题目描述
You are given four positive integers $x_0,x_1,a,b$. And you know $x_i=a\cdot x_{i-1}+b\cdot x_{i-2}$ for all $i\ge 2$.

Given two positive integers n, and MOD, please calculate $x_n$ modulo MOD.

Does the problem look simple? Surprise! The value of n may have many many digits!

输入描述:
The input contains two lines.
The first line contains four integers $x_0,x_1,a,b$ $(1\le x_0,x_1,a,b\le 10^9)$
The second line contains two integers $n$, MOD $(1\le n<10^{(10^6)},10^9<MOD\le 2\times 10^9,nhasnoleadingzero)$.

输出描述:
Print one integer representing the answer.

示例1
输入
1 1 1 1
10 1000000001
输出
89
说明
The resulting sequence x is Fibonacci sequence. The 11-th item is 89.

示例2
输入
1315 521 20185 5452831
9999999999999999999999999999999999999 1000000007
输出
914730061

题目大意：
已知公式$x_i=a\cdot x_{i-1}+b\cdot x_{i-2}$
第一行输入$x_0,x_1,a,b$
第二行输入$n,p$
求$x_n\%p$的值

解题思路：
由于此题$n$的数据范围很大，而且又知道公式，所以可以很容易想到是矩阵快速幂的裸题，但由于$n$的范围很大，所以需将二进制的快速幂写法改为十进制。
由于$x_i=a\cdot x_{i-1}+b\cdot x_{i-2}$
所以
$\left|\begin{array}{cc}{x}_{n+1}& x_n\\ 0& 0\end{array}\right|=\left|\begin{array}{cc}{x}_0& x_1\\ 0& 0\end{array}\right|\ast {\left|\begin{array}{cc}a& 1\\ b& 0\end{array}\right|}^n$
因此直接敲矩阵快速幂即可。

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
struct node
{ll arr[2][2];node() {ms(arr);}
};
ll p;
node mul(node a,node b) {node c;for(int i=0;i<2;i++) {for(int j=0;j<2;j++) {for(int k=0;k<2;k++) {c.arr[i][j]=(c.arr[i][j]+(a.arr[i][k]*b.arr[k][j])%p)%p;}}}return c;
}
char str[1001000];
node b1,base;
int len;
ll quickpow() {for(int i=len-1;i>=0;i--) {int x=(int)(str[i]-'0');node nape2;nape2.arr[0][0]=base.arr[0][0];nape2.arr[0][1]=base.arr[0][1];nape2.arr[1][0]=base.arr[1][0];nape2.arr[1][1]=base.arr[1][1];while(x) {if(x&1) b1=mul(b1,nape2);nape2=mul(nape2,nape2);x>>=1;}node nape1=mul(base,base);base=mul(base,base);base=mul(base,base);base=mul(base,base);base=mul(nape1,base);}return b1.arr[0][1]%p;
}
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);//ios::sync_with_stdio(0),cin.tie(0); ll x0,x1,a,b;scanf("%I64d %I64d %I64d %I64d",&x0,&x1,&a,&b);scanf("%s",str);scanf("%I64d",&p);base.arr[0][0]=a;base.arr[0][1]=1LL;base.arr[1][0]=b;base.arr[1][1]=0;b1.arr[0][0]=x1;b1.arr[0][1]=x0;b1.arr[1][0]=0;b1.arr[1][1]=0;len=strlen(str);printf("%I64d\n",quickpow());return 0;
}