无聊的军官

//分析不难发现数据中一定包含多个环，只需求出每个环的最小公倍数即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
long long  i,j,n,m,num[100003];
long long des,ans;
long long minn[100003];
void dfs(long long  x,long long t)
{
if (x==des&&t!=0)
     {
      minn[des]=t;
      return ;
     }
    dfs(num[x],t+1);
}
long long gcd (long long x,long long y)
{
long long r=x%y;
while (y!=0)
{
r=x%y;
x=y;
y=r;
}
return x;
}
int main()
{
freopen("officer.in","r",stdin);
freopen("officer.out","w",stdout);
scanf("%I64d",&n);
for (i=1;i<=n;i++)
scanf("%I64d",&num[i]);
for (i=1;i<=n;i++)//其实没必要每个都搜，搜过的环标记一下就好了，鉴于自己懒的改就这样吧
{
des=i;
dfs(i,0);
}
ans=minn[1];
for (i=2;i<=n;i++)
{
long long k=gcd(ans,minn[i]);
ans=ans*(minn[i]/k);
}
printf("%I64d\n",ans);
}