wa on t4

4
Time: 46 ms, memory: 0 KB
Verdict: WRONG_ANSWER
Input
20
7 0 9 0 5 0 15 0 0 1 17 0 11 19 0 0 3 0 13 0
Participant's output
8
Jury's answer
15
Checker comment
wrong answer 1st numbers differ - expected: '15', found: '8'

#include <bits/stdc++.h>
using namespace std;
int n;
int a[101] = {3}, two, one, ans = 0;
bool book[101];
int main() 
{
	cin >> n;
	if (n % 2 == 1) two = (n - 1) / 2, one = two + 1;
	else       one = n / 2, two = n / 2;
	for (int i = 1; i <= n; i++) 
	{
		int x;
		cin >> x;
		book[x] = 1;
		if (x == 0)   		a[i] = 0;
		else if (x % 2 == 1) a[i] = 1, one--;//只保留他的奇偶性，
		else if (x % 2 == 0) a[i] = 2, two--;//计算剩余可放置的奇数和偶数的数量
		if (i != 1 && a[i] != 0 && a[i - 1] != 0 && a[i] != a[i - 1]) ans++;
	}														//本身存在的一对数
	int l = 0, r = 0;
	while (l <= n && r <= n) 
	{
		l++;
		if (a[l] != 0) continue;
		r = l;
		int flag = 0;
		while (r <= n) 
		{
			r++;
			if (a[r] != 0) 
			{
				flag = 1;
				break;
			}
		}//这一部分计算0的区间，l位于第一个0，r位于最后一个零的后面

		int num0 = abs(l - r);//连续零的数量
		int st = a[l - 1], ed = a[r];//零的前面和后面的奇偶性
		
		//下面计算每一个0的区间的特点，并计算答案
		if (flag == 0) //当前零的区间是最后一个，
		{			   //且当前零的区间的最后一个零是整个数列的最后一个。
			if (a[n] == 0) 
			{
				if (st % 2 == 1) 
				{
					if (one >= num0) one -= num0;
					else ans++;
				} 
				else 
				{
					if (two >= num0) two -= num0;
					else ans++;
				}
				break;
			}
		} 
		else if (st == 3) 
		{
			//当前零的区间是第一个，
			//且当前零的区间的第一个零是整个数列的第一个。
			if (ed % 2 == 1) 
			{
				if (one >= num0) one -= num0;
				else ans++;
			} 
			else 
			{
				if (two >= num0) two -= num0;
				else ans++;
			}
		} 
		else if (st != ed) ans++;//当前0的区间前后奇偶性不相同
		else if (st == ed) //当前0的区间前后奇偶性相同
		{
			if (st == 1) 
			{
				if (one >= num0) one -= num0;
				else ans++;
			} 
			else 
			{
				if (two >= num0) two -= num0;
				else ans += 2;
			}
		}
		l = r;
	}
	cout << ans;
	return 0;
}