兩個矩陣，排成線性序列，若逆序對奇偶性相同，則可以互相轉化矩陣

註意：0不算入內，

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

typedef long long ll;

const ll nil=20000000000;
const int maxn=500000+10;
ll L[maxn/2],R[maxn/2];
int n;
ll a[maxn];
int top;

ll megersort(int l,int r,int mid,ll a[]){
   int n1=mid-l;
   
 
int n2=r-mid;
   ll cnt=0;
   int i,j;
   for (i=0;i<n1;i++) L[i]=a[l+i];
   for (j=0;j<n2;j++) R[j]=a[mid+j];
   L[n1]=nil;
   R[n2]=nil;
   i=j=0;
   for (int k=l;k<r;k++){
        if(L[i]<=R[j]){
            a[k]=L[i++];
        }
        else {
            a[k]=R[j++];
            cnt=((cnt%2
 
)+(mid+j-k-1)%2)%2;
        }
   }
  return cnt%2;
}

ll meger(int l,int r,ll a[]){
   if(l+1<r){
    int mid=(l+r)>>1;
    ll v1=meger(l,mid,a);
    ll v2=meger(mid,r,a);
    ll v3=megersort(l,r,mid,a);
    return (v1%2+v2%2+v3%2)%2;
   }
   else return 0;
}

int main(){
    while(scanf("%d
 
",&n)!=EOF){
    memset(R,0,sizeof(R));
    top=-1;
    int x;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
          scanf("%d",&x);
          if(x!=0) a[++top]=x;
        }
    }
    ll ans1=meger(0,top+1,a);
    memset(a,0,sizeof(a));
    memset(L,0,sizeof(L));
    memset(R,0,sizeof(R));
    top=-1;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++){
          scanf("%d",&x);
          if(x!=0) a[++top]=x;
        }
    }
    ll ans2=meger(0,top+1,a);
    if(ans1%2==ans2%2) printf("TAK\n");
    else printf("NIE\n");
    //printf("%lld %lld\n",ans1,ans2);
    }
return 0;
}

奇數碼問題（逆序對求解）