hdoj How Many Equations Can You Find【DFS】

最新推荐文章于 2019-03-16 09:05:43 发布

原创最新推荐文章于 2019-03-16 09:05:43 发布 · 361 阅读

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本文介绍了一种使用递归方法来解决给定数字字符串转换为算术表达式的问题，目标是找到所有可能的加减组合使得最终计算结果等于预设的目标值。通过具体的示例和代码实现展示了如何有效地解决这个问题。

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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): 709 Accepted Submission(s): 462

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

用递归枚举所有情况，若满足条件，数目+1，最后输出即可

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
__int64 n;
int num,m,len;
char s[15];
void dfs(int x,__int64 ans)
{
	if(x==len)
	{
		if(ans == n)
			num++;
		return ;
	}
	__int64 a = 0;
	for(int i = x; i < len; i++)
	{
		a = a*10+s[i]-'0';
		dfs(i+1, a+ans);
		if(x!=0)
		dfs(i+1, ans-a);
	}
} 
int main()
{
	while(~scanf("%s",s))
	{
		scanf("%I64d",&n);
		len = strlen(s);
		num = 0;
		dfs(0,0);
		printf("%d\n",num);
	}
	return 0;
}