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

\[ \begin{aligned} f_i &=a_i\left(f_{i+1}+g_{i+1}\times b_i\times f_{i+1}+g_{i+1}\times b_i\times g_{i+1}\times b_i\times f_{i+1}+...\right)\\ &=a_if_{i+1}\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\ &=a_if_{i+1}{1\over 1-b_i\times g_{i+1}}\\ g_i &=b_i+a_ig_{i+1}a_i+a_ig_{i+1}b_ig_{i+1}a_i+...\\ &=b_i+a_i^2g_i\sum\limits_{k=0}^\infty (b_i\times g_{i+1})^k\\ &=b_i+a_i^2g_{i+1}{1\over 1-b_ig_{i+1}} \end{aligned} \]

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
    
     #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 go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)template
      
       inline bool cmax(T&a,const T&b){return a
       
        
         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;}