http://uoj.ac/contest/35/problem/244

对其他人来说好简单的一道题，我当时却不会做TWT

注定滚粗啊

题解很好的~

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

const int N = 100003;

ll in() {

	ll k = 0, fh = 1; char c= getchar();

	for (; c < '0' || c > '9'; c = getchar())

		if (c == '-') fh = -1;

	for (; c >= '0' && c <= '9'; c = getchar())

		k = k * 10 + c - 48;

	return k * fh;

}

int n;

ll a[N], f[N], ans, now;

int main() {

	scanf("%d", &n);

	for (int i = 0; i <= n; ++i) a[i] = in();

	f[n] = a[n];

	now = a[n];

	ans = a[n] * (4 * n + 1);

	for (int i = n - 1; i >= 0; --i) {

		f[i] = f[i + 1] + now + a[i];

		if (a[i] < now) {

			now = a[i];

			ans = min(ans, f[i] * 2 + now * (4 * i - 1));

		}

	}

	printf("%lld\n", ans);

	return 0;

}