[洛谷P5137]polynomial

题目大意：求：
$$
\sum\limits_{i=0}^na^{n-i}b^i\pmod{p}
$$
$T(T\leqslant10^5)$组数据，$a,b,n,p\leqslant10^{18}​$

题解：$\sum\limits_{i=0}^na^{n-i}b^i=\dfrac{a^{n+1}-b^{n+1}}{a-b}$，然后$a-b$可能在$\pmod p$下没有逆元或者干脆是$0$。

出题人给了一个递归讨论$n$奇偶性的做法。（出题人在题解中各种表达他的毒瘤）

这边讲一个矩阵快速幂的。

令$f_n=\sum\limits_{i=0}^na^{n-i}b^i$

考虑$f_n\to f_{n+1}$，发现$f_{n+1}=af_n+b^{n+1}$，于是就可以愉快地矩阵快速幂啦。转移矩阵：
$$
\left[
\begin{matrix}
a&0\\
1&b
\end{matrix}
\right]
$$

把$[f_n,b^{n+1}]$左乘转移矩阵就可以得到$[f_{n+1},b_{n+2}]$，为了方便，可以把向量写成矩阵，然后发现若初始矩阵如下时：
$$
\left[
\begin{matrix}
0&0\\
1&b
\end{matrix}
\right]
$$
转移矩阵、状态矩阵右上角一定为$0$，就可以减少常数啦！

卡点：无

 

C++ Code：

#include <cstdio>
#include <cctype>
namespace std {
	struct istream {
#define M (1 << 22 | 3)
		char buf[M], *ch = buf - 1;
		inline istream() { fread(buf, 1, M, stdin); }
		inline istream& operator >> (int &x) {
			while (isspace(*++ch));
			for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
			return *this;
		}
		inline istream& operator >> (long long &x) {
			while (isspace(*++ch));
			for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
			return *this;
		}
#undef M
	} cin;
	struct ostream {
#define M (1 << 20 | 3)
		char buf[M], *ch = buf - 1;
		inline ostream& operator << (long long x) {
			if (!x) {*++ch = '0'; return *this;}
			static int S[20], *top; top = S;
			while (x) {*++top = x % 10 ^ 48; x /= 10;}
			for (; top != S; --top) *++ch = *top;
			return *this;
		}
		inline ostream& operator << (const char x) {*++ch = x; return *this;}
		inline ~ostream() { fwrite(buf, 1, ch - buf + 1, stdout); }
#undef M
	} cout;
}

int Tim;
long long n, a, b, mod;

inline void reduce(long long &x) { x += x >> 63 & mod; }
inline long long mul(long long x, long long y) {
	long long res = x * y - static_cast<long long> (static_cast<long double> (x) * y / mod + 0.5) * mod;
	return res + (res >> 63 & mod);
}

struct Matrix {
	long long s00, s10, s11;
	Matrix() { }
	Matrix(long long __s00, long long __s10, long long __s11) : s00(__s00), s10(__s10), s11(__s11) { }
#define M(l, r) mul(s##l, rhs.s##r)
	inline void operator *= (const Matrix &rhs) {
		static long long __s00, __s10, __s11;
		__s00 = M(00, 00);
		reduce(__s10 = M(10, 00) + M(11, 10) - mod);
		__s11 = M(11, 11);
		s00 = __s00, s10 = __s10, s11 = __s11;
	}
#undef M
} ;

long long calc(long long n) {
	a %= mod, b %= mod;
	Matrix base(a, 1, b), res(0, 1, b);
	for (; n; n >>= 1, base *= base) if (n & 1) res *= base;
	return res.s10;
}

int main() {
	std::cin >> Tim;
	while (Tim --> 0) {
		std::cin >> n >> a >> b >> mod;
		std::cout << calc(n) << '\n';
	}
	return 0;
}