首页 > 代码库 > 杭电2266 How Many Equations Can You Find【DFS】
杭电2266 How Many Equations Can You Find【DFS】
How Many Equations Can You Find
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 892 Accepted Submission(s): 590
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
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lcy
AC代码：
1 #include<cstring> 2 #include<cstdio> 3 using namespace std; 4 #define LL long long 5 LL num,n,len; 6 char str[15]; 7 void DFS(LL x, LL sum) 8 {//x是取的字符串的长度, sum是当前的值 9 if(x == len)//当取得字符串长度==字符串总串长度时 10 {11 if(sum==n)//当前结果== n时 12 num++;13 return ; 14 }15 LL k=0;//核心的东东 短短的几句 却是指数级的搜索量啊16 for(int i=x;i<len;i++)//这里刚开始写成了从0开始，哎合适能搜到头啊17 {18 k=k*10+str[i]-‘0‘;19 20 //加减两种可能 21 DFS(i+1, sum+k);22 if(x != 0)//第一个数字前不能加“-”号23 DFS(i+1, sum-k);24 }25 }26 int main()27 {28 while(scanf("%s %lld",str,&n)!=EOF)29 {30 len=strlen(str);31 num = 0;32 DFS(0,0);33 printf("%lld\n", num);34 }35 return 0;36 }
杭电2266 How Many Equations Can You Find【DFS】
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