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
题目大意
将数字字符串分成一部分，选择正负号，然后相加，不过要注意，第一个子串必须是正的，（题目要求）。
思路
由此可以知道，每一步下接两步，加或则减，想到了二叉树和深度优先搜索。
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#include <bits/stdc++.h>
using namespace std;
int n;
string s;
//int vis[13][13]; 是一直往后推，便不需要标记了，自然不需要回溯了
int sum;
void dfs(int x,int res)
{
int j,k,ans;
if(x==s.length()){
if(res==n)sum++;
return ;
}
for(j=x;j<s.length();j++){ // 从当前位置出发，所构成的子字符串
ans=0;
for(k=x;k<=j;k++){
ans=ans*10+s[k]-'0';
}
dfs(j+1,res+ans);
if(x!=0)dfs(j+1,res-ans); // 第一个子串，不能为负
}
}
int main()
{
while(cin>>s>>n){
sum=0;
dfs(0,0);
cout << sum << endl;
}
return 0;
} |