Codeforces 524C Idempotent functions

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

C. Idempotent functions

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Some time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1, 2, ..., n} is such function , that for any the formula g(g(x)) = g(x) holds.

Let's denote as f(k)(x) the function f applied k times to the value x. More formally, f(1)(x) = f(x), f(k)(x) = f(f(k - 1)(x)) for each k > 1.

You are given some function . Your task is to find minimum positive integer k such that functionf(k)(x) is idempotent.

Input

In the first line of the input there is a single integer n (1 ≤ n ≤ 200) — the size of function f domain.

In the second line follow f(1), f(2), ..., f(n) (1 ≤ f(i) ≤ n for each 1 ≤ i ≤ n), the values of a function.

Output

Output minimum k such that function f(k)(x) is idempotent.

Sample test(s)

input

4
1 2 2 4

output

input

3
2 3 3

output

input

3
2 3 1

output

题目大意：给定一种运算f，对于输入的数组来说，一步操作，f(x) = a[x]，两步操作，f^2(x) = a[a[x]]....倘若每次进行k步操作之后，得到的都是同一个数，即k为符合条件的步数。求出能令所有数都符合条件的最小步数。

初步我们可以很快想到，如果能求出每个数的符合条件的最小步数，然后求出这些步数的最小公倍数即可。但是需要注意的是，对于一开始并没有直接进入循环的情况，即"6"字形的情况，我们要分开求出多余的长度len[i]和循环节长度step[i]，选出最长的MAXlen，ans即为MAXlen对所有step[i]的最小公倍数的上取整。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAXN 500
typedef long long LL;
int n;
int a[MAXN];
LL step[MAXN];
int vis[MAXN], v[MAXN];
int main() {LL ans, maxNum = 0;scanf("%d", &n);for(int i = 1; i <= n; i++)scanf("%d", &a[i]);for(int i = 1; i <= n; i++) {memset(vis, 0, sizeof(vis));stack<int> s;int j;for(j = i; !vis[j]; j = a[j]) {vis[j] = 1;s.push(j);}LL num = 1;while(s.top() != j) {s.pop(); num++;}s.pop();step[i] = num;maxNum = max(maxNum, (LL)s.size());}ans = step[1];for(int i = 1; i <= n; i++) {ans = ans / __gcd(ans, step[i]) * step[i];}LL temp = maxNum / ans;if(maxNum % ans || maxNum == 0) temp++;ans = temp * ans;printf("%I64d\n", ans);return 0;
}

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