【HDU】2266 - How Many Equations Can You Find（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): 896 Accepted Submission(s): 594

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

Author

dandelion

Source

HDU 8th Programming Contest Online

刚看到题，以为有什么技巧，算了一下复杂度，3^11并不是很高，那就dfs硬怼吧。

代码如下：

#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
char s[20];
LL n;
int l;
int ans;
void dfs(LL sum,LL num,int pos,int op)		//当前和，已积累数字，位置，积累数字正负 
{
	if (pos == l)		//递归终止条件 
	{
		if (n == sum + num * op)
			ans++;
		return;
	}
	dfs(sum,num*10+s[pos]-'0',pos+1,op);		//无符号 
	dfs(sum+num*op,s[pos]-'0',pos+1,1);		//加号
	dfs(sum+num*op,s[pos]-'0',pos+1,-1);		//减号 
}
int main()
{
	while (~scanf ("%s %lld",s,&n))
	{
		ans = 0;
		l = strlen(s);
		dfs(0,s[0]-'0',1,1);
		printf ("%d\n",ans);
	}
	return 0;
}