Acwing 2074. Countdown simulation

Title Description

https://www.acwing.com/problem/content/description/2076/
The general meaning of the title is $T$ Group sequence , The sequence length of each group is different , The length will be given each time , Give a number at the same time $K$, Let you find out how many consecutive $[K,k-1,k-2,...,1]$.
The title is not difficult. , Direct simulation is fine , Let's look at my code ideas .

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Code

#include<iostream>
using namespace std;
const int N=2e5+10;
int num[N];
int t,n,k;
int main()
{
    
    scanf("%d",&t);
    int cnt=1;  // Record group number , from 1 Start 
    while(t--)
    {
    
        scanf("%d%d",&n,&k);
        for(int i=0;i<n;i++) scanf("%d",&num[i]);
        
        int count=0,i,j;
        for(i=0;i<n;)
        {
    
            if(num[i]==k)   // Find a number in the sequence equal to k
            {
    
                if(n-i<k) break;// At this time, if the remaining sequence length is less than k, There is no need to look for it directly , There will be no more qualified 
                // Judge whether it is countdown 
                bool flag=true;
                for(j=i;j<=i+k-1;j++)  
                    if(num[j]!=k-(j-i))
                    {
    
                        flag=false;
                        break;
                    }
                if(flag)  // It's a countdown 
                {
    
                    count++;
                    i=i+k-1+1;   // hold i Update to the next countdown 
                }
                else   // Not counting down 
                    i=j;   // hold i Update to the first non conformance 
            }
            else
                i++;
        }
        printf("Case #%d: %d\n",cnt++,count);
    }
    return 0;
}

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