CodeForces - 1173C

The meaning of problems

Total 2n cards, there are n n Zhang and Zhang empty card marked with the numbers 1-n cards, and now you have got n cards, cards pile of n cards, you can take a card from your hand to put under the deck, and then removed the top of a card deck. The minimum number of operations required to the deck plate 1-n in the order of arrangement

Thinking

Two cases:

Ⅰ 1 in the sequence number and sequence directly to the card

Must satisfy the condition: 1 in the sequence number 1

　　　　　　2. From a strictly increasing sequence start (check1 ())

　　　　　　3. turn any digital numbers are in the hands of (check2 ())

Answer: number to be added, i.e., nb [n]

Does not satisfy Ⅰ Ⅱ

The answer: By adding time to time added n + 1 digits

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define M (1000000007)
#define N (1000010)
int n,mx,ans,a[N],b[N],id[N];
bool out[N];

bool check1(){
    for (int i=id[1]+1;i<=n;i++) 
        if (b[i]-b[i-1]!=1) return 0;
    return 1;
}

bool check2(){
    for (int i=1;i<=id[1]-1;i++)
        if (b[i]!=0&&b[i]-b[n]<=i) return 0;
    return 1;
}

int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),out[a[i]]=1;
    for (int i=1;i<=n;i++){
        scanf("%d",&b[i]); id[b[i]]=i;
    }
    if (id[1]!=0&&check1()&&check2()){
        printf("%d\n",n-b[n]);
        return 0;
    }
    for (int i=1;i<=id[1];i++) out[i]=1; 
    ans=id[1];
    for (int i=id[1]+1;i<=n;i++) if (b[i]!=0)
        mx=max(mx,(i-id[1])-(b[i]-1));
    printf("%d\n",ans+mx+n);
}