D. Finals in arithmetic
题目连接：
http://www.codeforces.com/contest/625/problem/D
Description
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.
Sample Input
4
Sample Output
2
Hint
题意
给你一个数k，要求你找到一个数x，使得x+反着的x = k
比如33,你就可以找21，因为21+12=33
不允许前导0
题解：
贪心，我们先不管前导0这个条件，我们如果sum[i]==sum[n-i-1]的话，ans[i]=(sum[i]+1)/2,ans[n-i-1]=sum[i]/2就好了
如果不相等的话，我们应该怎么呢？我们需要考虑进位
要么从后一位进1，要么从前一位退10回来，就这两种，讨论一下就好了
注意165这种数据，1开头的
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
char s[maxn];
char ans[maxn];
int sum[maxn];
int n;
int check()
{
for(int i=0;i<n/2;)
{
int l=i,r=n-1-i;
if(sum[l]==sum[r])
i++;
else if(sum[l]==sum[r]+1||sum[l]==sum[r]+11)//考虑从后面进了一位,11 = 1+10
{
sum[l]--;
sum[l+1]+=10;
}
else if(sum[l]==sum[r]+10)//考虑从R前面退一位
{
sum[r-1]--;
sum[r]+=10;
}
else return 0;
}
if(n%2==1)
{
if(sum[n/2]%2==1||sum[n/2]>18||sum[n/2]<0)return 0;
ans[n/2]=sum[n/2]/2+'0';
}
for(int i=0;i<n/2;i++)
{
if(sum[i]>18||sum[i]<0)return 0;
ans[i]=(sum[i]+1)/2+'0';
ans[n-i-1]=(sum[i])/2+'0';
}
return ans[0]>'0';
}
int main()
{
scanf("%s",s);
n=strlen(s);
for(int i=0;i<n;i++)
sum[i]=s[i]-'0';
if(check())return puts(ans);
if(s[0]=='1'&&n>1)//首位为1的时候
{
for(int i=0;i<n;i++)
sum[i]=s[i+1]-'0';
sum[0]+=10;
n--;
if(check())puts(ans);
else printf("0");
}
else
printf("0");
}