A very hard Aoshu problem

A very hard Aoshu proble

Problem Description

Aoshu is very popular among primary school students. It is mathematics, but much harder than ordinary mathematics for primary school students. Teacher Liu is an Aoshu teacher. He just comes out with a problem to test his students:
Given a serial of digits, you must put a '=' and none or some '+' between these digits and make an equation. Please find out how many equations you can get. For example, if the digits serial is "1212", you can get 2 equations, they are "12=12" and "1+2=1+2". Please note that the digits only include 1 to 9, and every '+' must have a digit on its left side and right side. For example, "+12=12", and "1++1=2" are illegal. Please note that "1+11=12" and "11+1=12" are different equations.

 

Input

There are several test cases. Each test case is a digit serial in a line. The length of a serial is at least 2 and no more than 15. The input ends with a line of "END".

 

Output

For each test case , output a integer in a line, indicating the number of equations you can get.

 

Sample Input

1212

12345666

1235

END

 

Sample Output

2

2

0

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

 

由于n小于=15，故暴力枚举即可。又由于只有数字1~9组成，无需考虑几个相邻的0的情况。很简单的一道水题，我们用二进制来枚举。水过。

 

 

 

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
char str[30];
int get(int i,int j,int x,int cnt)
{
    int ret,a,b,c;
    int s=i;
    ret=a=0;
    for(a=0;a<cnt;a++)
    {
        if((1<<a)&x)
        {
            c=0;
            for(b=s;b<=i+a;b++)
            {
                c=c*10+str[b]-'0';
            }
            s=i+a+1;
            ret+=c;
            
        }
    }
    c=0;
    for(b=s;b<=j;b++)
        c=c*10+str[b]-'0';
    ret+=c;
    return ret;
}
int main()
{
    int i,j,k,ans,a,b,c,len;
    while(scanf("%s",str)&&str[0]!='E')
    {
        len=strlen(str);
        ans=0;
        for(i=1;i<len;i++)
        {
            for(j=0;j<(1<<(i-1));j++)
            {

                for(k=0;k<(1<<(len-i-1));k++)
                {
                    a=get(0,i-1,j,i-1);
                    b=get(i,len-1,k,len-i-1);
                    if(a==b)
                    {
                        ++ans;
                    }
                }

            }
        }
        cout<<ans<<endl;
    }
    return 0;
}