BZOJ2976 : [Poi2002]出圈游戏

首先模拟一遍得到n个同余方程，然后用扩展欧几里得求出最小的可行解即可，时间复杂度$O(n^2)$。

 

#include<cstdio>
#define N 30
int 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;
}