A very hard Aoshu problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1507    Accepted Submission(s): 1029


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

 


Source

​​2012 ACM/ICPC Asia Regional Jinhua Online​​

 


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【分析】

乍一看以为是数位dp...不过其实没那么复杂，因为每个字符串长度最大只有15,而且只有一个=，所以直接暴搜就可以，这里有个小优化，预处理一下[i,j]这个数字是多少，然后把整个字符串分成两段，暴搜左边的能出现的数字在右边能不能出现就可以了。

【代码】

#include <cstdio>  
#include <cstring>  
char s[100];  
int len,ans;  
int f[100][100];  
void findd(int deep,int sum,int summ)  
{  
    if(deep==len)  
    {  
        if(sum==summ) ans++;  
        return;  
    }  
    for(int i=deep;i<len;i++)  
        findd(i+1,sum+f[deep][i],summ);  
}  
void find(int deep,int end,int sum)  
{  
    if(deep==end) findd(end,0,sum);  
    for(int i=deep;i<end;i++)  
        find(i+1,end,sum+f[deep][i]); 
}  
int main()  
{  
/*  FILE *fp1, *fp2;
    fp1 = fopen("sample.in","r");
    fp2 = fopen("sample.out","w");
 */
    while(~fscanf(fp1,"%s",s))  
    {  
        if(s[0]=='E') break;  
        memset(f,0,sizeof(f));  
        len=strlen(s);  
        ans=0;  
        for(int i=0;i<len;i++)  
            for(int j=i;j<len;j++)    
                for(int k=i;k<=j;k++)  
                    f[i][j]=f[i][j]*10+s[k]-48;  
        for(int i=1;i<len;i++)  
            find(0,i,0);  
        fprintf(fp2,"%d\n",ans);  
    }  
/*  fclose(fp1);
    fclose(fp2);
*/    return 0;  
}