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。
1 #pragma GCC optimize(3) 2 #pragma GCC optimize(2) 3 #include <map> 4 #include <set> 5 #include <array> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <cstdio> 10 #include <cstring> 11 #include <sstream> 12 #include <iostream> 13 #include <stdlib.h> 14 #include <algorithm> 15 #include <unordered_map> 16 17 using namespace std; 18 19 typedef long long ll; 20 typedef pair<int, int> PII; 21 22 #define Time (double)clock() / CLOCKS_PER_SEC 23 24 #define sd(a) scanf("%d", &a) 25 #define sdd(a, b) scanf("%d%d", &a, &b) 26 #define slld(a) scanf("%lld", &a) 27 #define slldd(a, b) scanf("%lld%lld", &a, &b) 28 29 const int N = 1e5 + 10; 30 const int mod = 998244353; 31 32 ll gcd(ll a, ll b) 33 { 34 return b ? gcd(b, a % b) : a; 35 } 36 37 ll exgcd(ll a, ll b, ll &x, ll &y) 38 { 39 if (!b) 40 { 41 x = 1; 42 y = 0; 43 return a; 44 } 45 int d = exgcd(b, a % b, y, x); 46 y -= (a / b) * x; 47 return d; 48 } 49 50 int n; 51 ll k; 52 ll a[N], vis[N]; 53 vector<int> v; 54 55 int main() 56 { 57 #ifdef ONLINE_JUDGE 58 #else 59 freopen("/home/jungu/code/in.txt", "r", stdin); 60 // freopen("/home/jungu/code/out1.txt", "w", stdout); 61 // freopen("/home/jungu/code/out.txt","w",stdout); 62 #endif 63 ios::sync_with_stdio(false); 64 cin.tie(0), cout.tie(0); 65 66 cin >> n >> k; 67 68 for (int i = 1; i <= n; i++) 69 { 70 cin >> a[i]; 71 } 72 73 ll len = 0; 74 75 for (int i = 1; i <= n; i++) 76 { 77 78 if (!vis[i]) 79 { 80 v.clear(); 81 int j = i; 82 while (!vis[j]) 83 { 84 vis[j] = 1; 85 v.push_back(j); 86 j = a[j]; 87 } 88 len = (int) v.size(); 89 ll mid, c; 90 ll g = exgcd(k, len, mid, c); 91 mid = (mid % len + len) % len; 92 for (int j = 0; j < len; j++) 93 { 94 a[v[j]] = v[(mid + j) % len]; 95 } 96 } 97 } 98 99 for (int i = 1; i <= n; i++) 100 { 101 cout << a[i] << " "; 102 } 103 cout << endl; 104 105 return 0; 106 }