CF385E 翻

[[math]] [[matrices]] [[fast_exponentiation]]
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Consider ignoring the modulo, then calculate at the same time.

\[\begin{aligned} dx&=x+y+dx+t\\ dy&=x+y+dy+t\\ x&=dx+2x+y+t\\ y&=dy+x+2y+t\\ t&=t+1 \end{aligned} \]

Matrix fast exponentiation

\[\begin{aligned} dx=1*dx+0*dy+1*x+1*y+1*t+0*1\\ dy=0*dx+1*dy+1*x+1*y+1*t+0*1\\ x=1*dx+0*dy+2*x+1*y+1*t+0*1\\ y=0*dx+1*dy+1*x+2*y+1*t+0*1\\ t=0*dx+0*dy+0*x+0*y+1*t+1*1\\ 1=0*dx+0*dy+0*x+0*y+0*t+1*1\end{aligned} \]

We can use matrices to describe it.

\[\left[\begin{array}{} dx\\dy\\x\\y\\t\\1 \end{array} \right] \left[\begin{array}{} 1&0&1&0&0&0\\ 0&1&0&1&0&0\\ 1&1&2&1&0&0\\ 1&1&1&2&0&0\\ 1&1&1&1&1&0\\ 0&0&0&0&1&1 \end{array} \right]=\left[\begin{array}{} x+y+dx+t\\x+y+dy+t\\dx+2x+y+t\\dy+x+2y+t\\t+1\\1 \end{array} \right] \]

Notice the 1.

Then we can use matrix fast exponentiation.

It consists of fast exponentiation, multiplication and square.

Multiplication is a multiplication of matrices of (n,n) and (1,n), whose answer is a matrix of (n,1)

Square is a multiplication of two matrices of (n,n), whose answer is also a matrix of (n,n)

Fast exponentiation is the same of the fast exponentiation of numbers.

Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,x,y,dx,dy,t,ret;
ll _[6][6]={
{1,0,1,0,0,0},
{0,1,0,1,0,0},
{1,1,2,1,0,0},
{1,1,1,2,0,0},
{1,1,1,1,1,0},
{0,0,0,0,1,1}
},o[6][6],a[6],b[6];
void mult(){
	int i,j;
	for(i=0;i<6;i++)
		for(b[i]=0,j=0;j<6;j++)(b[i]+=_[j][i]*a[j])%=n;
	for(i=0;i<6;i++)a[i]=b[i];
}
void square(){
	int i,j,k;
	for(i=0;i<6;i++)
		for(j=0;j<6;j++)
			for(o[i][j]=0,k=0;k<6;k++)
				(o[i][j]+=_[i][k]*_[k][j])%=n;
	for(i=0;i<6;i++)
		for(j=0;j<6;j++)_[i][j]=o[i][j];
}
void fpw(ll a){
	for(;a;a>>=1,square())if(a&1)mult();
}
int main(){
	scanf("%lld%lld%lld%lld%lld%lld",&n,&x,&y,&dx,&dy,&t);
	if(t==0)printf("%lld %lld",x,y),exit(0);
	dx=(dx%n+n)%n;dy=(dy%n+n)%n;
	a[0]=dx,a[1]=dy,a[2]=x,a[3]=y;a[5]=1;
	fpw(t);
	if(a[2]==0)a[2]=n;
	if(a[3]==0)a[3]=n;
	printf("%lld %lld",a[2],a[3]);
}