hdu 2266(How Many Equations Can You Find)搜索、递归
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
想说的是,搜索和递归的精髓真心还没有掌握,还没到达运用自如的地步!
常常惊叹于递归的强大,难以驾驭啊!
本题给出我们一个数字串,可以在适当的位置添加“+”或者“-”号,使得运算的结果等于给出的n;
这题膜拜一下小媛在努力。。。。。。
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 __int64 num,n,len; 6 char str[15]; 7 void DFS(__int64 x,__int64 sum) 8 { 9 if(x==len) 10 { 11 if(sum==n) 12 13 num++; 14 return ; 15 16 } 17 __int64 k=0;//核心的东东 短短的几句 却是指数级的搜索量啊 18 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 19 { 20 k=k*10+str[i]-'0'; 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 27 28 int main() 29 { 30 while(scanf("%s %I64d",str,&n)!=EOF) 31 { 32 len=strlen(str); 33 num=0; 34 DFS(0,0); 35 printf("%I64d\n",num); 36 } 37 return 0; 38 } 39