Finals in arithmetic CodeForces - 625D （构造）

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

思路就是构造，容易发现假如没有进位的产生，那么，头尾对称，是个回文字序列，关键是有进位，所以需要处理进位的问题， 还有一个坑点就是最高有进位的问题。

#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)//因为存在--和+10的操作，容易出现超出范围的不合法情况
            break;
        if(l==r)//中间位置必须为偶数
        {
            if(a[l]%2)
                break;
            a[l]/=2;
        }
        else
        {
            if(a[l]<10)
            {
                if(l==0)//最高为特判不能为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;
}