HDU - 5920 Ugly Problem

HDU - 5920 Ugly Problem

这个题限时训练的时候是3次尝试但是没有做出来的，所以我压根就没看（……可以说是非常诚实的了），后来找题解看了看，其实是一个分奇偶的大数减，但是把……我写着写着写迷糊了，就抄了一个交的。

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≤s≤101000).
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

抄的代码的来源https://blog.csdn.net/qq_25576697/article/details/76850938

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm> 
#include<cstring>
#include<string>
#define N 1010
#define ll long long
//#define INF 1e18
ll const INF = 1e18;
using namespace std;
char a[N],b[N],c[N],ans[N][N]; 
int len,num;
int main()
{
    int t,i,j,k,m,ta,tb,n,tc;
    bool flag,jw;
    scanf("%d",&t);
    for(k=1;k<=t;k++){
        scanf("%s",a);
        len=strlen(a);
        num=0;
        while(len>=1){
            flag=0;
            b[len>>1]=a[len>>1];
            for(i=len/2-1;i>=0;i--){
                if(a[i]!=a[len-1-i]&&!flag){
                    flag=1;
                    b[i]=b[len-1-i]=a[i]<a[len-1-i]?a[i]:a[len-1-i];
                }
                else{
                    b[i]=b[len-1-i]=a[i];
                }
            }
            b[len]=0;
            if(!flag){
                strcpy(ans[num++],a);
                break;
            }
            if(b[0]=='0'){
                if(a[0]=='1'){
                    strcpy(ans[num++],"1\0");
                    for(i=0;i<len-1;i++){
                        ans[num][i]='9';
                    }
                    ans[num++][len-1]=0;
                }
                else{
                    m=a[0]-'0';
                    ans[num][0]=(char)(11-m+'0');
                    ans[num++][1]=0;
                    ans[num][0]=ans[num][len-1]=(char)(a[0]-1);
                    for(i=1;i<len-1;i++){
                        ans[num][i]='9';
                    }
                    ans[num++][len]=0;
                }
                b[0]=a[0];
                for(i=1;i<len;i++){
                    b[i]='0';
                }
                b[len]=0;
            }
            else{
                strcpy(ans[num++],b);
            }
            jw=false;
            for(i=len-1;i>=0;i--){
                ta=a[i]-'0';
                tb=b[i]-'0';
                n=(10+a[i]-b[i]-jw)%10;
                c[i]=n+'0';
                jw=(a[i]-b[i]-jw)<0;
            }
            c[len]=0;
            tc=0;
            while(c[tc]=='0')
            tc++;
            strcpy(a,c+tc);
            len=strlen(a);
        }
        printf("Case #%d:\n%d\n",k,num);
        for(int i=0;i<num;i++)
        cout<<ans[i]<<endl;
    }
    return 0;
}