aim) ans++;
return;
}
int t=0;
for(int i=cur;i<n;i++){
t=t*10+num[i]-‘0‘;                         //其实就是暴力枚举出所有的+-情况
dfs(i+1,sum+t);
if(cur>0) dfs(i+1,sum-t);
}
}

int main()
{
while(cin>>num>>aim/*scanf("%s%d",num,&aim)!=EOF*/){
n=strlen(num);
ans=0;
dfs(0,0);
printf("%d\n",ans);
}
return 0;
}

例如 1234

首先