Ugly Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0

Special Judge

Problem Description

Everyone hates ugly problems.

You are given a positive integer. You must represent that number by sum of palindromic numbers.

A palindromic number is a positive integer such that if you write out that integer as a string in decimal without leading zeros, the string is an palindrome. For example, 1 is a palindromic number and 10 is not.

 

 

Input

In the first line of input, there is an integer T denoting the number of test cases.

For each test case, there is only one line describing the given integer s (

1\leq s \leq 10^{1000}).

 

 

Output

For each test case, output “Case #x:” on the first line where x is the number of that test case starting from 1. Then output the number of palindromic numbers you used, n, on one line. n must be no more than 50. en output n lines, each containing one of your palindromic numbers. Their sum must be exactly s.

 

 

Sample Input

2

18

1000000000000

 

 

Sample Output

Case #1:

2

9

9

Case #2:

2

999999999999

1

 

Hint

9 + 9 = 18 999999999999 + 1 = 1000000000000

 

题目连接：

---

题意：将一个数拆成n（n<=50）个回文数的和。

代码：

#include
         
          #include
          
           #include
           
            using namespace std;char s[2000];int ans[100][2000];int gg[100];int sub(int t,int len){    int i;    for(i=len-1; i>=0; i--)    {        s[i]=s[i]-'0'-ans[t][i];        if(s[i]<0)        {            s[i]+=10+'0';            s[i-1]-=1;        }        else s[i]+='0';    }    for(i=0; i
            
             s[j]-'0') sign=0;                    else if(ans[t][j]
             
              =0&&ans[t][i]<0)                    {                        ans[t][i]+=10;                        ans[t][i-1]-=1;                        i--;                    }                    for(i=pre,j=len-1; i<=j; i++,j--)                        ans[t][j]=ans[t][i];                    if(ans[t][pre]==0)                    {                        ans[t][len-1]=9;                        gg[t]=pre+1;                    }                }                pre=sub(t,len);                t++;            }            printf("Case #%d:\n",asd);            printf("%d\n",t);            for(i=0; i