hdu 2266（How Many Equations Can You Find）搜索、递归

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

 

 

想说的是，搜索和递归的精髓真心还没有掌握，还没到达运用自如的地步！

常常惊叹于递归的强大，难以驾驭啊！

本题给出我们一个数字串，可以在适当的位置添加“+”或者“-”号，使得运算的结果等于给出的n；

这题膜拜一下小媛在努力。。。。。。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
__int64 num,n,len;
char str[15];
void DFS(__int64 x,__int64 sum)
{
    if(x==len)
    {
        if(sum==n)
        
            num++;
            return ;
        
    }
    __int64 k=0;//核心的东东  短短的几句  却是指数级的搜索量啊
    for(int i=x;i<len;i++)//这里刚开始写成了从0开始，哎合适能搜到头啊
    {
        k=k*10+str[i]-'0';
        DFS(i+1,sum+k);
        if(x!=0)//第一个数字前不能加“-”号
            DFS(i+1,sum-k);
    }
}


int main()
{
    while(scanf("%s %I64d",str,&n)!=EOF)
    {
        len=strlen(str);
        num=0;
        DFS(0,0);
        printf("%I64d\n",num);
    }
    return 0;
}