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] 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].
In the second example Vova can put a brick vertically on part 3 to make the wall [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].
In the third example the wall is already complete.
题解：如果相邻的两堆差值为2的倍数则可以成对消去，这个过程可以用栈来模拟，最后判断如果剩余堆的数量大于一，输出NO即可
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
int main() {
stack<int>s;
int n;
cin>>n;
for(int t=0; t<n; t++) {
int k;
scanf("%d",&k);
if(s.empty()) {
s.push(k);
} else {
if(abs(s.top()-k)%2==1) {
s.push(k);
} else {
s.pop();
}
}
}
if(s.size()>1)
cout<<"NO"<<endl;
else {
cout<<"YES"<<endl;
}
return 0;
}
