[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e4 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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[];
 int minv;

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

 int main() {
     //freopen("in.txt", "r", stdin);
     int n;
     cin >> n;
     rep_up1(i, n) {
         sint(f[i]);
     }
     LL ans = ;
     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 ;
 }