HDU2266How Many Equations Can You Find(深搜)

最新推荐文章于 2019-11-11 18:18:12 发布

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本文介绍了一种通过在给定的数字串中插入加号或减号来形成等式的算法。该算法使用深度优先搜索遍历所有可能的组合，并统计等式结果等于目标整数N的方案数量。输入为一个由数字组成的字符串及一个目标值N，输出为符合条件的不同等式的数量。

How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 641 Accepted Submission(s): 421

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

#include <stdio.h>  
#include <string.h>   
#define ll long long  
int len;  
char s[15];  
ll n,ans;  
  
void dfs(int sum,int id)  
{  
    if(id==len)  
    {  
        if(sum==n)  
            ans++;  
        return;  
    }  
    ll tmp=0;  
    for(int i=id;i<len;i++)  
    {  
        tmp=tmp*10+(s[i]-'0');  
        dfs(sum+tmp,i+1);  
        if(id)  
            dfs(sum-tmp,i+1);  
    }  
}  
int main()  
{  
   while(~scanf("%s%lld",s,&n))  
   {  
       ans=0;  
       len=strlen(s);  
       dfs(0,0);  
       printf("%lld\n",ans);  
   }  
}