题面
题解
把\(a_i\)和\(b_i\)都变成小数的形式，记\(f_i\)表示\(1\)单位的光打到第\(i\)个玻璃上，能从第\(n\)个玻璃下面出来的光有多少，记\(g_i\)表示能从第\(i\)块玻璃反射出来的光有多少，，递推式的话，我们枚举一下这束光在\(i\)和\(i+1\)块玻璃之间反射了几次就可以了
\[\begin{aligned}
f_i
&=a_i\left(f_{i+1}+g_{i+1}\times b_i\times f_{i+1}+g_{i+1}\times b_i\times g_{i+1}\times b_i\times f_{i+1}+...\right)\\
&=a_if_{i+1}\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\
&=a_if_{i+1}{1\over 1-b_i\times g_{i+1}}\\
g_i
&=b_i+a_ig_{i+1}a_i+a_ig_{i+1}b_ig_{i+1}a_i+...\\
&=b_i+a_i^2g_i\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\
&=b_i+a_i^2g_{i+1}{1\over 1-b_ig_{i+1}}
\end{aligned}
\]
f_i
&=a_i\left(f_{i+1}+g_{i+1}\times b_i\times f_{i+1}+g_{i+1}\times b_i\times g_{i+1}\times b_i\times f_{i+1}+...\right)\\
&=a_if_{i+1}\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\
&=a_if_{i+1}{1\over 1-b_i\times g_{i+1}}\\
g_i
&=b_i+a_ig_{i+1}a_i+a_ig_{i+1}b_ig_{i+1}a_i+...\\
&=b_i+a_i^2g_i\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\
&=b_i+a_i^2g_{i+1}{1\over 1-b_ig_{i+1}}
\end{aligned}
\]
然而问题来了，如果\(b_i=0\)的特殊情况该怎么办？
我们发现在这种情况下，手玩出来的和代入柿子计算的值似乎是一样的？
所以直接代入柿子并没有问题的说……
递推即可
//minamoto
#include<bits/stdc++.h>
#define R register
#define inline __inline__ __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=5e5+5,P=1e9+7,inv=570000004;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
return res;
}
int a[N],b[N],f[N],g[N],res,n,tmp;
int main(){
// freopen("testdata.in","r",stdin);
n=read();
fp(i,1,n)a[i]=read(),b[i]=read(),a[i]=mul(a[i],inv),b[i]=mul(b[i],inv);
f[n]=a[n],g[n]=b[n];
fd(i,n-1,1){
tmp=ksm(P+1-mul(g[i+1],b[i]),P-2);
f[i]=1ll*a[i]*f[i+1]%P*tmp%P;
g[i]=add(b[i],1ll*a[i]*a[i]%P*g[i+1]%P*tmp%P);
}
printf("%d\n",f[1]);
return 0;
}