[codeforces-542-C]YY？

链接：http://codeforces.com/problemset/problem/542/C

题意：对一个函数f(x)，定义域[1,n], 令f(k,x) = f(f(f(f...f(x))))(共迭代k次)。求最小的k，使得f(k, x) 满足 g(g(x)) = g(x)的性质，也就是f(k,f(k,x)) = f(k,x)(x属于[1,n])。

思路：方法虽然简单，但比较巧妙，值得一做。由于f(k,x)只与f的迭代次数有关，而f数组一开始是给定的，所以不妨对所有的x属于[1,n]单独进行处理。考虑f(1,x),f(2,x),f(3,x)...,f(k,x)组成的序列，因为f(k,f(k,x)) = f(2k, x)，所以转化为求序列中第k位置的数Ak，使得Ak = A2k。由于序列的后一项是前一项进行一次f()操作得到的，所以必存在循环节，且循环节长度小于等于n，不妨令序列从P0位置开始循环，循环节长度为K0，则这种情况下的k的取值集合就是{sK0, (s+1)K0, (s+2)K0, ...}，sK0为大于等于P0的最小数。然后对x取遍[1,n]，求k的取值集合的交集元素的最小值便是答案，但显然不能这样求。由于对自变量的每个值，k的取值总是那种情况下的循环节K0的倍数，不妨先对所有的K0求一下最小公倍数，然后再调整结果使得它满足大于等于所有的P0，调整的时候只需每次加上所有K0的最小公倍数，直到满足条件。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 3e4 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

int f[207];
int minv;

int find(int x) {
    int used[207], c = 0;
    mem0(used);
    while (!used[f[x]]) {
        c ++;
        used[x = f[x]] = c;
    }
    max_update(minv, used[f[x]]);
    return c - used[f[x]] + 1;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int n;
    cin >> n;
    rep_up1(i, n) {
        sint(f[i]);
    }
    LL ans = 1;
    rep_up1(i, n) {
        LL tmp = find(i);
        ans = ans / gcd(ans, tmp) * tmp;
    }
    LL tmp = ans;
    while (ans < minv) ans += tmp;
    cout << ans << endl;
    return 0;
}