【Codeforces Round #440 (Div. 2) B】Maximum of Maximums of Minimums

在这里输入题意

【题解】

k=1的时候就是最小值, k=2的时候,暴力枚举分割点。 k=3的时候,最大值肯定能被“独立出来”,则直接输出最大值

【代码】

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

const int N = 1e5;

int a[N+10],n,k,premi[N+10],aftermi[N+10];

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    scanf("%d%d",&n,&k);
    for (int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    if (k==1)
    {
        int ans = a[1];
        for (int i = 2;i <= n;i++)
            ans = min(ans,a[i]);
        printf("%d
",ans);
    }else if (k==2)
    {
        premi[1] = a[1];
        for (int i = 2;i <= n;i++)
            premi[i] = min(premi[i-1],a[i]);
        aftermi[n] = a[n];
        for (int i = n-1;i >= 1;i--)
            aftermi[i] = min(aftermi[i+1],a[i]);
        int ans = max(premi[1],aftermi[2]);
        for (int i = 2;i <= n-1;i++)
            ans = max(ans,max(premi[i],aftermi[i+1]));
        printf("%d
",ans);
    }else {
        int ans = a[1];
        for (int i = 1;i <= n;i++)
            ans = max(ans,a[i]);
        printf("%d
",ans);
    }
    return 0;
}