luogu P5151 HKE与他的小朋友

嘟嘟嘟


看到(i)变成了(A_i)，我突然想起了置换这个东西。于是马上到网上学了一遍轮换乘法。
手模后发现轮换乘法满足结合律，但不满足交换律。
于是就可以快速幂啦。
需要注意的是每一次相乘是(O(n))的，因此总复杂度为(O(n log n))。
代码一看就懂

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<map>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, k, a[maxn];

int tp[maxn], ret[maxn];
In void mul(int* ret, int* a)
{
  for(int i = 1; i <= n; ++i) tp[i] = ret[a[i]];
  for(int i = 1; i <= n; ++i) ret[i] = tp[i];
}
In void quickpow(int* a, int b)
{
  for(int i = 1; i <= n; ++i) ret[i] = i;
  for(; b; b >>= 1, mul(a, a))
    if(b & 1) mul(ret, a);
}

int main()
{
  n = read(); k = read();
  for(int i = 1; i <= n; ++i) a[i] = read();
  quickpow(a, k);
  for(int i = 1; i <= n; ++i) tp[ret[i]] = i;
  for(int i = 1; i <= n; ++i) write(tp[i]), space; enter;
  return 0;
}