D. Let‘s Go Hiking

D. Let’s Go Hiking

解题思路：对于A来说，A只能往下寻找。对于B来说，B只能往上寻找。所以A只能选择某个顶点，如果选择一个单调递增或者单调递减的，A选择了之后，B就可以直接拦截。此时A要选择一个单调递增或者单调递减长度最长的，要不然B选择的长度会是最长的，那么B就必胜。只有在单调长度最长且为奇数，并且顶点两边单调性相反，A是必胜。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double lf;
typedef unsigned long long ull;
typedef pair<ll,int>P;
const int inf = 0x7f7f7f7f;
const ll INF = 1e16;
const int N = 1e5+10;
const ull base = 131;
const ll mod =  1e9+7;
const double PI = acos(-1.0);
const double eps = 1e-4;

inline int read(){
    
    int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){
    
    if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
    
    x=x*10+ch-'0';ch=getchar();}return x*f;}
inline string readstring(){
    
    string str;char s=getchar();while(s==' '||s=='\n'||s=='\r'){
    
    s=getchar();}while(s!=' '&&s!='\n'&&s!='\r'){
    
    str+=s;s=getchar();}return str;}
int random(int n){
    
    return (int)(rand()*rand())%n;}
void writestring(string s){
    
    int n = s.size();for(int i = 0;i < n;i++){
    
    printf("%c",s[i]);}}
ll fast_power(ll a,ll p){
    
    
    ll ans = 1;
    while(p){
    
    
        if(p&1) ans = (ans*a)%mod;
        p >>= 1;
        a = (a*a)%mod;
    }
    return ans;
}


int a[N];
int l[N],r[N];

int main(){
    
    
    //srand((unsigned)time(NULL));
    //freopen(  "out.txt","w",stdout);
    int n = read();
    for(int i = 1;i <= n;i++){
    
    
        a[i] = read();
    }
    for(int i = 1;i <= n;i++){
    
    
        if(a[i] > a[i-1]){
    
    
            l[i] = l[i-1]+1;
        }else l[i] = 1;
    }
    for(int i = n;i >= 1;i--){
    
    
        if(a[i] > a[i+1]){
    
    
            r[i] = r[i+1]+1;
        }else r[i] = 1;
    }
    int ma = 0,num = 0,id = 0;
    for(int i = 1;i <= n;i++){
    
    

        if(ma < l[i] || ma < r[i]){
    
    
            ma = max(l[i],r[i]);
            num = 1;
            id = i;
        }else if(ma == l[i] || ma == r[i])num++;
    }
    if(num==1&&ma%2&&l[id]==r[id]){
    
    
        puts("1");
    }else puts("0");
    return 0;
}