2019 牛客暑期多校訓練營 第五場 C generator 1 10進制矩陣快速冪

題意:   xi​=a⋅xi−1​+b⋅xi−2​ for all i≥2   求第n項取mod的值 1<n<10^(10^6)   10^9<mod<2*10^9

題解：求出 1， 2， 4， 8，次冪接下來就可以求出 10，20，40，80，100 次冪

#include <bits/stdc++.h>
#include <cstring>
#define ll long long
using namespace std;
ll mod;
struct Mat {
	int n,m;
	ll mem[3][3];
	Mat(int n, int m) {
        this->n = n, this->m = m;
        memset(mem, 0, sizeof mem);
        return;
    }
	Mat operator*(const Mat &a)const{
        Mat c(n,a.m);
        for (int i = 1; i <= c.n; i++) {
            for (int j = 1; j <= c.m; j++) {
                for (int k = 1; k <= m; k++)
                    c.mem[i][j] += mem[i][k] * a.mem[k][j];
                c.mem[i][j] %= mod;
            }
        }
        return c;
    }
};
Mat solve(Mat a,Mat c,char *s){
	Mat b = c;
	for(int i=strlen(s)-1;i>=0;i--){
		int cur = s[i] - '0';
		while(cur--) b = b*a;
		Mat tmp = a*a;
		a = tmp*tmp;
		a = a*a*tmp;
	}
	return b;
}

char s[10000000];
int main() {
	ll x0,x1,a,b;
	ll tmp;
	Mat m1(2,2),m2(2,2);
	scanf("%lld %lld %lld %lld",&x0,&x1,&a,&b);
	scanf("%s %lld",s,&mod);
	m1.mem[1][1] = 0;
	m1.mem[1][2] = b;
	m1.mem[2][1] = 1;
	m1.mem[2][2] = a;
	m2.mem[1][1] = x0;
	m2.mem[2][1] = 0;
	m2.mem[1][2] = x1;
	m2.mem[2][2] = 0;
	m1 = solve(m1,m2,s);
	printf("%lld",m1.mem[1][1]%mod);
	return 0;
}
/*
1 1 2 2
1 8

*/