HDU：2266 How Many Equations Can You Find（深搜+DFS）

How Many Equations Can You Find

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 913    Accepted Submission(s): 608

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

 

Author

dandelion

 

Source

HDU 8th Programming Contest Online

 

Recommend

lcy

题目大意：给你一个字符串里面都是数字，在这些数字中两两之间加上任意个加减号，可加可不加，问能否得到答案ans，输出方法数。

解题思路：深搜，看代码吧，很简单。

代码如下：

#include <cstdio>#include <cstring>#define LL long longLL cnt;LL size;char s[20];LL ans;void dfs(LL pos,LL sum){if(pos==size)//扫完式子，则结束 {if(sum==ans)//扫完的时候判断是否等于ans {cnt++;}return ;}LL tmp=0;//中间变量，存pos点后的数的可能情况，因为+、-号可加可不加，下面的数可能是个位、十位、百位。。。//用long long是因为题目中说数字最多12位，tmp最多可能存11位了，爆 int了。。 for(LL i=pos;i<size;i++){tmp=tmp*10+s[i]-'0';dfs(i+1,sum+tmp);//注意是i if(pos!=0)//第一位前面不能加减号 {dfs(i+1,sum-tmp);}}}int main(){while(scanf("%s %lld",s,&ans)!=EOF){size=strlen(s);cnt=0;dfs(0,0);printf("%lld\n",cnt);}return 0;}