【Codeforces Round #440 (Div. 2) B】Maximum of Maximums of Minimums
2017-10-15 19:47
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【链接】 我是链接,点我呀:)
【题意】
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\n",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 = a ; 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\n",ans); }else { int ans = a[1]; for (int i = 1;i <= n;i++) ans = max(ans,a[i]); printf("%d\n",ans); } return 0; }
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