2016中国大学生程序设计竞赛（长春）-重现赛 1010Ugly Problem 回文数 模拟

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：http://acm.hdu.edu.cn/showproblem.php?pid=5920

---

题意：将一个数拆成n（n<=50）个回文数的和。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char s[];
int ans[][];
int gg[];
int sub(int t,int len)
{
    int i;
    for(i=len-; i>=; i--)
    {
        s[i]=s[i]-''-ans[t][i];
        if(s[i]<)
        {
            s[i]+=+'';
            s[i-]-=;
        }
        else s[i]+='';
    }
    for(i=; i<len; i++)
        if(s[i]!='') break;
    return i;
}
int main()
{
    int i,j,t,T;
    while(scanf("%d",&T)!=EOF)
    {
        getchar();
        for(int asd=; asd<=T; asd++)
        {
            scanf("%s",s);
            memset(ans,,sizeof(ans));
            int pre=,len=strlen(s);
            t=;
            while(pre<len)
            {
                int sign=;
                for(i=pre,j=len-; i<=j; i++,j--)
                {
                    ans[t][i]=ans[t][j]=s[i]-'';
                    if(ans[t][j]>s[j]-'') sign=;
                    else if(ans[t][j]<s[j]-'')sign=;
                }
                gg[t]=pre;
                if(sign==)
                {
                    i--;
                    ans[t][i]-=;
                    while(i>=&&ans[t][i]<)
                    {
                        ans[t][i]+=;
                        ans[t][i-]-=;
                        i--;
                    }
                    for(i=pre,j=len-; i<=j; i++,j--)
                        ans[t][j]=ans[t][i];
                    if(ans[t][pre]==)
                    {
                        ans[t][len-]=;
                        gg[t]=pre+;
                    }
                }
                pre=sub(t,len);
                t++;
            }
            printf("Case #%d:\n",asd);
            printf("%d\n",t);
            for(i=; i<t; i++)
            {
                for(j=gg[i]; j<len; j++)
                    printf("%d",ans[i][j]);
                printf("\n");
            }
        }
    }
    return ;
}