Help me in solving AOCP10 problem

help
,
1

My issue

My code

// Update the code below to solve the problem

#include <stdio.h>

int main() {
    int t,count=8;
    scanf("%d", &t);
    
    for (int i = 0; i < t; i++) {
        int n;
        scanf("%d", &n);
        int A[n];
        for (int j = 0; j < n; j++) {
            scanf("%d", &A[j]);
        }
        
        for(int j=0; j<n;j++){
            if(A[j]==6 || A[j]==7 || A[j]==13 || A[j]==14 || A[j]==20 || A[j]==21 || A[j]==27 || A[j]==28)
            continue;  
            
            else
            count++;
        }
    
        printf("%d\n",count);
    }
    return 0;
    }

Learning course: C for problem solving - 2
Problem Link: CodeChef: Practical coding for everyone

2

Your code is incorrect because it does not reset the count variable to 8 after each iteration, for each case, there will be 8 holidays (due to sundays and saturdays) but your code doesn’t take this into account, so for next case its taking the already incremented count value to be the new value rather than starting from 8.

#include <stdio.h>

int main() {
    int t;
    scanf("%d", &t);

    while (t--) {
        int count = 8; // Reset count for each test case
        int n;
        scanf("%d", &n);
        int A[n];
        for (int j = 0; j < n; j++) {
            scanf("%d", &A[j]);
        }

        for(int j = 0; j < n; j++){
            if(A[j] == 6 || A[j] == 7 || A[j] == 13 || A[j] == 14 || A[j] == 20 || A[j] == 21 || A[j] == 27 || A[j] == 28)
                continue;
            else
                count++;
        }

        printf("%d\n", count);
    }
    return 0;
}