[Luogu5550] Chino number of columns

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I can not write a qwq

Title Description

Chino given n n number A_1 ... A_N A . 1 . . . A n , given a constant S , m, she will turn these n the number n do k group operations, each operation comprising the following step:

1.swap(as​,am​)(交换a_s,a_m​)

2. 2 to n number of both a forward translation (the first . 1 1 moved to the n-th positions)

Chino want to know, k the set of operations, which n number n are how much?

Orz yky, dyh, wjk, jjy, cxr, gsy, cpy, zcy, tyz, yy, Hz, zhr YG,

Input Format

The first row, four numbers, n-, S, m, K

The next line n number, representing A_1, A_2 ... A_N A . 1 , A 2 . . . A n

Output Format

An output line, n- number, representing A_1, A_2 ... A_N A . 1 , A 2 . . . A n-

Sample input and output

Input # 1copy

4 1 2 3
1 2 3 4

Output # 1copy

1 2 3 4

Description / Tips

All figures are in long long or less

Ideas:

　　Matrix acceleration recursive, configured two matrices, a memory swap operation, a shift operation memory.

　　If the exchange 1 and 2:

　　　　　　0 1 0 0

　　　　　　1 0 0 0

　　　　　　0 0 0 1

　　　　　　0 0 0 1

　　Shift operation, all moved forward one,

　　　　　    0 1 0 0

　　　　　　0 0 1 0

　　　　　　0 0 0 1

　　　　　　1 0 0 0　　　　

Fast power to solve the matrix K power, namely the number of operations:

KSM JZ (JZ A, int B) 
{ 
	JZ ANSS; 
	Memset (anss.c, 0, the sizeof (anss.c)); 
	for (int I =. 1; I <= n-; I ++) anss.c [I] [ I] =. 1; 
	for (; B; B >> =. 1, a = a * a) 
	{ 
		IF (B &. 1) ANSS = a * ANSS; 
// a = a * a; was wrong, because previously calculated 
	} 
	ANSS return; 
}

Overloaded multiplication sign:

struct jz{
	int c[100][100];
}f,base,l1,l2;
int n,m,s,k;
jz operator * (const jz &a,const jz &b)
{
	jz lin;
	for(int i=1;i<=80;i++)
		for(int j=1;j<=80;j++)
		{
			lin.c[i][j]=0;
			for(int k=1;k<=80;k++)
			{
				lin.c[i][j]+=(a.c[k][j] * b.c[i][k]);
			}
		}
	return lin;
}

　Code:

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<cstring>
#define int long long
using namespace std;
struct jz{
	int c[100][100];
}f,base,l1,l2;
int n,m,s,k;
jz operator * (const jz &a,const jz &b)
{
	jz lin;
	for(int i=1;i<=80;i++)
		for(int j=1;j<=80;j++)
		{
			lin.c[i][j]=0;
			for(int k=1;k<=80;k++)
			{
				lin.c[i][j]+=(a.c[k][j] * b.c[i][k]);
			}
		}
	return lin;
}
void dy(jz x)//调试用的，可以忽略
{
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=n;++j)
        cout<<x.c[i][j]<<" ";
        cout<<endl;
    }
}
jz ksm(jz a,int b)
{
	jz anss;
	memset(anss.c,0,sizeof(anss.c));
	for(int i=1;i<=n;i++) anss.c[i][i]=1;
	for(;b;b>>=1,a=a*a)
	{
		if(b&1)anss=a*anss;
//		a=a*a;
	}
	return anss;
}
signed main()
{
	scanf("%lld%lld%lld%lld",&n,&s,&m,&k);
	for(int i=1;i<=n;i++) scanf("%lld",&f.c[i][1]);
	for(int i=1;i<=n;i++) if(i!=s&&i!=m)l1.c[i][i]=1;
	l1.c[s][m]=l1.c[m][s]=1;
	for(int i=1;i<=n-1;i++)l2.c[i][i+1]=1;
	l2.c[n][1]=1;
	base=l1*l2;
	base=ksm(base,k);
	f=f*base;
	for(int i=1;i<=n;i++)printf("%lld ",f.c[i][1]);
	cout<<endl;
	dy(l1);
	cout<<endl;
	dy(l2);
	return 0;
}

　　

Thanks to the help of wlj_dy