Codeforces 771-div2 C (trouble, permutation is not very good)

subject
The question : Given a length of n Of permutation, It is stipulated that for the two tuples with reverse order relationship (i,j), One side . How many connected blocks are there in the graph .（n<=2e5）
Ideas : Used a tree array + Union checking set , also T 了 , Delete the tree array and search the set .
  The solution only maintains a maximum , Finish my practice . Let's think about how to form a connected block , There will be no future in this position < Number of maximum values , As a connecting block, I can't claim an edge from the next connecting block , It's one of its own .
  say concretely , For the first i A place , If the maximum value of maintenance is just ==i, There is no way to extend the border , Number of connected blocks ++.
Time complexity : O(n)
Code :

#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;
void solve()
{
    
   read(n);
   int ans = 0;
   int mx = 0;
   for(int x,i=1;i<=n;++i)
   {
    
   	   read(x); 
   	   mx = max(mx,x);
   	   ans += mx==i;
   }
   write(ans); puts("");
}
signed main(void)
{
      
   // T = 1;
   // OldTomato; cin>>T;
   read(T);
   while(T--)
   {
    
   	 solve();
   }
   return 0;
}