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A very hard Aoshu problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1528    Accepted Submission(s): 1048

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

题意：给定一个字符串，要求插入一个=号和若干个+号使式子成立。

思路：枚举等号的位置，依次将等号左边的所有情况计算出来同时计算等号右边的看是否匹配。

# include 
    
     # include 
     
      char s[20];int num[20][20], ans, len;void init(){    for(int i=1; i<=len; ++i)    {        num[i][i] = s[i]-'0';        for(int j=i+1; j<=len; ++j)            num[i][j] = num[i][j-1]*10 + s[j] - '0';    }}void dfsr(int pos, int leftsum, int sum){    if(pos > len)    {        if(sum == leftsum)            ++ans;        return;    }    for(int i=pos; i<=len; ++i)        dfsr(i+1, leftsum, sum+num[pos][i]);}void dfsl(int equ, int pos, int sum){    if(pos > equ)        dfsr(equ+1, sum, 0);    for(int i=pos; i<=equ; ++i)        dfsl(equ, i+1, sum+num[pos][i]);}int main(){    while(~scanf("%s",s+1))    {        if(s[1] == 'E')            break;        len = strlen(s+1);        memset(num, 0, sizeof(num));        ans = 0;        init();        for(int i=1; i<=len; ++i)            dfsl(i, 1, 0);        printf("%d\n",ans);    }    return 0;}