3240: [Noi2013]矩阵游戏

3240: [Noi2013]矩阵游戏

Time Limit: 10 Sec   Memory Limit: 256 MB
Submit: 1304   Solved: 556
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Description

婷婷是个喜欢矩阵的小朋友,有一天她想用电脑生成一个巨大的n行m列的矩阵(你不用担心她如何存储)。她生成的这个矩阵满足一个神奇的性质：若用F[i][j]来表示矩阵中第i行第j列的元素，则F[i][j]满足下面的递推式:

F[1][1]=1
F[i,j]=a*F[i][j-1]+b (j!=1)
F[i,1]=c*F[i-1][m]+d (i!=1)
递推式中a,b,c,d都是给定的常数。

现在婷婷想知道F[n][m]的值是多少,请你帮助她。由于最终结果可能很大，你只需要输出F[n][m]除以1,000,000,007的余数。

Input

一行有六个整数n,m,a,b,c,d。意义如题所述

Output

包含一个整数，表示F[n][m]除以1,000,000,007的余数

Sample Input

3 4 1 3 2 6

Sample Output

85

HINT

样例中的矩阵为：

1 4 7 10

26 29 32 35

76 79 82 85

1<=N,M<=10^1000 000,a<=a,b,c,d<=10^9

Source

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矩阵乘法。。。变态的矩阵乘法 大概你无法用高精完成它
首先矩阵乘法满足结合律，于是只要把m-1个a矩阵和一个b矩阵乘出，再拿这个东西n次方，得到矩阵乘以原来的东西，在往上走一步即可 不过奇妙的是这东西可以用费马小定理消次方从而避免高精,不过学长建议我写10进制快速幂。。然而T了(一定是我姿势不够优美)。。。。然后然后就写了个费马小定理的(矩阵A的1E9+7次方刚好等于矩阵A，然而当矩阵A的a系数等于1时次方变为1E9+8)具体可以手算证明。。。。4.6s

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;

typedef long long LL;
const LL mo = 1000000007;
const int maxn = 1E6 + 10;

struct data{
	LL a[2][2];
	data () {
		memset(a,0,sizeof(a));
		for (int i = 0; i < 2; i++) a[i][i] = 1;
	}
	
	data operator * (const data &b) {
		data c;
		memset(c.a,0,sizeof(c.a));
		for (int i = 0; i < 2; i++)
			for (int j = 0; j < 2; j++)
				for (int l = 0; l < 2; l++)
					c.a[i][j] = (c.a[i][j] + a[i][l]*b.a[l][j]%mo)%mo;
		return c;
	}
	
	bool operator == (const data &b) const {
		for (int i = 0; i < 2; i++)
			for (int j = 0; j < 2; j++)
				if (a[i][j] != b.a[i][j])
					return 0;
		return 1;
	}
};

char n[maxn],m[maxn];
LL a,b,c,d,s[2],ans[2],pn,pm;
int ln,lm;

data ks(data now,LL y)
{
	data ret;
	for (; y; y >>= 1LL) {
		if (y & 1) ret = ret*now;
		now = now*now;
	}
	return ret;
}

data kn(data now)
{
	data ret;
	for (int i = 0; i < ln; i++) {
		//ret = ret*ks(now,n[i]-'0');
		for (int j = '0'; j < n[i]; j++) ret = ret*now;
		now = ks(now,10);
		//for (int i = 0; i < 10; i++) now = now*now;
	}
	return ret;
}

data km(data now)
{
	data ret;
	for (int i = 0; i < lm; i++) {
		//ret = ret*ks(now,m[i]-'0');
		for (int j = '0'; j < m[i]; j++) ret = ret*now;
		now = ks(now,10);
	}
	return ret;
}

LL ksm(LL x,LL y)
{
	LL ret = 1;
	for (; y; y >>= 1LL) {
		if (y & 1) ret = ret*x%mo;
		x = x*x%mo;
	}
	return ret;
}

int main()
{
	#ifdef YZY
		   freopen("yzy.txt","r",stdin);
	#endif
	
	scanf("%s",&n); scanf("%s",&m);
	cin >> a >> b >> c >> d; 
	
	ln = strlen(n); lm = strlen(m);
	//for (int i = 0; i < ln/2; i++) swap(n[i],n[ln-1-i]);
	//for (int i = 0; i < lm/2; i++) swap(m[i],m[lm-1-i]);
	data A,B; A.a[1][0] = b; A.a[0][0] = a; B.a[0][0] = c; B.a[1][0] = d;
	int pm = c == 1?mo:mo-1;
	int pn = a == 1?mo:mo-1;
	LL N,M; N = M = 0;
	for (int i = 0; i < ln; i++) N = (N*10LL%pn + 1LL*(n[i]-'0'))%pn;
	for (int i = 0; i < lm; i++) M = (M*10LL%pm + 1LL*(m[i]-'0'))%pm;
	data x = ks(A,(M-1+pm)%pm);
	x = x*B;
	data y = ks(x,N);
	
	s[0] = s[1] = 1; 
	for (int i = 0; i < 2; i++)
		for (int j = 0; j < 2; j++)
			ans[i] = (ans[i] + s[j]*y.a[j][i]%mo)%mo;
	LL ANS = ans[0];
	ANS = (ANS - d + mo)%mo;
	ANS = ANS*ksm(c,mo-2LL)%mo;
	cout << ANS;
	
	return 0;
}