很明显"当前还未打印的数字中的最大数字"压栈时,就应该把弹栈打印
此时一定是字典序最大,如果最后栈内还有值,就全部出栈打印即可

\

#ifdef debug
#include <time.h>
#include "/home/majiao/mb.h"
#endif

#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <math.h>

#define MAXN (1000005)
#define ll long long int
#define INF (0x7f7f7f7f)
#define fori(lef, rig) for(int i=lef; i<=rig; i++)
#define forj(lef, rig) for(int j=lef; j<=rig; j++)
#define fork(lef, rig) for(int k=lef; k<=rig; k++)
#define QAQ (0)

using namespace std;

#ifdef debug
#define show(x...) \
do { \
    cout << "\033[31;1m " << #x << " -> "; \
    err(x); \
} while (0)

void err() { cout << "\033[39;0m" << endl; }
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }
#endif

#ifndef debug
namespace FIO {
    template <typename T>
    void read(T& x) {
        int f = 1; x = 0;
        char ch = getchar();

        while (ch < '0' || ch > '9') 
            { if (ch == '-') f = -1; ch = getchar(); }
        while (ch >= '0' && ch <= '9') 
            { x = x * 10 + ch - '0'; ch = getchar(); }
        x *= f;
    }
};
using namespace FIO;
#endif


int n, m, Q, K, a[MAXN];
stack<int> stk;

int main() {
#ifdef debug
    freopen("test", "r", stdin);
    clock_t stime = clock();
#endif
    scanf("%d ", &n);
    int now = n;
    for(int i=1, x; i<=n; i++) {
        scanf("%d ", &x);
        if(x == now) //测试数据比较水,一开始少了now--都能AC
            a[++m] = x, now --;
        else stk.push(x);
    }
    while(!stk.empty()) {
        a[++m] = stk.top();
        stk.pop();
    }
    for(int i=1; i<=n; i++)
        printf("%d%c", a[i], i==n?'\n':' ');






#ifdef debug
    clock_t etime = clock();
    printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC);
#endif 
    return 0;
}