time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
As you have noticed, there are lovely girls in Arpa’s land.
People in Arpa’s land are numbered from 1 to n. Everyone has exactly one crush, i-th person’s crush is person with the number crushi.
Someday Arpa shouted Owf loudly from the top of the palace and a funny game started in Arpa’s land. The rules are as follows.
The game consists of rounds. Assume person x wants to start a round, he calls crushx and says: “Oww…wwf” (the letter w is repeated t times) and cuts off the phone immediately. If t > 1 then crushx calls crushcrushx and says: “Oww…wwf” (the letter w is repeated t - 1 times) and cuts off the phone immediately. The round continues until some person receives an “Owf” (t = 1). This person is called the Joon-Joon of the round. There can’t be two rounds at the same time.
Mehrdad has an evil plan to make the game more funny, he wants to find smallest t (t ≥ 1) such that for each person x, if x starts some round and y becomes the Joon-Joon of the round, then by starting from y, x would become the Joon-Joon of the round. Find such t for Mehrdad if it’s possible.
Some strange fact in Arpa’s land is that someone can be himself’s crush (i.e. crushi = i).
Input
The first line of input contains integer n (1 ≤ n ≤ 100) — the number of people in Arpa’s land.
The second line contains n integers, i-th of them is crushi (1 ≤ crushi ≤ n) — the number of i-th person’s crush.
Output
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
Examples
input
4
2 3 1 4
output
3
input
4
4 4 4 4
output
-1
input
4
2 1 4 3
output
1
Note
In the first sample suppose t = 3.
If the first person starts some round:
The first person calls the second person and says “Owwwf”, then the second person calls the third person and says “Owwf”, then the third person calls the first person and says “Owf”, so the first person becomes Joon-Joon of the round. So the condition is satisfied if x is 1.
The process is similar for the second and the third person.
If the fourth person starts some round:
The fourth person calls himself and says “Owwwf”, then he calls himself again and says “Owwf”, then he calls himself for another time and says “Owf”, so the fourth person becomes Joon-Joon of the round. So the condition is satisfied when x is 4.
In the last example if the first person starts a round, then the second person becomes the Joon-Joon, and vice versa.
【题目链接】:http://codeforces.com/contest/742/problem/C
【题解】
先用个while 搞出所有的环的长度;
需要注意的就是；
环的长度如果小于等于2；则t为什么值都是可以的；所以不用管这种环；
然后啊长度大于2的环搞出来就好；
当然对于长度大于2的环还有特殊的情况；
就是这个长度为偶数；
那么这个人可以从最左边到中间，然后再到最右边(然后又回到1);这种情况我一开始漏掉了；真的是蠢死了；
显然这样可以让t更小一点;
所以对于长度大于2的偶数直接除2；
然后对上面需要处理的长度；求他们的最小公倍数就可以了；
每次都差一点，大概就是我的缺陷吧。坚持不下去?
【完整代码】
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
void rel(LL &r)
{
r = 0;
char t = getchar();
while (!isdigit(t) && t!='-') t = getchar();
LL sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void rei(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)&&t!='-') t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
const int MAXN = 100+30;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n;
int a[MAXN];
bool flag[MAXN];
vector <int> g;
map <int,int> dic;
int main()
{
//freopen("F:\rush.txt","r",stdin);
scanf("%d",&n);
rep1(i,1,n)
cin>>a[i];
memset(flag,false,sizeof(flag));
rep1(i,1,n)
if (!flag[i])
{
flag[i] = true;
int s = i,t = a[i],dd = 1;
while (!flag[t])
{
flag[t] = true;
t = a[t];
dd++;
}
if (t!=s)
{
puts("-1");
return 0;
}
if (dd>2)
{
if ((dd&1)==0)
dd/=2;
g.pb(dd);
}
}
LL ans;
int len = g.size();
if (len==0)
ans = 1;
else
{
LL temp = g[0];
rep1(i,1,len-1)
{
LL temp1 = g[i];
LL gg = __gcd(temp,temp1);
temp = (temp*temp1)/gg;
}
ans = temp;
}
cout << ans << endl;
return 0;
}