Reversion Count

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<time.h>
#include<iostream>
#include<ctype.h>
#include<map>
#include<set>
#include<algorithm>
#include<stdlib.h>
#include<queue>
#include<stack>
using namespace std;
string a,b,c;
string sub(string a, string b)
{
    string c;
    bool ok = 0;
    int len1 = a.length();
    int len2 = b.length();
    int len = max(len1, len2);
    for(int i = len1; i < len; i++)//短的串前边加0方便后边计算
        a = "0" + a;
    for(int i = len2; i < len; i++)
        b = "0" + b;
    if(a < b)//判断正负
    {
        string temp = a;
        a = b;
        b = temp;
        ok = 1;
    }
    for(int i = len - 1; i >= 0; i--)
    {
        if(a[i] < b[i])
        {
            a[i - 1] -= 1;
            a[i] += 10;
        }
        char temp = a[i] - b[i] + ‘0‘;
        c = temp + c;//此处不能写为c+temp因为是倒着算，需要正着存，顺序写错就不对了，自己可以试几组样例
    }
    int pos = 0;
    while(c[pos] == ‘0‘ && pos < len) pos++;//去除前导0
    if(pos == len) return "0";
    if(ok) return "-" + c.substr(pos);
    return c.substr(pos);
}

string  chufa(string s,int x)
{
    int jj=0,fla=0;
    string ans="";
    for(int i=0; i<s.size(); i++)
    {
        jj=jj*10+s[i]-‘0‘;
        if(jj>=x)
        {
            fla=1;
            ans+=jj/x+‘0‘;
            jj%=x;
        }
        else
        {
            if(fla==1)//在运算过程中不够除商0；
                ans+=‘0‘;
        }
    }
    return ans;
}
int main()
{
    int n,i,j;
    while(cin>>a)
    {
        set<char>qq;
        b=a;
        reverse(a.begin(),a.end());
        c=sub(b,a);
        if(c[0]==‘-‘)//题目不要求正负，可以删除再计算
            c.erase(0,1);
        if(c.size()==1)
        {
            printf("YES\n");
            continue;
        }
        string ans=chufa(c,9);

        for(int i=0; i<ans.size(); i++)
        {
            qq.insert(ans[i]);
        }
        if(qq.size()==1)
            printf("YES\n");
        else
            printf("NO\n");
    }
}