谁能用c帮我解决一下这个问题，通过了必采纳

DEV会爆内存

#include <bits/stdc++.h>
using namespace std;
#define MAXN 100010
int main() {
    int n;
    cin >> n;
    int a[MAXN];
    int b[MAXN];
    int c[MAXN] = {0};
    int d[MAXN] = {0};
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    int e = -1;
    int f = 0;
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        if (b[i] != 0) {
            c[b[i]] = i + 2;
            d[b[i]] = i + 2 - b[i];
            if (i + 2 - b[i] >= f) {
                f = i + 2 - b[i];
            }
        }
        if (b[i] == 1) {
            e = i;
        }
    }
    if (e == -1) {
        int g = 0;
        int h = 0;
        int k = d[1]+n;
        for (int i = 2; i <= n; i++) {
            if (d[i] > h) {
                g = 1;
                h = d[i];
                k = d[i] + n;
            }
        }
        if (g == 0) {
            cout << n << endl;
            return 0;
        } else {
            cout << k << endl;
            return 0;
        }
    }
    int l = 1;
    for (int i = 0; i < n - e; i++) {
        if (b[i + e] != i + 1) {
            l = 0;
            break;
        }
    }
    if (l == 0) {
        int h = d[1];
        int k = d[1] + n;
        for (int i = 2; i <= n; i++) {
            if (d[i] > h) {
                h = d[i];
                k = d[i] + n;
            }
        }
        cout << k << endl;
        return 0;
    } else {
        int g = 0;
        int m = n - e + 1;
        for (int i = m, j = 1; i <= n; i++, j++) {
            if (c[i] > j) {
                g = 1;
            }
        }
        if (g == 0) {
            cout << e << endl;
            return 0;
        } else {
            cout << e + n + 1 << endl;
            return 0;
        }
    }
}

#include <iostream>
#include <stdio.h>
#include <algorithm>

#define MAXN 200010

using namespace std;

int main()
{
    ios::sync_with_stdio(false);
    //freopen("E://test.txt", "r", stdin);
    long n;
    cin >> n;

    long Hand[MAXN];//手牌
    long Desk[MAXN];//桌子上的牌

    long JiPaiQi[MAXN] = {0};//记牌器用于记录还需要多少步才能被取出来
    long Cha[MAXN] = {0};//差值用于计算该牌
    for (int i = 0; i < n; i++)
    {
        cin >> Hand[i];
    }

    long firstdes = -1;//数字为1的具体位置
    long BigCha = 0;

    for (long i = 0; i < n; i++)
    {//此处的Desk[i]代表该牌的值
        cin >> Desk[i];
        if (Desk[i] != 0)
        {
            JiPaiQi[Desk[i]] = i + 2;//记录着几号牌距离能被使用还有几步
            Cha[Desk[i]] = i + 2 - Desk[i];
            //步数减去当前牌号是一个差值，假设1号牌在0号位置，那么需要2步才能取出来
            //则最终的步数会是2-1再加上完整的一轮，是为了寻找启动点而设计的
            if (i + 2 - Desk[i] >= BigCha)
            {
                BigCha = i + 2 - Desk[i];
            }
        }
        if (Desk[i] == 1)
        {
            firstdes = i;
        }
    }

    if (firstdes == -1)
    {//1在手里，要么是从启动点开始，要么是直接轮着放下去就好了
        int flag = 0;
        long QiDongDian = 0;
        long bushu = Cha[1]+n;
        for (long i = 2; i <= n; i++)
        {//对所有的地下的牌进行遍历，如果出现无法及时取出的牌则从该启动点开始
            if (Cha[i] > QiDongDian)
            {
                flag = 1;
                QiDongDian = Cha[i];
                bushu = Cha[i] + n;
            }
        }


        if (flag == 0)
        {//全部可以及时的取出答案便是n
            cout << n << endl;
            return 0;
        }
        else
        {//无法及时取出的情况，从启动点开始
            cout << bushu << endl;
            return 0;
        }
    }

    int op = 1;//判定此时1后面及其以后是否紧跟着2，3等
    for (long i = 0; i < n - firstdes; i++)
    {
        if (Desk[i + firstdes] != i + 1)
        {
            op = 0;
            break;
        }
    }

    if (op == 0)
    {//后方未进行紧跟
        //cout<<"hh";
        long QiDongDian = Cha[1];
        long bushu = Cha[1] + n;

        for (long i = 2; i <= n; i++)
        {
            if (Cha[i] > QiDongDian)
            {
                QiDongDian = Cha[i];
                bushu = Cha[i] + n;
            }
        }
        /*for (int i = 1; i <= n; i++)
        {
            cout << Cha[i] << " " << JiPaiQi[i] << endl;
        }*/
        cout << bushu << endl;
        return 0;
    }
    else
    {//后方紧跟的情况，需要判定是否能取到所有的数字
        //cout<<"OK"<<endl;
        int flag = 0;
        //long QiDongDian = JiPaiQi[1];
        long QiBuShuZi = n - firstdes + 1;


        for (long i = QiBuShuZi, j = 1; i <= n; i++, j++)
        {//对所有的地下的牌进行遍历，如果出现无法及时取出的牌则从该启动点开始
            if (JiPaiQi[i] > j)
            {
                flag = 1;
                //QiDongDian = JiPaiQi[i];
            }
        }


        if (flag == 0)
        {//全部可以及时的取出答案便是n
            cout << firstdes << endl;
            return 0;
        }
        else
        {//无法及时取出的情况，从启动点开始
            cout << firstdes + n + 1 << endl;
            return 0;
        }
    }

}

我谔谔

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
signed main() {
    ll n;
    cin >> n;
    ll a[285714];
    ll b[285714];
    ll c[285714] = {0};
    ll d[285714] = {0};
    for (ll i = 0; i < n; i++) {
        cin >> a[i];
    }
    ll e = -1;
    ll f = 0;
    for (ll i = 0; i < n; i++) {
        cin >> b[i];
        if (b[i] != 0) {
            c[b[i]] = i + 2;
            d[b[i]] = i + 2 - b[i];
            if (i + 2 - b[i] >= f) {
                f = i + 2 - b[i];
            }
        }
        if (b[i] == 1) {
            e = i;
        }
    }
    if (e == -1) {
        ll g = 0;
        ll h = 0;
        ll k = d[1]+n;
        for (ll i = 2; i <= n; i++) {
            if (d[i] > h) {
                g = 1;
                h = d[i];
                k = d[i] + n;
            }
        }
        if (g == 0) {
            cout << n << endl;
            return 0;
        } else {
            cout << k << endl;
            return 0;
        }
    }
    ll l = 1;
    for (ll i = 0; i < n - e; i++) {
        if (b[i + e] != i + 1) {
            l = 0;
            break;
        }
    }
    if (l == 0) {
        ll h = d[1];
        ll k = d[1] + n;
        for (ll i = 2; i <= n; i++) {
            if (d[i] > h) {
                h = d[i];
                k = d[i] + n;
            }
        }
        cout << k << endl;
        return 0;
    } else {
        ll g = 0;
        ll m = n - e + 1;
        for (ll i = m, j = 1; i <= n; i++, j++) {
            if (c[i] > j) {
                g = 1;
            }
        }
        if (g == 0) {
            cout << e << endl;
            return 0;
        } else {
            cout << e + n + 1 << endl;
            return 0;
        }
    }
}