題意：\(n\)個賭場，每個賭場有\(p_{i}\)的勝率，如果贏了就走到下一個賭場，輸了就退回上一個賭場，規定\(1\)號賭場的上一個是\(0\)號賭場，\(n\)號賭場的下一個是\(n + 1\)號賭場，一旦到達\(0\)或\(n + 1\)號賭場就相當於退出賭局了。
定義統治區間\([l, r]\)為從第\(l\)個賭場開始，到達第\(r + 1\)個賭場，且在過程中不經過\([1, l - 1]\)的賭場。維護2種操作：
1，修改一個賭場的勝率
2，詢問統治\([l, r]\)的概率
題解：
設\(f_{i}\)表示從\(x\)能走到\(n\)的概率，則有：
\[f_{i} = f_{i - 1}(1 - p_{i}) + f_{i + 1} \cdot p_{i}\]

上面是求詢問區間\([1, n - 1]\)時的答案，也就是\(1\)到\(n\)的概率。
替換一下，同理可得，在詢問區間\([l, r]\)時，也就是要求\(l\)

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 101000
#define ac 500000

int n, q, w;
int l[ac], r[ac];
double ans, go;
double tree[ac], p[AC], t[AC], sum[ac];

struct node{
    double x, w;
};

inline int read()
{
    int x = 0;char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

inline void update(int x){
    int ll = x * 2, rr = ll + 1;
    sum[x] = sum[ll] * sum[rr];
    tree[x] = tree[ll] + tree[rr] * sum[ll];
}

void build(int x, int ll, int rr)
{
    l[x] = ll, r[x] = rr;
    if(ll == rr) {tree[x] = sum[x] = t[ll]; return ;}
    int mid = (ll + rr) >> 1;
    build(x * 2, ll, mid), build(x * 2 + 1, mid + 1, rr);
    update(x);
}

void change(int x, int go, double w)
{
    if(l[x] == r[x]){sum[x] = tree[x] = w; return ;}
    int mid = (l[x] + r[x]) >> 1;
    (go <= mid) ? change(x * 2, go, w) : change(x * 2 + 1, go, w);
    update(x);
}

node find(int x, int ll, int rr)
{
    if(l[x] == ll && r[x] == rr) return (node){tree[x], sum[x]};
    int mid = (l[x] + r[x]) >> 1;
    if(rr <= mid) return find(x * 2, ll, rr);
    else if(ll > mid) return find(x * 2 + 1, ll, rr);
    else 
    {
        node now = find(x * 2, ll, mid), y = find(x * 2 + 1, mid + 1, rr);
        now.x = now.x + y.x * now.w, now.w = now.w * y.w;//要更新now.w!!!
        return now;
    }
}

void pre()
{
    n = read(), q = read();
    for(R i = 1; i <= n; i ++)
    {
        double a = read(), b = read();
        p[i] = a / b;
    }
    for(R i = 1; i <= n; i ++) t[i] = (1 - p[i]) / p[i];
}

void work()
{
    int opt, a, b, x;
    for(R i = 1; i <= q; i ++)
    {
        opt = read();
        if(opt == 1)
        {
            x = read(), a = read(), b = read(), go = 1.0 * a / b;
            go = (1 - go) / go, change(1, x, go);
        }
        else
        {
            a = read(), b = read();
            node x = find(1, a, b);
        //  printf("%lf\n", x.x);
            printf("%.10lf\n", 1 / (1 + x.x));
        }
    }
}

int main()
{
    freopen("in.in", "r", stdin);
    pre();
    build(1, 1, n);
    work();
    fclose(stdin);
    return 0;
}