Ugly Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2531 Accepted Submission(s): 834
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
Hint
9 + 9 = 18 999999999999 + 1 = 1000000000000
题意 ： 将一个长度为1000以内的大数分解为50个以内的回文数的和
弹性，每次将当前剩余和贪心长度的一半，然后做成不大于当前剩余和的回文串，然后减去，然后继续更新。
注意特判10，100，110，120，。。。190等情况，且0不能算为回文数
例如
n = 1000000
那么先取100 ， 反转成001 和 000 比，如果比000大，就减一 变成了99
做成回文串 99 0 99 ，1000000减去 99099 得 900901
取 900 反转成 009比 901 小， 所以取900，做成回文串 900009，减去 得 892
取 8 反转成8比2大，所以减一变成7，所以 做成回文串797，减去得 95
取9 反转成9比5大，所以减一变成8，所以做成回文串88，减去得7
7已经是回文串了，直接放入
所以 1000000 = 99099 + 900009 + 797 + 88 + 7
如果最后剩下10，100，110的话需要特判一下，不然会死循环，试着理解一下。
import java.util.*;
import java.math.*;
public class Main
{
public static void main(String args[])
{
int T;
Scanner cin = new Scanner(System.in);
T = cin.nextInt();
cin.nextLine();
for(int t = 1; t <= T; t++)
{
String str;
String str1 ,str2, str3;
//str = ""+t;
str = cin.nextLine();
BigInteger x,y,z;
Vector vec = new Vector();
int len = str.length();
while(len > 1)
{
if(str.compareTo("10") == 0) {vec.addElement(9);str = "1";len = 1;continue;}
if(len == 3 && str.charAt(0) == '1' && str.charAt(2) == '0')
{
vec.addElement("99");
str = new BigInteger(str).subtract(new BigInteger("99")).toString();
len = str.length();
continue;
}
x = new BigInteger(str);
str1 = str.substring(0, len/2);
str2 = str.substring(len-len/2,len);
// System.out.println(str);
// System.out.println(str1);
// System.out.println(str2);
str1 = (String)new StringBuffer(str1).reverse().toString();
y = new BigInteger(str.substring(0, len/2));
if(str1.compareTo(str2) > 0) {y = y.subtract(BigInteger.ONE);}
str3 = y.toString();
if(len % 2 == 1) {
str3 += str.charAt(len/2);
}
str3 += new StringBuffer(y.toString()).reverse().toString();
vec.addElement(str3);
str = x.subtract(new BigInteger(str3)).toString();
len = str.length();
}
if(len == 1 && str.charAt(0) != '0') vec.addElement(str);
System.out.println("Case #"+ t +":");
System.out.println(vec.size());
for(int i = 0; i < vec.size(); i++)
System.out.println(vec.elementAt(i));
}
cin.close();
}
}