Description
There are n casinos lined in a row. If Memory plays at casino \(i\), he has probability \(p_{i}\) to win and move to the casino on the right \((i + 1)\) or exit the row (if \(i = n\)), and a probability \(1 - p_{i}\) to lose and move to the casino on the left \((i - 1\)) or also exit the row (if \(i = 1\)).
We say that Memory dominates on the interval \(i \dots j\) if he completes a walk such that,\(\bullet\)He starts on casino \(i\).\(\bullet\)He never looses in casino \(i\).\(\bullet\)He finishes his walk by winning in casino \(j\). Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process Now Memory has some requests, in one of the following forms:\(1 i a b\): Set \(p_{i} = \frac{a}{b}\).\(2 l r\): Print the probability that Memory will dominate on the interval \(l \dots r\), i.e. compute the probability that Memory will first leave the segment \(l \dots r\) after winning at casino \(r\), if she starts in casino \(l\). It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. \(p_{i} \le p_{i + 1}\) for all \(i\) from \(1\) to \(n - 1\). Please help Memory by answering all his requests!
Input
The first line of the input contains two integers \(n\) and \(q(1 \le n, q \le 100 000)\), — number of casinos and number of requests respectively.
The next n lines each contain integers \(a_{i}\) and \(b_{i}\) \((1 \le a{i} < b_{i} \le 10^{9})\) — is the probability \(p_{i}\) of winning in casino \(i\). The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n). It's guaranteed that there will be at least one query of type \(2\), i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.
Output
Print a real number for every request of type \(2\) — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed \(10^{-4}\).
Namely: let's assume that one of your answers is \(a\), and the corresponding answer of the jury is \(b\). The checker program will consider your answer correct if \(\mid a - b \mid \le 10^{ - 4}\).
Sample Input
3 13
1 3 1 2 2 3 2 1 1 2 1 2 2 1 3 2 2 2 2 2 3 2 3 3 1 2 2 3 2 1 1 2 1 2 2 1 3 2 2 2 2 2 3 2 3 3
Sample Output
0.3333333333
0.2000000000 0.1666666667 0.5000000000 0.4000000000 0.6666666667 0.3333333333 0.2500000000 0.2222222222 0.6666666667 0.5714285714 0.6666666667
对于区间\(l \dots r\)，我们用\(f\)记录成功离开区间的概率，\(g\)记录从\(r\)出发最后到\(r+1\)，没有离开过区间的概率。\(f_{1},g_{1}\)为\(l \dots mid\)的\(f,g\)值，\(f_{2},g_{2}\)为\(mid+1 \dots r\)的\(f,g\)值。合并方程：
\[f = f_{1}f_{2}+f_{1}(1-f_{2})g_{1}f_{2}+\cdots=\frac{f_{1}f_{2}}{1-(1-f_{2})g_{1}}\]\[g = g_{2}+(1-g_{2})g_{1}f_{2}+(1-g_{2})g_{1}(1-f_{2})g_{1}f_{2}+\cdots=g_{2}+\frac{(1-g_{2})g_{1}f_{2}}{1-(1-g_{2})g_{1}}\] 线段树维护下。#include#include #include using namespace std;typedef long double ld;#define maxn (400010)int N,Q,A[maxn],B[maxn],lef[maxn]; ld g[maxn],f[maxn];struct node { ld f,g; };inline int gi(){ int f = 1,ret = 0; char ch; do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-'); if (ch == '-') f = -1,ch = getchar(); do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9'); return f*ret;}inline void build(int now,int l,int r){ if (l == r) { lef[l] = now; g[now] = f[now] = (ld)A[l]/(ld)B[l]; return; } int mid = (l+r)>>1; build(now<<1,l,mid); build(now<<1|1,mid+1,r); f[now] = (f[now<<1]*f[now<<1|1])/(1-g[now<<1]*(1-f[now<<1|1])); g[now] = g[now<<1|1]+(1-g[now<<1|1])*g[now<<1]*f[now<<1|1]/(1+(f[now<<1|1]-1)*g[now<<1]);}inline node query(int now,int l,int r,int ql,int qr){ if (l == ql&&r == qr) return (node){ f[now],g[now] }; int mid = (l+r)>>1; if (qr <= mid) return query(now<<1,l,mid,ql,qr); else if (ql > mid) return query(now<<1|1,mid+1,r,ql,qr); else { node a,b,ret; a = query(now<<1,l,mid,ql,mid); b = query(now<<1|1,mid+1,r,mid+1,qr); ret.f = (a.f*b.f)/(1-a.g*(1-b.f)); ret.g = b.g+(1-b.g)*a.g*b.f/(1+(b.f-1)*a.g); return ret; }}int main(){ freopen("E.in","r",stdin); freopen("E.out","w",stdout); scanf("%d %d",&N,&Q); for (int i = 1;i <= N;++i) A[i] = gi(),B[i] = gi(); build(1,1,N); while (Q--) { int opt = gi(); if (opt == 1) { int now = lef[gi()],a = gi(),b = gi(); f[now] = g[now] = (ld)a/(ld)b; for (now >>= 1;now;now >>= 1) { f[now] = (f[now<<1]*f[now<<1|1])/(1-g[now<<1]*(1-f[now<<1|1])); g[now] = g[now<<1|1]+(1-g[now<<1|1])*g[now<<1]*f[now<<1|1]/(1+(f[now<<1|1]-1)*g[now<<1]); } } else { int l = gi(),r = gi(); printf("%.10lf\n",(double)query(1,1,N,l,r).f); } } fclose(stdin); fclose(stdout); return 0;}