Codeforces Round #527 (Div. 3) D2.Great Vova Wall (Version 2)-优快云博客

本文解析了GreatVovaWall算法问题，通过使用栈结构进行分析，判断是否可以通过特定操作使数组所有元素相等。详细介绍了算法流程及实现代码。

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

Vova can only use

Vova can put bricks only horizontally on the neighbouring parts of the wall of equal height. It means that if for some

Note that Vova can't put bricks vertically.

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

The second line contains

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

In the second example Vova can put no bricks in the wall.

In the third example the wall is already complete.

题意

　　给出一组数，可以让连续的两个同时加一，能用这种操作使得最后数组的数相同。

分析

　　用栈一个一个判断就行了。

　　1、当前元素小于栈顶元素肯定不行。

　　2、最后栈为空肯定可以。

　　3、最后栈不空并且有一个元素，栈顶元素小于数组最大值，并且不是最后一个元素肯定不行，否则可以

　　4、最后栈里有大于一个的元素。

///  author:Kissheart  ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=2e5+10;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
stack<int>s;
int n;
int a[MAX],maxn;
int main()
{
    while(!s.empty()) s.pop();

    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);

    for(int i=1;i<=n;i++)
    {
        maxn=max(maxn,a[i]);
        if(s.empty())
        {
            s.push(a[i]);
            continue;
        }
        //printf("%d %d\n",s.top(),a[i]);
        if(s.top()==a[i])
            s.pop();
        else if(s.top()<a[i])
        {
            printf("NO\n");
            return 0;
        }
        else
            s.push(a[i]);
    }
    if(s.empty())
        printf("YES\n");
    else if(s.size()==1)
    {
        if(s.top()<maxn && maxn!=a[n])
            printf("NO\n");
        else
            printf("YES\n");
    }
    else
        printf("NO\n");
    return 0;
}