2. 程式人生 > >CodeForces Round #527 (Div3) D2. Great Vova Wall (Version 2)

CodeForces Round #527 (Div3) D2. Great Vova Wall (Version 2)

阿新 • • 發佈：2018-12-27

http://codeforces.com/contest/1092/problem/D2

 

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 

a">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 only horizontally on the neighbouring parts of the wall of equal height. It means that if for some 

i">ii the current height of part ii is 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 
n">nn of it).

Note that Vova can't put bricks vertically.

Vova is a perfectionist, so he considers the wall completed when:

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 
NO
input Copy 
2
10 10
output Copy 
YES
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 no bricks in the wall.

In the third example the wall is already complete.

程式碼：

#include <bits/stdc++.h>
using namespace std;

int N;
stack<int> s;

int main() {
    scanf("%d", &N);
    bool flag = true;
    int maxx = INT_MIN;
    for(int i = 0; i < N; i ++) {
        int x;
        scanf("%d", &x);
        maxx = max(maxx, x);
        if(s.empty()) s.push(x);
        else {
            if(s.top() < x)
                flag = false;
            else if(s.top() == x)
                s.pop();
            else s.push(x);
        }
    }

    if(s.size() > 1) flag = false;
    else if(s.size() == 1 && s.top() < maxx) flag = false;

    if(flag) printf("YES\n");
    else printf("NO\n");
    return 0;
}

　　emmmm  兩個牆的高度如果一樣的話可以通過橫著放增加到任意高度按照這個運用棧 好像是除了做運算第一次用棧正經的寫題

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