solution of this problem in C language

include

include

int cmpfunc(const void *a, const void b) { return ((int*)a - (int)b); }

int lower_bound(int *arr, int n, int val) { int l = 0, r = n; while (l < r) { int mid = (l + r) / 2; if (arr[mid] < val) l = mid + 1; else r = mid; } return l; }

void printMedian(int *arr, int n) { if (n == 0) { printf("Wrong!\n"); return; } if (n % 2 == 1) { printf("%d\n", arr[n/2]); } else { long long sum = (long long)arr[n/2 - 1] + (long long)arr[n/2]; if (sum % 2 == 0) printf("%lld\n", sum/2); else printf("%.1f\n", sum/2.0); } }

void median(int N, char (*s)[3], int *x) { int *arr = malloc(N * sizeof(int)); int size = 0;

for (int i = 0; i < N; i++) {
    if (s[i][0] == 'a') {
        int pos = lower_bound(arr, size, x[i]);
        for (int j = size; j > pos; j--) arr[j] = arr[j-1];
        arr[pos] = x[i];
        size++;
        printMedian(arr, size);

    } else { 
        if (size == 0) {
            printf("Wrong!\n");
            continue;
        }
        int pos = lower_bound(arr, size, x[i]);
        if (pos == size || arr[pos] != x[i]) {
            printf("Wrong!\n");
            continue;
        }
        for (int j = pos; j < size-1; j++) arr[j] = arr[j+1];
        size--;
        printMedian(arr, size);
    }
}
free(arr);

}

int main() { int N; scanf("%d", &N); char s[N][3]; int x[N]; for (int i = 0; i < N; i++) { scanf("%s %d", s[i], &x[i]); } median(N, s, x); return 0; }