广西大学oj1510(类Fibonacci数列求值)

运用了矩阵快速幂的方法

#include<iostream>
#include<cstring>
typedef long long ll;
const int mod = 1000000007;
using namespace std;

ll base[3][3];
ll ans[3][3];
ll temp[3][3];
//初始化
void NO(ll a,ll b)
{
	memset(base, 0, sizeof base);
	memset(ans, 0, sizeof ans);
	base[1][1] = a;
	base[1][2] = 1;
	base[2][1] = b;
	ans[1][1] = 1;
	ans[1][2] = 0;
}
//矩阵乘法
void mul(ll a[3][3], ll b[3][3])
{
	memset(temp, 0, sizeof temp);
	for (ll i = 1; i <= 2; i++)
		for (ll j = 1; j <= 2; j++)
			for (ll k = 1; k <= 2; k++)
				temp[i][j] = (temp[i][j] + a[i][k] * b[k][j] % mod) % mod;
	for (ll i = 1; i <= 2; i++)
		for (ll j = 1; j <= 2; j++)
			a[i][j] = temp[i][j] % mod;
}
void fun(ll n)
{
	while (n) {
		if (n & 1)
			mul(ans, base);
		mul(base, base);
		n >>= 1;
	}
}
int main()
{
	ll n, a, b;
	cin >> n >> a >> b;
	if (n <= 1) {
		cout << n;
		return 0;
	}
	NO(a, b);
	fun(n - 1);
	ll tmp = ans[1][1] % mod;
	cout << tmp;
}