【題解】LuoGu5323：[BJOI2019]光線

原題傳送門
這是一道很有趣的題目

大家都會非常直覺的想出$dp$
$f_i$表示從第$i$塊玻璃射出的光有多少
$g_i$表示從第$i$塊玻璃往前面反射回去的光有多少
得到轉移方程
$f_i=a_if_{i-1}+b_ig_{i+1}$
$g_i=b_if_{i-1}+a_ig_{i+1}$
又有
$f_n=a_nf_{n-1}$
$g_n=b_nf_{n-1}$
感覺有點奇怪，高斯消元？
不用這麼麻煩，發現我們最重要的是$f_n$，$g_i$是沒有用的，嘗試消掉$g_i$
令
$F_i=\frac{f_i}{f_{i-1}}$
$G_i=\frac{g_i}{f_{i-1}}$
又有
$F_n=a_n$
$G_n=b_n$
$F_iG_{i+1}=\frac{g_{i+1}}{f_{i-1}}$
推導
$f_i=a_if_{i-1}+b_ig_{i+1}$
$F_i=a_i+b_i\frac{g_{i+1}}{f_{i-1}}=a_i+b_iF_iG_{i+1}$
$F_i=\frac{a_i}{1-b_iG_{i+1}}$
繼續
$g_i=b_if_{i-1}+a_ig_{i+1}$
$G_i=b_i+a_i\frac{g_{i+1}}{f_{i-1}}=b_i+a_iF_iG_{i+1}$
先把$F_i,G_i$求出來之後
$f_i=f_{i-1}F_i$推得答案

Code：

#include <bits/stdc++.h>
#define maxn 500010
#define LL long long
using namespace std;
const LL qy = 1000000007;
int n;
LL a[maxn], b[maxn], f[maxn], F[maxn], G[maxn], inv100;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

LL ksm(LL n, LL k){
	if (!k) return 1;
	LL sum = ksm(n, k >> 1);
	sum = sum * sum % qy;
	if (k & 1) sum = sum * n % qy;
	return sum;
}

int main(){
	n = read(), inv100 = ksm(100, qy - 2);
	for (int i = 1; i <= n; ++i) a[i] = read() * inv100 % qy, b[i] = read() * inv100 % qy;
	F[n] = a[n], G[n] = b[n];
	for (int i = n - 1; i; --i)
		F[i] = a[i] * ksm((1LL - b[i] * G[i + 1] % qy + qy) % qy, qy - 2) % qy,
		G[i] = (b[i] + a[i] * F[i] % qy * G[i + 1] % qy) % qy;
	f[0] = 1;
	for (int i = 1; i <= n; ++i) f[i] = F[i] * f[i - 1] % qy;
	printf("%lld\n", f[n]);
	return 0;
}