2020牛客多校 第二场 J

题意：

给定序列长度n，已知原排列为${1， 2， 3， 4，... ，n}$，将它置换$k$次后得到序列$a$，问将它置换一次的序列是什么

思路：

首先找出序列中所有轮换，每一个轮换即为一个环，长度记为$len$。显而易见$A$序列为环置换$k % len$次的结果，而最终答案为每个环置换一次的结果，即置换$len * y + 1$次的结果（$y$未知）。

可令一次操作置换$t = k % len$次，设进行$x$次操作会得到最终结果。

由此可推出$ t * x = y * len + 1$，即$ t * x + len * y = 1$，欧几里得对$x$求最小正整数解即可。

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
#include <array>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 1e5 + 10;
const int mod = 998244353;

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y -= (a / b) * x;
    return d;
}

int n;
ll k;
ll a[N], vis[N];
vector<int> v;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out1.txt", "w", stdout);
    // freopen("/home/jungu/code/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    cin >> n >> k;

    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }

    ll len = 0;

    for (int i = 1; i <= n; i++)
    {

        if (!vis[i])
        {
            v.clear();
            int j = i;
            while (!vis[j])
            {
                vis[j] = 1;
                v.push_back(j);
                j = a[j];
            }
            len = (int) v.size();
            ll mid, c;
            ll g = exgcd(k, len, mid, c);
            mid = (mid % len + len) % len;
            for (int j = 0; j < len; j++)
            {
                a[v[j]] = v[(mid + j) % len];
            }
        }
    }

    for (int i = 1; i <= n; i++)
    {
        cout << a[i] << " ";
    }
    cout << endl;

    return 0;
}