[Luogu P5550] Chino's sequence [matrix multiplication]

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Problem solving ideas

Matrix multiplication

Consider constructing an operation matrix CC $\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]$

First, deal with the special cases s and m, if $s=1，m=2$，
A=$\left[\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$

When operating numbers, only forward operations

A=$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& 1& 1\end{array}\right]$

Then the problem is converted to $CC\ast A^k$

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Code

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
const long long INF=1000000007;
long long n,s,m,k;
struct c{
    
    
	long long n,m;
	long long a[100][100];
}A,B,CC;
c operator *(c A,c B){
    
    
	c C;
	C.n=A.n,C.m=B.m;
	for(int i=1;i<=C.n;i++)
		for(int j=1;j<=C.m;j++)
			C.a[i][j]=0;
	for(int k=1;k<=A.m;k++)
		for(int i=1;i<=C.n;i++)
			for(int j=1;j<=C.m;j++)
				C.a[i][j]=(C.a[i][j]+(A.a[i][k]*B.a[k][j])%INF)%INF;
	return C;
}
void poww(long long x){
    
    
	if(x==1)
	{
    
    
		B=A;
		return;
	}
	poww(x/2);
	B=B*B;
	if(x&1)
		B=B*A; 
}
int main(){
    
    
	scanf("%lld%lld%lld%lld",&n,&s,&m,&k);
	CC.n=1,CC.m=n;
	A.n=n,A.m=n;
	for(int i=1;i<=n;i++)
	{
    
    
		scanf("%lld",&CC.a[1][i]);
		if(i!=s&&i!=m)
		{
    
    
			if(i-1>0)
				A.a[i][i-1]=1;
			else 
				A.a[i][n]=1;
		}
	}		
	if(m-1>0) A.a[s][m-1]=1;
	else A.a[s][n]=1;
	if(s-1>0) A.a[m][s-1]=1;
	else A.a[m][n]=1;
	poww(k);
	CC=CC*B;
	for(int i=1;i<=n;i++)
		printf("%lld ",CC.a[1][i]);
}
/*
4 1 2 3
1 2 3 4
*/