III способ

        int k = 0, m = n-r-1;
        for (int i = 0; i < (m - k + 1) / 2; i++) {
            int t = a[k + i];
            a[k + i] = a[m - i];
            a[m - i] = t;
        }

        k = n-r; m = n-1;
        for (int i = 0; i < (m - k + 1) / 2; i++) {
            int t = a[k + i];
            a[k + i] = a[m - i];
            a[m - i] = t;
        }

        k = 0; m = n-1;
        for (int i = 0; i < (m - k + 1) / 2; i++) {
            int t = a[k + i];
            a[k + i] = a[m - i];
            a[m - i] = t;
        }