[数列通项 矩阵快速幂] BZOJ4002 [JLOI2015]有意义的字符串

传送门：http://blog.csdn.net/popoqqq/article/details/45148309

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define P 7528443412579576937LL
using namespace std;
typedef unsigned long long ll;

inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(ll &x){
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

inline ll Mul(ll a,ll b){
	ll ret=0;
	for (;b;(a+=a)%=P,b>>=1)
		if (b&1)
			(ret+=a)%=P;
	return ret;
}

struct Matrix{
	ll a[2][2];
	Matrix(ll _=0,ll __=0,ll ___=0,ll ____=0) {
		a[0][0]=_; a[0][1]=__; a[1][0]=___; a[1][1]=____;
	}
}A;

const Matrix I(1,0,0,1);

inline Matrix operator *(const Matrix &A,const Matrix &B){
	Matrix ret;
	for (int i=0;i<2;i++)
		for (int j=0;j<2;j++)
			for (int k=0;k<2;k++)
				(ret.a[i][j]+=Mul(A.a[i][k],B.a[k][j]))%=P;
	return ret;
}

inline Matrix Pow(Matrix a,ll b){
	Matrix ret=I;
	for (;b;a=a*a,b>>=1)
		if (b&1)
			ret=ret*a;
	return ret;
}

ll B,D,N,an;

int main()
{
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(B); read(D); read(N);
	if (N==0) return printf("1\n"),0;
	A=Matrix(B,(((D-B*B)/4)%P+P)%P,1,0);
	A=Pow(A,N-1);
	(an=Mul(A.a[0][0],B)+Mul(A.a[0][1],2))%=P;
	if (B!=D*D && ~N&1) an=(an==0)?P-1:an-1;
	printf("%lld\n",an);
	return 0;
}