poj 2369 Permutations 置换

题目链接

给一个数列， 求这个数列置换成1， 2， 3....n需要多少次。

就是里面所有小的置换的长度的lcm。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
int a[], vis[];
int gcd(int x, int y) {
    return y?gcd(y, x%y):x;
}
int lcm(int x, int y) {
    return x/gcd(x, y)*y;
}
int main()
{
    int n;
    while(cin>>n) {
        for(int i = ; i<n; i++){
            scanf("%d", &a[i]);
            a[i]--;
        }
        int ans = ;
        for(int i = ; i<n; i++) {
            if(!vis[i]) {
                int tmp = i, len = ;
                while(!vis[tmp]) {
                    vis[tmp] = ;
                    len++;
                    tmp = a[tmp];
                }
                ans = lcm(ans, len);
            }
        }
        cout<<ans<<endl;
    }
    return ;
}