题目链接：
Ugly Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 363 Accepted Submission(s): 134
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
题意：
给一个大数,然后让你找到不超过50个回文数的和为这个数;
思路：
每次找前边取一半减一,然后再复制另一半,这样很快就好了，手写了了一个大数减法,写成了智障了都;
AC代码：
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL;
typedef unsigned long long ULL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
} const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e4+120;
const int maxn=1e3+220;
const double eps=1e-12; int n,k,cnt;
char s[maxn];
int a[maxn]; struct Big
{
int a[maxn],leng;
}ans[maxn],temp;
Big fun(Big A,Big B)
{
Big C;
mst(C.a,0);
for(int i=A.leng;i<maxn;i++)A.a[i]=0;
for(int i=B.leng;i<maxn;i++)B.a[i]=0;
int hi=max(A.leng,B.leng),lc=0;
for(int i=0;i<hi;i++)
{
if(A.a[i]<B.a[i])
{
A.a[i+1]--;
A.a[i]+=10;
C.a[i]=A.a[i]-B.a[i];
}
else C.a[i]=A.a[i]-B.a[i];
}
for(int i=0;i<=hi;i++)
{
if(C.a[i]<0)
{
C.a[i+1]--;
C.a[i]+=10;
}
}
int flag=1;
for(int i=hi+1;i>=0;i--)
{
if(C.a[i]>0){lc=i;flag=1;break;}
}
if(flag)C.leng=lc+1;
else C.leng=0;
return C;
} Big newBig(Big A)
{
int lenth=A.leng;
Big B,C,D,E;
mst(B.a,0);B.leng=0;
mst(C.a,0);C.leng=0;
mst(D.a,0);D.leng=0;
mst(E.a,0);E.leng=0;
B.leng=lenth;
if(lenth%2==1)
{
int mid=lenth/2;
for(int i=lenth-1;i>=mid;i--)B.a[i]=A.a[i];
for(int i=mid-1;i>=0;i--)B.a[i]=D.a[i]=0;
D.a[mid]=1;
D.leng=mid+1;
E=fun(B,D);
if(E.leng==A.leng-1)
{
C.leng=E.leng;
for(int i=0;i<C.leng;i++)C.a[i]=9;
}
else
{
C.leng=A.leng;
for(int i=C.leng-1;i>=mid;i--)C.a[i]=E.a[i];
for(int i=mid-1;i>=0;i--)C.a[i]=E.a[2*mid-i];
}
}
else
{
int mid=lenth/2;
for(int i=lenth-1;i>=mid;i--)B.a[i]=A.a[i];
for(int i=mid-1;i>=0;i--)B.a[i]=D.a[i]=0;
D.a[mid]=1;D.leng=mid+1;
E=fun(B,D);
if(E.leng!=lenth)
{
C.leng=E.leng;
for(int i=C.leng-1;i>=0;i--)C.a[i]=9;
}
else
{
C.leng=lenth;
for(int i=C.leng-1;i>=mid;i--)C.a[i]=E.a[i];
for(int i=mid-1;i>=0;i--)C.a[i]=E.a[2*mid-i-1];
}
}
return C;
}
void solve()
{
int lenth=temp.leng;
Big A;
while(lenth)
{
if(lenth==2&&temp.a[1]==1)
{
A.leng=1;
A.a[0]=9;
}
else if(lenth==1)
{
if(temp.a[0]==0)break;
A.leng=1;
A.a[0]=temp.a[0];
ans[++cnt]=A;
break;
}
else A=newBig(temp);
ans[++cnt]=A;
temp=fun(temp,A);
lenth=temp.leng;
}
}
int main()
{
int t,Case=0;
read(t);
while(t--)
{
printf("Case #%d:\n",++Case);
scanf("%s",s);
int len=strlen(s),num=0;
cnt=0;
for(int i=len-1;i>=0;i--)temp.a[num++]=s[i]-'0';
temp.leng=len;
solve();
printf("%d\n",cnt);
for(int i=1;i<=cnt;i++)
{
for(int j=ans[i].leng-1;j>=0;j--)printf("%d",ans[i].a[j]);
printf("\n");
}
}
return 0;
}