P7385 「EZEC-6」跳一跳 题解

请先阅读Ecrade_的文章中Subtask4,5的思路
题目:P7385 「EZEC-6」跳一跳
此题数据范围极大，为了优化时间复杂度，采用矩阵快速幂
根据Ecrade_的推理可知:
$\{\begin{array}{l}S[n]=1\times S[n-1]+f^{{}^{\prime }}[n]+(2b{)}^n\\ f^{{}^{\prime }}[n+1]=0\times S[n-1]+(a+b)\times f^{{}^{\prime }}[n]+a\times (2b{)}^n\\ (2b{)}^{n+1}=0\times S[n-1]+0\times f^{{}^{\prime }}[n]+2b\times (2b{)}^n\end{array}$
得式子:
$$\left[\begin{array}{c}S[n]\\ f[n+1]\\ (2b{)}^{n+1}\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 0& a+b& a\\ 0& 0& 2b\end{array}\right]\times \left[\begin{array}{c}S[n-1]\\ f^{{}^{\prime }}[n]\\ (2b{)}^n\end{array}\right]$$
$$={\left[\begin{array}{ccc}1& 1& 1\\ 0& a+b& a\\ 0& 0& 2b\end{array}\right]}^{n-1}\times \left[\begin{array}{c}S[1]\\ f^{{}^{\prime }}[2]\\ (2b{)}^2\end{array}\right]$$
设$\left[\begin{array}{ccc}1& 1& 1\\ 0& a+b& a\\ 0& 0& 2b\end{array}\right]$为$A$,则有$$\left[\begin{array}{c}S[n]\\ f[n+1]\\ (2b{)}^{n+1}\end{array}\right]=A^n\times \left[\begin{array}{c}a+b\times 2\\ (a+b)\times a+a\times b\times 2\\ 4b^2\end{array}\right]$$
用$B$储存$A^n$最终得到式子:
$$S[n]=B[1][1]\ast (a+b\times 2)+B[1][2][(a+b)\times a+a\times b\times 2]+B[1][2]\times 4b^2$$

Code:

#include
#define ll unsigned long long 
using namespace std;
const ll p=1000000007;
inline ll Read(){
     
	ll dx=0,fh=1;
	char c=getchar();
	while(c>'9'||c<'0'){
     
		if(c=='-')	fh=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
     
		dx=dx*10+c-'0';
		c=getchar();
	}
	return dx*fh;
}
ll quick_pow(ll x,ll y){
     
	ll r=1,base=x,yy=y;
	while(y){
     
		if(y&1)	r=(r*base)%p;
		base=(base*base)%p;
		y>>=1;
	}
	return r%p;
}
ll n,a,b,m,spe;
struct Matricx{
     
	ll val[5][5];
};
Matricx Mul(Matricx x,Matricx y){
     
	Matricx Ans;
	for(int i=1;i<=4;++i)
		for(int j=1;j<=4;++j){
     
			Ans.val[i][j]=0;
			for(int k=1;k<=4;++k)	Ans.val[i][j]=(Ans.val[i][j]+((x.val[i][k]%p)*(y.val[k][j]%p))%p)%p;
			Ans.val[i][j]%=p;
		}
	return Ans;
}
Matricx quickpow(Matricx x,ll y){
     
	Matricx r,base=x;
	for(int i=1;i<=4;++i){
     
		for(int j=1;j<=4;++j)	r.val[i][j]=0;
		r.val[i][i]=1;
	}
	while(y){
     
		if(y&1) r=Mul(r,base);
		base=Mul(base,base);
		y>>=1;
	}
	return r;
}
Matricx A,B;
int main(){
     
	n=Read(),a=Read(),b=Read(),m=Read();
	a=a*570000004%p,b=b*570000004%p;
	for(int i=1;i<=m;++i){
     
		ll spe_num=Read(),spe_score=Read();
		spe=(spe+quick_pow(a+b,spe_num-1)*b%p*spe_score%p)%p;
	}
	A.val[1][1]=1,A.val[1][2]=1,A.val[1][3]=1;
	A.val[2][1]=0,A.val[2][2]=a+b,A.val[2][3]=a;
	A.val[3][1]=0,A.val[3][2]=0,A.val[3][3]=2*b;
	B=quickpow(A,n-1);
	ll s1=(a+b*2)%p,f2=( ((a+b)*a)%p + (a*b*2)%p  )%p;
	printf("%u\n",(((B.val[1][1]*s1%p + B.val[1][2]*f2%p)%p + B.val[1][3]*((4*b*b)%p)%p)%p +spe%p)%p);
	return 0;
}

记住打完代码后一定要检查一下每个式子有没有漏取模！本人因为这玩意从早上九点查到了下午三点半！