C++14(g++5.4) 解法, 执行用时: 80ms, 内存消耗: 1612K, 提交时间: 2019-12-29 14:56:57

#include <iostream>
#include <stack>
#include <algorithm>
using namespace std;

int main()
{
  int i, n, tmp;
  long long h[200000], ans, ansh, ret;
  stack<int> s;
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) scanf("%lld", &h[i]);
    ans = 0;
    i = 0;
    while (1) {
      if (i < n && (s.empty() || h[s.top()] <= h[i])) s.push(i++);
      else {
	tmp = s.top(); s.pop();
        ret = h[tmp] * (s.empty() ? i : (i - s.top() - 1));
        if (ret > ans) {
          ans =	ret;
          ansh = h[tmp];
        }
      }
      if (i >= n && s.empty()) break;
    }
    printf("%lld %lld\n", ansh, ans);
  }
  return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 61ms, 内存消耗: 4660K, 提交时间: 2020-07-17 16:02:31

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

int t,h[100005],S[100005];
int main()
{
	long long i,j,n,ans1,ans2;
	while(~scanf("%lld",&n))
	{
		for(i=0;i<n;i++)scanf("%d",&h[i]);
		ans1=ans2=t=0,S[0]=-1;
		for(i=0;i<n;i++)
		{
			if(!t||(t&&h[i]>=h[S[t]]))S[++t]=i;
			else
			{
				while(t&&h[i]<h[S[t]])
				{
					j=S[t--];
					if(ans2<(i-S[t]-1)*h[j])ans2=(i-S[t]-1)*h[j],ans1=h[j];
				}
				S[++t]=i;
			}
		}
		while(t)
		{
			j=S[t--];
			if(ans2<(i-S[t]-1)*h[j])ans2=(i-S[t]-1)*h[j],ans1=h[j];
		}
		printf("%lld %lld\n",ans1,ans2);
	}
}