D1. Great Vova Wall (Version 1)
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Vova's family is building the Great Vova Wall (named by Vova himself). Vova's parents, grandparents, grand-grandparents contributed to it. Now it's totally up to Vova to put the finishing touches.
The current state of the wall can be respresented by a sequence aa of nn integers, with aiai being the height of the ii-th part of the wall.
Vova can only use 2×12×1 bricks to put in the wall (he has infinite supply of them, however).
Vova can put bricks horizontally on the neighboring parts of the wall of equal height. It means that if for some ii the current height of part iiis the same as for part i+1i+1, then Vova can put a brick there and thus increase both heights by 1. Obviously, Vova can't put bricks in such a way that its parts turn out to be off the borders (to the left of part 11 of the wall or to the right of part nn of it).
The next paragraph is specific to the version 1 of the problem.
Vova can also put bricks vertically. That means increasing height of any part of the wall by 2.
Vova is a perfectionist, so he considers the wall completed when:
- all parts of the wall has the same height;
- the wall has no empty spaces inside it.
Can Vova complete the wall using any amount of bricks (possibly zero)?
Input
The first line contains a single integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of parts in the wall.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the initial heights of the parts of the wall.
Output
Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero).
Print "NO" otherwise.
Examples
input
Copy
5
2 1 1 2 5
output
Copy
YES
input
Copy
3
4 5 3
output
Copy
YES
input
Copy
2
10 10
output
Copy
YES
input
Copy
3
1 2 3
output
Copy
NO
Note
In the first example Vova can put a brick on parts 2 and 3 to make the wall [2,2,2,2,5][2,2,2,2,5] and then put 3 bricks on parts 1 and 2 and 3 bricks on parts 3 and 4 to make it [5,5,5,5,5][5,5,5,5,5].
In the second example Vova can put a brick vertically on part 3 to make the wall [4,5,5][4,5,5], then horizontally on parts 2 and 3 to make it [4,6,6][4,6,6] and then vertically on part 1 to make it [6,6,6][6,6,6].
In the third example the wall is already complete.
题意：一个墙有n个部分，每部分的高度为 ai,现在有 2*1 和 1*2 的砖块去填这个墙，使墙的高度都相同。
分析：这是一个规律题，比赛的时候没有想出来，就是
如果相邻墙的奇偶性相同时候，可以发生同一变化
如果只对单独一个部分的话，奇偶性不会发生变化
与就是说奇偶相同的相邻墙可以删去，他们可以达到容易一个变化。
然后我们可以用一个栈了，删除奇偶性相同的元素，为什么要用栈呢？
我们看下面的例子：
2 1 1 2
我们先2 1 1
我们先删去1 1，然后加入 2，2 2 也就删去了。
#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
#define N 100+5
#define rep(i,n) for(int i=0;i<n;i++)
#define sd(n) scanf("%d",&n)
#define sll(n) scanf("%lld",&n)
#define pd(n) scanf("%d\n",n)
#define pll(n) scanf("%lld\n",n)
#define MAX 26
typedef long long ll;
const ll mod=1e6;
ll n,m;
ll a;
ll b[N];
stack<int> s;
int main()
{
scanf("%lld",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld",&a);
a=a&1;//a[i]%2
if(s.empty())
{
s.push(a);
}
else if(a==s.top())
{
s.pop();
}
else
s.push(a);
}
if(s.size()>1)
printf("NO\n");
else
printf("YES\n");
return 0;
}
