Codeforces 771 div2 B (no one FST, refers to himself)

subject
The question : Given length is n Array of a, It is allowed to exchange two adjacent and odd numbers any time , Determine whether the array can be made non descending .
Ideas : Violent simulation , I feel a little silly indeed ,2222211111, achieve n Fang . sure T, But I didn't expect the positive solution at that time , Buried .
As long as the parity is different, it can be exchanged , We don't need to care about different parity , It can be changed anyway .
How to judge no solution ? Two numbers with the same parity , And there is an inverse relationship , Direct mail .
Therefore, the current last odd number can be maintained dynamically 、 Even values , If there is an inverse relationship , unsolvable . Otherwise , Through several exchanges, the preceding numbers can be arranged in order .
Time complexity : O(n)
Code :

// Problem: B. Odd Swap Sort
// Contest: Codeforces - Codeforces Round #771 (Div. 2)
// URL: https://codeforces.com/contest/1638/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round()  rounding  ceil()  Rounding up  floor()  Rounding down 
// lower_bound(a.begin(),a.end(),tmp,greater<ll>())  First less than or equal to 
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){
     return x&(-x);}
template<typename T>void write(T x)
{
    
    if(x<0)
    {
    
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
    
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){
    if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){
    x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
int last[2];
void solve()
{
    
    read(n);
    bool flag = true;
    last[0] = last[1] = 0;
    for(int x,i=1;i<=n;++i)
    {
    
    	read(x);
    	if(x < last[x&1]) flag = false;
    	last[x&1] = x; 	
    }
    if(flag) puts("Yes");
    else puts("No");
}
signed main(void)
{
      
   // T = 1;
   // OldTomato; cin>>T;
   read(T);
   while(T--)
   {
    
   	 solve();
   }
   return 0;
}