BZOJ2976: [Poi2002] out ring game

First, we get the n congruence equation again, then we can find the smallest feasible solution by expanding Euclid, the time Complexity $o (n^2) $.

#include <cstdio> #define N 30int n,i,j,k,x,y,a[n],b[n],d[n],ans;namespace solve{int flag=1,k=1,m=0,d,x,y;int EXGCD (int a,int b,int&x,int&y) {  if (!b) return x=1,y=0,a;  int D=EXGCD (b,a%b,x,y), t=x;  return x=y,y=t-a/b*y,d;} void Add (int a,int r) {  if (!flag) return;  D=EXGCD (k,a,x,y);  if ((r-m)%d) {Flag=0;return;}  X= (x* (r-m)/d+a/d)% (A/D), y=k/d*a,m= ((x*k+m)%y)%y;  if (m<0) m+=y;  K=y;} int ans () {  if (!flag) return 0;  return m?m:k;}} int main () {  scanf ("%d", &n);  for (i=1;i<=n;i++) a[i]=i,scanf ("%d", &x), b[x]=i;  for (y=1,i=n;i>1;i--) {    x=b[n-i+1];    Solve::add (I, ((a[x]-a[y]+1)%i+i)%i);    for (d[x]=1,k=0,j=1;j<=n;j++) if (!d[j]) a[j]=++k;    For (Y=x;d[j];) if ((++y) >n) Y=1;  }  if (Ans=solve::ans ()) printf ("%d", ans), Else puts ("NIE");  return 0;}

　　

BZOJ2976: [Poi2002] out ring game