So, we're dealing with participants applying for either Fizzbuzz or Blind C. Let's represent the set of participants for Blind C as $$$S1$$$ and for Fizzbuzz as $$$S2$$$. Now, the participants who applied for either event form a set, let's call it $$$S$$$. The question boils down to finding the number of participants who applied for both events.

In the world of sets, there's this handy formula:

$$$[ N(A \cup B) = N(A) + N(B) - N(A \cap B) ]$$$

In our scenario, $$$N(A)$$$ is the number of participants for Blind C, $$$N(B)$$$ is for Fizzbuzz, and $$$N(A \cup B)$$$ is the total participants in both events. So, our answer would be the sum of participants in $$$S1$$$ and $$$S2$$$, minus the total participants in both sets. Clear as mud?

#include <stdio.h>

int main() {
    
     // Total participants in both events (N(A ∪ B))
    int totalParticipantsBothEvents;
    scanf("%d", &totalParticipantsBothEvents);

    // Number of participants for Fizzbuzz (N(B))
    int participantsFizzbuzz;
    scanf("%d", &participantsFizzbuzz);
    
    // Number of participants for Blind C (N(A))
    int participantsBlindC;
    scanf("%d", &participantsBlindC);

    // Calculate the number of participants who applied for both events (N(A ∩ B))
    int participantsBothEvents = participantsBlindC + participantsFizzbuzz - totalParticipantsBothEvents;

    printf("%d\n", participantsBothEvents);

    return 0;
}

The task is to determine the minimum rotation needed to align a plane on the runway. It boils down to finding the minimum angle between the plane and the $$$x$$$-axis. To calculate this angle, take the modulo of the angle by $$$360^\circ$$$ since the angle repeats its value after a full circle.

Now, adjust the angle based on the quadrant.

That's the key to finding the answer. Easy, right?

#include <stdio.h>

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

    // Take modulo to handle angle repetition after 360°
    angle %= 360;

    // Adjust the angle based on the quadrant
    int minimumRotation;
    if (angle >= 0 && angle < 90) {
        minimumRotation = angle;
    } else if (angle >= 90 && angle < 180) {
        minimumRotation = 180 - angle;
    } else if (angle >= 180 && angle < 270) {
        minimumRotation = angle - 180;
    } else {
        minimumRotation = 360 - angle;
    }

    printf("%d\n", minimumRotation);

    return 0;
}

Solution

#include <stdio.h>

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

    int arr[n];
    int cnt_odd = 0;

    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);

        if (arr[i] % 2 == 1) {
            cnt_odd++;
        }
    }

    if (cnt_odd % 2 == 0) {
        printf("YES\n");
    } else {
        printf("NO\n");
    }

    return 0;
}

Solution

#include <stdio.h>

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

    int v[32];
    for (int i = 0; i < 32; i++) {
        v[i] = (x >> i) & 1;
    }

    int a = 0, b = 0;
    int flag = 0;

    for (int i = 0; i < 32; i++) {
        if (v[i] == 1) {
            if (!flag) {
                b = 1 << i;
                flag = 1;
            } else {
                a += 1 << i;
            }
        }
    }

    if (a == 0 || b == 0) {
        printf("-1\n");
    } else {
        printf("%d %d\n", a, b);
    }

    return 0;
}

Solution

#include <stdio.h>
#include <limits.h>

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int max(int a, int b){
    if(a>b){
        return a;
    }
    
    else{
        return b;
    }
}

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

    int arr[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }

    int pre[n + 1];
    int suf[n + 1];
    
    for(int i=0; i<=n; i++){
        pre[i]=0;
        suf[i]=0;
    }

    for (int i = 1; i < n + 1; i++) {
        pre[i] = gcd(pre[i - 1], arr[i - 1]);
    }
    for (int i = n - 1; i >= 0; i--) {
        suf[i] = gcd(suf[i + 1], arr[i]);
    }

    int ans = INT_MAX;
    for (int i = 1; i < n; i++) {
        if (pre[i] == suf[i]) {
            ans = (ans < (max(i, n - (i + 1)))) ? ans : (max(i, n - (i + 1)));
        }
    }

    if (ans == INT_MAX) {
        printf("-1\n");
        return 0;
    }

    printf("%d\n", ans + 1);

    return 0;
}