MXN Puzzle-Reverse order of "n*m the accessibility of digital problems"

MXN Puzzlesol:

\ (n*m\) the feasibility of a certain situation to reach another situation
It is possible to determine the parity of the number by reverse order.
Let's write the number of this \ (n*m-1\) in a column.

• Considering the odd behavior, the odd-numbered case is the same as the number of odd parity
  The space movement will not change the number of reverse order, and the upper and lower movement is equivalent to change the number of total (m-1\) in reverse order number of the column, because \ (m-1\) is even, so the number of reverse before and after the odd parity is unchanged.
• Consider the behavior of odd, listed as an even number of cases (reverse order number + before and after the status of space lines of the difference) parity of the same time can reach
  Moving up and down will change the number of inverse pairs (m-1\) number, and \ (m-1\) is odd, when the top and bottom moved the even line, the parity is not changed, the odd line is moved up and down, parity changes.

• If the behavior is even, ibid.

  AC CODE:

Source code#include<cstdio> #include <cstring> #include <algorithm>using namespace Std;int read () {    int X=0,f=1;char ch= '; while (ch> ' 9 ' | |    ch< ' 0 ') {if (ch== '-') F=-1;ch=getchar ();}    while (ch<= ' 9 ' &&ch>= ' 0 ') {x= (x<<3) + (x<<1) + (ch^ ' 0 '); Ch=getchar ();} return x*f;} const int N = + 2;int n,m;int a[n*n],tot;int tr[n*n];int lowbit (int x) {return x&-x;}    void Modify (int p,int k) {for (int i=p;i<=tot;i+=lowbit (i)) {tr[i]+=k;    }}int query (int p) {int ans=0;    for (int i=p;i;i-=lowbit (i)) {ans+=tr[i]; } return ans;    int main () {//Freopen ("Data.in", "R", stdin);//Freopen ("Sol.out", "w", stdout);        while (1) {n=read (); M=read ();        if (!n&&!m) break;        memset (tr,0,sizeof (tr));        Tot=0;int POS;                for (int i=1;i<=n;i++) {for (int j=1;j<=m;j++) {int x;x=read ();                if (x) a[++tot]=x;            else pos=i; }} int muL=n*m;        int ans=0;            for (int i=tot;i;i--) {ans+=query (a[i]-1);        Modify (a[i],1);        } if (m%2==0) {ans+=n-pos;        } if (ans%2==0) printf ("yes\n");    else printf ("no\n"); } return 0;}

MXN Puzzle-Reverse order the accessibility of the N*M digital problem scenario