codeforces 1313C1 Skyscrapers (easy version) 暴力枚举

对于每座高楼，两侧的楼不能同时高于这座楼
于是就是一个关于高楼的单峰函数

暴力枚举以每个位置作为最高点的答案
注意开 long long

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 100010;

int n;
ll a[maxn],b[maxn],ans[maxn];

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	n = read();
	for(int i=1;i<=n;i++) a[i] = read();
	
	ll ma = 0;
	for(int i=1;i<=n;i++){
		ll h = a[i],sh = a[i];
		b[i] = a[i];
		for(int j=i-1;j>=1;j--){
			if(a[j] >= h){
				b[j] = h;
				sh += b[j];
			}else{
				h = a[j];
				b[j] = h;
				sh += b[j];
			}
		}
		h = a[i];
		for(int j=i+1;j<=n;j++){
			if(a[j] >= h){
				b[j] = h;
				sh += b[j];
			}else{
				h = a[j];
				b[j] = h;
				sh += b[j];
			}
		}
		if(sh>ma){
			ma = sh;
			for(int j=1;j<=n;j++){
				ans[j] = b[j];
			}
		}
	}
	
	for(int i=1;i<=n;i++) printf("%lld ",ans[i]); printf("
"); 
	
	return 0;
}