解题思路

直接观察答案也好，思考性质也行，可以发现，先对输入数组排序后，如果跑到最大的，一定要最先输出最大才是最优，那么如果可以输出次大，那就继续输出次大，如果次大再最大前进栈，那么就没办法了，对于在后面的次大直接和最大一样比较就行，其余数按照STL栈输出即可。）可看代码，ε=ε=ε=┏(゜ロ゜;)┛

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e6 + 7;
int a[N], b[N];
stack<int> st;

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i)    b[i] = a[i] = read();
    sort(a + 1, a + 1 + n, greater<int>());
    int pos = 1;
    for (int i = 1; i <= n; ++i)
        if (a[pos] == b[i])    write(a[pos++]), putchar(32);
        else    st.push(b[i]);
    while (st.size())
        write(st.top()), st.pop(), putchar(32);
    return 0;
}