题解 [AGC031D] A Sequence of Permutations

传送门

自闭了啥也不会

• 置换以及类似的操作是有逆变换的，写成 \(p^{-1}\) 的形式很适合找规律（

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#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, k;
int p[N], q[N], t[N];

namespace force{
	void solve() {
		// cout<<"a 1: "; for (int i=1; i<=n; ++i) cout<<setw(2)<<p[i]<<' '; cout<<endl;
		// cout<<"a 2: "; for (int i=1; i<=n; ++i) cout<<setw(2)<<q[i]<<' '; cout<<endl;
		for (int i=3; i<=k; ++i) {
			for (int j=1; j<=n; ++j) t[p[j]]=q[j];
			// cout<<"a"<<setw(2)<<i<<": "; for (int j=1; j<=n; ++j) cout<<setw(2)<<t[j]<<' '; cout<<endl;
			for (int j=1; j<=n; ++j) p[j]=q[j];
			for (int j=1; j<=n; ++j) q[j]=t[j];
		}
		for (int i=1; i<=n; ++i) printf("%d ", q[i]);
		printf("\n");
	}
}

namespace task{
	struct perm{
		int p[N];
		inline int& operator [] (int t) {return p[t];}
		inline perm operator * (perm b) {
			perm ans;
			for (int i=1; i<=n; ++i) ans[i]=p[b[i]];
			return ans;
		}
		inline perm qinv() {
			perm ans;
			for (int i=1; i<=n; ++i) ans[p[i]]=i;
			return ans;
		}
	}p, q, L, invL, I, ans;
	inline perm qpow(perm a, ll b) {perm ans=I; for (; b; a=a*a,b>>=1) if (b&1) ans=ans*a; return ans;}
	void solve() {
		--k;
		for (int i=1; i<=n; ++i) I[i]=i;
		for (int i=1; i<=n; ++i) p[i]=::p[i];
		for (int i=1; i<=n; ++i) q[i]=::q[i];
		L=q*p.qinv()*q.qinv()*p;
		invL=p.qinv()*q*p*q.qinv();
		if (k%6==0) ans=qpow(L, k/6)*p*qpow(invL, k/6);
		else if (k%6==1) ans=qpow(L, k/6)*q*qpow(invL, k/6);
		else if (k%6==2) ans=qpow(L, k/6)*q*p.qinv()*qpow(invL, k/6);
		else if (k%6==3) ans=qpow(L, k/6)*q*p.qinv()*q.qinv()*qpow(invL, k/6);
		else if (k%6==4) ans=qpow(L, k/6)*q*p.qinv()*q.qinv()*p*q.qinv()*qpow(invL, k/6);
		else ans=qpow(L, k/6)*q*p.qinv()*q.qinv()*p*p*q.qinv()*qpow(invL, k/6);
		for (int i=1; i<=n; ++i) printf("%d%c", ans[i], " \n"[i==n]);
	}
}

signed main()
{
	freopen("b.in", "r", stdin);
	freopen("b.out", "w", stdout);

	n=read(); k=read();
	for (int i=1; i<=n; ++i) p[i]=read();
	for (int i=1; i<=n; ++i) q[i]=read();
	// force::solve();
	task::solve();

	return 0;
}