Finals in arithmetic CodeForces - 625D(구성)

Vitya is studying in the third grade. During the last math lesson all the pupils wrote on arithmetic quiz. Vitya is a clever boy, so he managed to finish all the tasks pretty fast and Oksana Fillipovna gave him a new one, that is much harder.
Let’s denote a flip operation of an integer as follows: number is considered in decimal notation and then reverted. If there are any leading zeroes afterwards, they are thrown away. For example, if we flip 123 the result is the integer 321, but flipping 130 we obtain 31, and by flipping 31 we come to 13.
Oksana Fillipovna picked some number a without leading zeroes, and flipped it to get number ar. Then she summed a and ar, and told Vitya the resulting value n. His goal is to find any valid a.
As Oksana Fillipovna picked some small integers as a and ar, Vitya managed to find the answer pretty fast and became interested in finding some general algorithm to deal with this problem. Now, he wants you to write the program that for given n finds any a without leading zeroes, such that a + ar = n or determine that such a doesn’t exist.
Input The first line of the input contains a single integer n (1 ≤ n ≤ 10100 000).
Output If there is no such positive integer a without leading zeroes that a + ar = n then print 0. Otherwise, print any valid a. If there are many possible answers, you are allowed to pick any.
Example Input 4 Output 2 Input 11 Output 10 Input 5 Output 0 Input 33 Output 21
사고방식은 구조이다. 만약에 진위가 발생하지 않았다면 머리와 꼬리가 대칭적이고 회문자 서열이다. 관건은 진위가 있기 때문에 진위를 처리해야 하는 문제, 그리고 가장 높은 진위가 있는 문제도 발견할 수 있다.

#include
#include
#include
#include
#include
#include
#include
#define maxx 100005
using namespace std;
char s[maxx];
int a[maxx];
int main()
{
    scanf("%s",s);
    int len=strlen(s);
    for(int i=0;i'0';
    int l=0,r=len-1;
    if(a[l]!=a[r])
    {
        a[l]--;
        a[l+1]+=10;
        if(!a[l])
            l++;
    }
    while(l<=r)
    {
        if(a[l]-a[r]>=10)
        {
            a[r]+=10;
            a[r-1]--;
        }
        if(a[l]-a[r]==1)
        {
            a[l]--;
            a[l+1]+=10;
        }
        if(a[l]!=a[r]||a[l]>18||a[l]<0)//    -- +10   ，              
            break;
        if(l==r)//         
        {
            if(a[l]%2)
                break;
            a[l]/=2;
        }
        else
        {
            if(a[l]<10)
            {
                if(l==0)//        0
                {
                    a[l]=1;
                    a[r]-=1;
                    if(a[r]<0)
                        break;
                }
                else
                    a[r]=0;

            }
            else
            {
                a[l]-=9;
                a[r]=9;
            }
        }
        l++,r--;
    }
    if(l<=r)
        cout<<0<else
    {
        if(a[0])
            cout<0];
for(int i=1;icout<return 0;
}