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

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
long long n,cnt,num;
int vis[20];
char s[20];
int len;
void dfs(int p,int num)
{
    if(p==len)
    {if(num==n)
    {
    cnt++;
    return;
    }}
    long long temp=0;
    for(int i=p;i<len;i++)//难点
    {
    temp=temp*10+(s[i]-'0');
    dfs(i+1,num+temp);
    if(p!=0)
    dfs(i+1,num-temp);
    }
}
int main()
{
    while(~scanf("%s%lld",s,&n))
    {
         len=strlen(s);
         cnt=0;
         //temp=0;
        dfs(0,0);
        printf("%d
",cnt);
    }
    return 0;
}