Va 716-Commedia dell & amp; #39; arte (replacement)

Va 716-Commedia dell & #39; arte (replacement)

Link to the question: Ultraviolet A 716-Commedia dell 'arte

Given a three-dimensional octal digital, 0 represents an empty position and asks if it can be sorted back.

Solution: for the case where n is an odd number, consider the reverse logarithm of the three-dimensional octal digital to consider the case where the State goes down, except 0, and the even number, consider transferring the position of 0 to (n, the reverse logarithm of the corresponding sequence after the n, n) position. If the reverse logarithm is an even number, the odd number is not allowed.

#include 
  
   #include 
   
    #include using namespace std;typedef long long ll;const int maxn = 1e6+5;int n, arr[maxn], t[maxn];int x, y, z;ll merge_sort (int l, int r, int* a, int* b) {    if (l == r)        return 0;    ll ret = 0;    int mid = (r + l) / 2;    int p = l, q = mid+1, mv = l;    ret = merge_sort(l, mid, a, b) + merge_sort(mid + 1, r, a, b);    while (p <= mid || q <= r) {        if (q > r || (p <= mid && a[p] < a[q]))            b[mv++] = a[p++];        else {            ret += mid - p + 1;            b[mv++] = a[q++];        }    }    for (int i = l; i <= r; i++)        a[i] = b[i];    return ret;}bool judge () {    if (n&1) {        ll ret = merge_sort(0, n*n*n-1, arr, t) - (z * n * n + x * n + y - 1);        return (ret&1) == 0;    }    while (z != n - 1) {        int p = z * n * n + x * n + y;        z++;        int q = z * n * n + x * n + y;        swap(arr[p], arr[q]);    }    while (x != n - 1) {        int p = z * n * n + x * n + y;        x++;        int q = z * n * n + x * n + y;        swap(arr[p], arr[q]);    }    while (y != n - 1) {        int p = z * n * n + x * n + y;        y++;        int q = z * n * n + x * n + y;        swap(arr[p], arr[q]);    }    ll ret = merge_sort(0, n*n*n-2, arr, t);    return (ret&1) == 0;}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &n);        for (int k = 0; k < n; k++) {            int Z = k * n * n;            for (int i = 0; i < n; i++) {                int X = i * n;                for (int j = 0; j < n; j++) {                    int tmp = Z + X + j;                    scanf("%d", &arr[tmp]);                    if (arr[tmp] == 0) {                        x = i; y = j; z = k;                    }                }            }        }        printf("%s\n", judge() ? "Puzzle can be solved." : "Puzzle is unsolvable.");    }    return 0;}