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
#include<stdio.h>
#include<string.h>
const int maxn = 10000+10;
typedef long long ll;
char s[maxn];
ll ans, len, n;
void DFS(int x, int y){
if (y == len){
if(x == n)
ans ++;
return ;
}
ll t = 0;
for (int i = y; i < len; i++){
t = t * 10 + (s[i]-'0');
DFS(x+t, i+1);
if (y != 0)
DFS(x-t, i+1);
}
}
/*
x是当前位置的和,y是当前位置的字符串长度
如果y == len, 意味着这是最后一个数字
判断此时位置的和是否与n相等。
t是当前位置之后截取的字符的值
每一次都遍历一次。
有一个条件就是y == 0的时候不能减
*/
int main()
{
while(scanf ("%s %lld", s, &n) != EOF){
ans = 0;
len = strlen(s);
DFS(0, 0);
printf("%lld\n", ans);
}
return 0;
}