传送门：http://codeforces.com/contest/1092/problem/D1
D1. Great Vova Wall (Version 1)
2 seconds
256 megabytes
standard input
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)?
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.
Print "YES" if Vova can complete the wall using any amount of bricks (possibly zero).
Print "NO" otherwise.
5
2 1 1 2 5
YES
3
4 5 3
YES
2
10 10
YES
3
1 2 3
NO
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.
题意概括：
给出 N 个起始得墙的高度。
可以在这些墙上面放 1*2 或者 2*1 的方块，问最后能否把所有墙变成同一高度。
要求中间不能有空隙。
解题思路：
可以推断出：
一、当前墙的高度为偶数时：
若墙的数量为奇数则只能继续加1*2的砖变换成为偶数的墙；
若墙的数量为偶数时，则可以添加2*1的转变成奇数的高度，或者添加1*2的砖变成偶数的高度。
二、当前墙的高度为奇数时：
若墙的数量为奇数，则只能保持奇数的高度。
若墙的数量为偶数时，则可以变换高度的奇偶性。
所以从上面的推断我们可以看出，奇数个奇偶性相同的的墙相连并不会改变原来的奇偶性，只有当偶数个奇偶性相同的墙相连才能改变墙的奇偶性。
所以我们顺序遍历，利用一个栈来实现偶数个奇偶性相同的墙相抵消，如果最后遍历完栈内元素>1,则说明有两个或者以上的无法抵消的墙，无解。否则有解。
AC code：
#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std; const int MAXN = 2e5+;
LL num[MAXN];
int stacks[MAXN], top;
int N; int main()
{
scanf("%d", &N);
for(int i = ; i <= N; i++){
cin >> num[i];
num[i]&=1LL;
}
top = ;
for(int i = ; i <= N; i++){
if(top == || stacks[top] != num[i])
stacks[++top] = num[i];
else top--;
}
if(top > ) puts("NO");
else puts("YES");
return ; }