How Many Equations Can You Find HDU

题目链接:点这里

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

题意:

    给你一个字符串s和一个数字n,你能有这个字符串能有多少方式组成这个数字

思路:

简单题,直接暴力深搜就可以出答案了

代码:

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;char s[20];int a[20][20];int n,m;int ans;void init(){    memset(a,0,sizeof(a));    m=strlen(s);    for(int i=0;i<m;++i){        a[i][i]=s[i]-'0';        for(int j=i+1;j<m;++j){            a[i][j]=a[i][j-1]*10+s[j]-'0';        }    }}void dfs(int i,int num){    if(i==m){        if(num==n)            ++ans;        return ;    }    for(int j=i;j<m;++j){        dfs(j+1,num+a[i][j]);        dfs(j+1,num-a[i][j]);    }}int main(){    while(scanf("%s %d",s,&n)!=EOF){        init();        ans=0;        m=strlen(s);        for(int i=0;i<m;++i){            dfs(i+1,a[0][i]);        }        cout<<ans<<endl;    }    return 0;}