Ugly Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1159    Accepted Submission(s): 409
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

≤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

題意：給定一個數字，找出一個或幾個迴文數字，使得這幾個迴文數字之和剛好等於給出的數字。迴文數字的數量不大於50！！！

字串長度len無論奇偶，i與len-1-i都是左右對稱的這點很受用

/*
思路：從中間開始向兩邊遍歷並逐位賦值給b，找到不對稱的地方時取較小的數，
每次找到的b都是小於a的較大回文數字，然後更新a，直到a的長度為1 
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N = 1010;  
char a[N],b[N],c[N],ans[N][N]; 
int len,num;
int main()
{
	int t;
	cin>>t;
	for(int k=1;k<=t;k++)
	{
		cin>>a;     //條件數字輸入 
		len=strlen(a);
		num=0;
		while(len>=1)
		{
			bool flag=0;
			b[len>>1]=a[len>>1];     //為奇數長度服務，因為在迴圈中奇數長度的中間字元沒有包含 
			for(int i=len/2-1;i>=0;i--)               //用這種方法可以很好的一起處理len為奇數或偶數的情況 
			{
				if(a[i]!=a[len-1-i]&&!flag)    //找到首個兩邊不相等的字元，並取小的 
				{
					flag=1;
					b[i]=b[len-1-i]=a[i]<a[len-1-i]?a[i]:a[len-1-i];
				}
				else                                //因為對輸出迴文數字的個數有要求所以要使得這些迴文數字儘可能大但是又不能比給定數字a大 
					b[i]=b[len-1-i]=a[i];
			}
			b[len]=0;
			if(!flag)      //如果本身已經是個迴文數字就退出while迴圈準備輸出 
			{
				strcpy(ans[num++],a);
				break;
			}
			if(b[0]=='0')        //特殊情況，b有前導0 
			{
				if(a[0]=='1')    //把a拆分為1,9999…9，還有餘下部分 
				{
					strcpy(ans[num++],"1\0");
					for(int i=0;i<len-1;i++)
						ans[num][i]='9';
					ans[num++][len-1]=0;
				}
				else        //根據a[0]不同把a拆分為8,2999…92，餘下部分  或7,3999…93，餘下部分   或…… 
				{
					int 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(int i=1;i<len-1;i++)
						ans[num][i]='9';
					ans[num++][len]=0;
				}
				b[0]=a[0];
				for(int i=1;i<len;i++)b[i]='0';
				b[len]=0;
			}           //沒有前導0，直接把b放到結果串中 
			else
			{
				strcpy(ans[num++],b);
			}
			//獲取c=a-b  並消除前導0存入a 
			bool jw=false;
			int ta,tb,n;
			for(int 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;
			int 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;
	}
}