HDU-2266-How Many Equations Can You Find【交叉DFS】

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How Many Equations Can You Find

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

Total Submission(s): 919    Accepted Submission(s): 613

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 321 1

 

Sample Output

181

大意：给你一个字符串，你可以在任意字符中间加入 “+” 或 “-”，使得最后的运算结果为 n

思路：从起始位置到最后位置的搜索，加、减交叉搜索，注意第一个字符前不能添减号（实际上是不能添任何符号的，但是添一个加号还是其本身，所以添不添加号无所谓了）

#include<cstdio>#include<algorithm>#include<cstring>#define LL long longusing namespace std;LL n,ans;int len;char str[50];void dfs(int id,LL sum){if(id==len) // 这里判 id==len是因为下面 dfs的是 i+1，如果改成 id==n-1，则字符串的最后一位就没计算了 {if(sum==n){ans++;}return ;}LL shu=0;for(int i=id;i<len;i++){shu=shu*10+str[i]-'0';dfs(i+1,sum+shu);if(id)dfs(i+1,sum-shu);}}int main(){while(~scanf("%s%lld",str,&n)){len=strlen(str);ans=0;dfs(0,0);printf("%lld\n",ans);}return 0;}