C. Nauuo and Cards
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nauuo is a girl who loves playing cards.
One day she was playing cards but found that the cards were mixed with some empty ones.
There are n cards numbered from 1 to n, and they were mixed with another n empty cards. She piled up the 2n cards and drew n of them. The n cards in Nauuo’s hands are given. The remaining n cards in the pile are also given in the order from top to bottom.
In one operation she can choose a card in her hands and play it — put it at the bottom of the pile, then draw the top card from the pile.
Nauuo wants to make the n numbered cards piled up in increasing order (the i-th card in the pile from top to bottom is the card i) as quickly as possible. Can you tell her the minimum number of operations?
Input
The first line contains a single integer n (1≤n≤2⋅105) — the number of numbered cards.
The second line contains n integers a1,a2,…,an (0≤ai≤n) — the initial cards in Nauuo’s hands. 0 represents an empty card.
The third line contains n integers b1,b2,…,bn (0≤bi≤n) — the initial cards in the pile, given in order from top to bottom. 0 represents an empty card.
It is guaranteed that each number from 1 to n appears exactly once, either in a1…n or b1…n.
Output
The output contains a single integer — the minimum number of operations to make the n numbered cards piled up in increasing order.
Examples
inputCopy
3
0 2 0
3 0 1
output
2
input
3
0 2 0
1 0 3
output
4
input
11
0 0 0 5 0 0 0 4 0 0 11
9 2 6 0 8 1 7 0 3 0 10
output
18
Note
Example 1
We can play the card 2 and draw the card 3 in the first operation. After that, we have [0,3,0] in hands and the cards in the pile are [0,1,2] from top to bottom.
Then, we play the card 3 in the second operation. The cards in the pile are [1,2,3], in which the cards are piled up in increasing order.
Example 2
Play an empty card and draw the card 1, then play 1, 2, 3 in order.
思路：
两种情况：
一：需要取牌的情况。
找到值与值的下标加一（从0 开始）的差的最大的值，也就是取走值最小，且下标最大的情况，是要取走的，但不一定要直接直接取到下标的位置，取到：下标+1-（值-1），比如，2在下标3的位置，那么取到下标2的位置就行，在将1放到最后面时，2 可以直接出去。总次 : 数就是牌的总数加值与值的下标加一（从0 开始）的差的最大的值。
二：不需要取牌的情况。
也就是从1的位置到最后一个值是升序并1+(n-1-下标)=最后一个值，判断1位置之前存不存在值与值的下标加一（从0 开始）的差是否有大于1的存在，存在输出n-最后一个值，不存在就输出第一种情况的值。
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define ll long long
#define mes(x,y) memset(x,y,sizeof(x))
using namespace std;
int main(){
std::ios::sync_with_stdio(false);
ll n,x,y,i,j,sum,flag,flag1=-1,flag2=-1,maxn1,maxn2;vector<ll>v;
while(cin>>n){
v.clear();sum=n;flag1=0;flag2=-1;flag=-1;maxn1=0;maxn2=0;
for(i=0;i<n;i++){
cin>>x;if(x!=0)flag1=1;
}
for(j=0;j<n;j++){
cin>>y;v.push_back(y);
if(y==1)flag=flag2=j;
}
for(j=0;j<n;j++){
if(v[j]<=i+1&&v[j]!=0){
y=j+1-(v[j]-1);
maxn1=max(maxn1,y);//找到值与值的下标加一（从0 开始）的差的最大的值，
if(j<flag2){
y=j-v[j]+3+v[n-1];
maxn2=max(maxn2,y);
}//判断1位置之前存不存在值与值的下标加一（从0 开始）的差是否有大于1的存在
}
}
while(v[flag2]+1==v[flag2+1]&&(flag2<n-1)){
flag2++;
}//从1的位置到最后一个值是升序
if(flag2==n-1){
if(flag==0){
cout<<"0"<<endl;continue;
} //存在手里没牌，但下面的牌已经排好的情况
if(flag1==1&&maxn2<=1){
cout<<n-v[n-1]<<endl;continue;
} //1+(n-1-下标)=最后一个值，判断1位置之前存不存在值与值的下标加一（从0 开始）的差是否有大于1的存在
}
cout<<sum+maxn1<<endl;//就输出第一种情况的值。
}
}
