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).
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.
If there is no t satisfying the condition, print -1. Otherwise print such smallest t.
4
2 3 1 4
3
4
4 4 4 4
-1
4
2 1 4 3
1
思路：每个点开始，因为x->y,y->x,那么这必定是一个环，所以就先找到环，然后假设环上不同的两点，相距x1，L-x1;
那么从一点开始到达另一点走x1+k1*L,那么另一点就为L - x1 + k2*L ;
这两个要相同，那么有X%L = x1&&X%L = L-x1;所以当L为偶数时最小为X=L/2;否则为X=L;
最后每个环求最小公倍数。
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<math.h> 6 #include<queue> 7 #include<stdlib.h> 8 #include<set> 9 #include<vector> 10 typedef long long LL; 11 using namespace std; 12 int ans[1005]; 13 LL cnt[1000005]; 14 int ma[105]; 15 bool flag[105]; 16 LL gcd(LL n,LL m); 17 int main(void) 18 { 19 int n; 20 scanf("%d",&n); 21 int i,j; 22 for(i = 1; i <= n; i++) 23 { 24 scanf("%d",&ans[i]); 25 } 26 memset(ma,-1,sizeof(ma)); 27 int id = -1; 28 for(i = 1; i <= n; i++) 29 { 30 int t = 0; 31 memset(flag,0,sizeof(flag)); 32 int v = ans[i]; 33 while(!flag[v]) 34 { 35 t++; 36 flag[v] = true; 37 cnt[v] = t; 38 v=ans[v]; 39 if(v == i) 40 {t++;ma[i] = t;break;} 41 } 42 } 43 int fl = 0;LL ak = 1; 44 for(i = 1; i <=n; i++) 45 { 46 if(ma[i]==-1) 47 fl = 1; 48 else 49 { 50 if(ma[i]%2==0) 51 ma[i]/=2; 52 LL ac = gcd(ma[i],ak); 53 ak = ak/ac*ma[i]; 54 } 55 } 56 if(fl)printf("-1 "); 57 else printf("%lld ",ak); 58 return 0; 59 } 60 LL gcd(LL n,LL m) 61 { 62 if(m == 0) 63 return n; 64 else return gcd(m,n%m); 65 }