Problem:
Score : 800 points

Problem Statement

For a positive integer n, we denote the integer obtained by reversing the decimal notation of n (without leading zeroes) by rev(n). For example, rev(123) = 321 and rev(4000) = 4.

You are given a positive integer D. How many positive integers N satisfy rev(N) = N + D?

Constraints

D is an integer.

1 ≤ D < 10^9

Input

Input is given from Standard Input in the following format:

D

Output

Print the number of the positive integers N such that rev(N) = N + D.

Sample Input 1

63

Sample Output 1

2

There are two positive integers N such that rev(N) = N + 63: N = 18 and 29.

Sample Input 2

75

Sample Output 2

0

There are no positive integers N such that rev(N) = N + 75.

Sample Input 3

864197532

Sample Output 3

1920

Solution in C++ (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
void read(int &x){
	char ch;bool ok;
	for(ok=0,ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-')ok=1;
	for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());if(ok)x=-x;
}
#define rg register
const int maxn=1e5+10;
int mm,m,n,a[60],f[60];
long long ans;
void dfs(int x,int n,long long sum){
	if(x==n/2+1){
		if(sum==m){
			long long now=a[1]>=0?(9-a[1]):(9+a[1]);
			for(rg int i=2;i<x;i++)now=now*(a[i]>=0?(10-a[i]):(10+a[i]));
			ans+=now;
		}
		return ;
	}
	for(rg int i=-9;i<10;i++){
		a[x]=i;
		dfs(x+1,n,sum+1ll*a[x]*(f[n-x]-f[x-1]));
	}
}
int main(){
	read(m),mm=m;f[0]=1;
	while(mm)n++,mm/=10;
    for(rg int i=1;i<=17;i++)f[i]=f[i-1]*10;
	for(rg int i=n;i<=17;i++)dfs(1,i,0);
	printf("%lld\n",ans);
}