cf 712E Memory and Casinos

题意：有一行$n(n leq 100000)$个方格，从左往右第$i$个方格的值为$p_i(p_i = frac{a}{b}, 1 leq a < b leq 1e9)$，有两种操作，一种是将某个方格的值更新为另一个分数表示的有理数，另一种操作是寻味区间$[l, r](l leq r)$的权值$w(l, r)$；$w(l, r)$如下定义：

方格在位置$i$有$p_i$的概率向右移动一格，有$1-p_i$的概率向左移动一格。$w(l, r)$表示方格初始位置在$l$并且以在位置$r$向右移动（下一个位置为$r+1$）为终结，移动过程始终不超出区间范围的概率值。

分析：对于任一区间$[l, r]$，设$f(i)$表示目前在位置$i$，在移动合法的情况下到达终结状态的概率值。那么显然有$f(i) = p_if(i + 1) + (1 - p_i)f(i - 1)$，注意边界情况是$f(l - 1) = 0$, 且$f(r + 1) = 1$，我们设$w(l, r) = f(l) = Delta$，那么可以得到递推关系$f(r + 1) = 1 = g(r + 1) + f(r - 1)$，其中$g(r + 1) = frac{prod_{i leq r - 1}(1 - p_i)}{prod_{i leq r}p_i} $，理论上我们可以用$g(i)$前缀和得到任意区间的和，用线段树分别维护奇数位置和偶数位置即可。然而，由于$g(i)$可能会非常大，以至于double存储失效，因此此方法并不可行。

用分类统计的方法来解，考虑小规模问题与大规模问题之间的联系，$[l, r]$中间一任意位置为$m$，讨论方格穿过$m$的次数（等比求和），于是可以得到具有局部可累加性质的递推关系。用线段上进行点维护和区间查询即可。单次询问复杂度$O(log(n))$。

code：

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <ctime>
#include <cmath>
#include <iostream>
#include <assert.h>
#pragma comment(linker, "/STACK:102400000,102400000")
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define mp std :: make_pair
#define st first
#define nd second
#define keyn (root->ch[1]->ch[0])
#define lson (u << 1)
#define rson (u << 1 | 1)
#define pii std :: pair<int, int>
#define pll pair<ll, ll>
#define pb push_back
#define type(x) __typeof(x.begin())
#define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
#define FOR(i, s, t) for(int i = (s); i <= (t); i++)
#define ROF(i, t, s) for(int i = (t); i >= (s); i--)
#define dbg(x) std::cout << x << std::endl
#define dbg2(x, y) std::cout << x << " " << y << std::endl
#define clr(x, i) memset(x, (i), sizeof(x))
#define maximize(x, y) x = max((x), (y))
#define minimize(x, y) x = min((x), (y))
using namespace std;
typedef long long ll;
const int int_inf = 0x3f3f3f3f;
const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
const int INT_INF = (int)((1ll << 31) - 1);
const double double_inf = 1e30;
const double eps = 1e-14;
typedef unsigned long long ul;
typedef unsigned int ui;
inline int readint() {
    int x;
    scanf("%d", &x);
    return x;
}
inline int readstr(char *s) {
    scanf("%s", s);
    return strlen(s);
}

class cmpt {
public:
    bool operator () (const int &x, const int &y) const {
        return x > y;
    }
};

int Rand(int x, int o) {
    //if o set, return [1, x], else return [0, x - 1]
    if (!x) return 0;
    int tem = (int)((double)rand() / RAND_MAX * x) % x;
    return o ? tem + 1 : tem;
}
ll ll_rand(ll x, int o) {
    if (!x) return 0;
    ll tem = (ll)((double)rand() / RAND_MAX * x) % x;
    return o ? tem + 1 : tem;
}

void data_gen() {
    srand(time(0));
    freopen("in.txt", "w", stdout);
    int kases = 1;
    //printf("%d
", kases);
    while (kases--) {
        ll sz = 1000;
        printf("%d %d
", sz, sz);
        FOR(i, 1, sz) {
            int x = Rand(1e2, 1);
            int y = Rand(1e9, 1);
            if (x > y) swap(x, y);
            printf("%d %d
", x, y);
        }
        FOR(i, 1, sz) {
            int o = Rand(2, 0);
            if (o) {
                printf("1 ");
                int pos = Rand(1000, 1);
                int x = Rand(1e9, 1), y = Rand(1e9, 1);
                if (x > y) swap(x, y);
                printf("%d %d %d
", pos, x, y);
            } else {
                printf("2 ");
                int x = Rand(1000, 1), y = Rand(1e3, 1);
                if (x > y) swap(x, y);
                printf("%d %d
", x, y);
            }
        }
    }
}

const int maxn = 1e5 + 10;
struct Seg {
    double l1, l2, r1, r2;
}seg[maxn << 2];
int n, q;
pii a[maxn];

void push_up(int u) {
    seg[u].l2 = seg[lson].l2 * seg[rson].l2 / (1 - seg[lson].r1 * seg[rson].l1);
    seg[u].l1 = seg[lson].l1 + seg[lson].l2 * seg[lson].r2 * seg[rson].l1 / (1 - seg[lson].r1 * seg[rson].l1);
    seg[u].r1 = seg[rson].r1 + seg[rson].r2 * seg[rson].l2 * seg[lson].r1 / (1 - seg[lson].r1 * seg[rson].l1);
    seg[u].r2 = seg[lson].r2 * seg[rson].r2 / (1 - seg[lson].r1 * seg[rson].l1);
}

double query1(int u, int l, int r, int L, int R);
double query3(int u, int l, int r, int L, int R);
double query4(int u, int l, int r, int L, int R);

double query(int u, int l, int r, int L, int R) {
    if (l == L && R == r) return seg[u].l2;
    int mid = (l + r) >> 1;
    if (R <= mid) return query(lson, l, mid, L, R);
    else if (L >= mid) return query(rson, mid, r, L, R);
    double lhs = query(lson, l, mid, L, mid), rhs = query(rson, mid, r, mid, R);
    double L1 = query1(rson, mid, r, mid, R), R1 = query3(lson, l, mid, L, mid);
    return lhs * rhs / (1. - L1 * R1);
}

double query3(int u, int l, int r, int L, int R) {
    if (l == L && r == R) return seg[u].r1;
    int mid = (l + r) >> 1;
    if (R <= mid) return query3(lson, l, mid, L, R);
    else if (L >= mid) return query3(rson, mid, r, L, R);
    double tem = query3(rson, mid, r, mid, R);
    double R2 = query4(rson, mid, r, mid, R);
    double R1 = query3(lson, l, mid, L, mid);
    double L2 = query(rson, mid, r, mid, R);
    double L1 = query1(rson, mid, r, mid, R);
    return tem + R2 * R1 * L2 / (1. - L1 * R1);
}

double query4(int u, int l, int r, int L, int R) {
    if (l == L && r == R) return seg[u].r2;
    int mid = (l + r) >> 1;
    if (R <= mid) return query4(lson, l, mid, L, R);
    else if (L >= mid) return query4(rson, mid, r, L, R);
    double lhs = query4(lson, l, mid, L, mid) * query4(rson, mid, r, mid, R);
    double rhs = query3(lson, l, mid, L, mid) * query3(rson, mid, r, mid, R);
    return lhs / (1. - rhs);
}

double query1(int u, int l, int r, int L, int R) {
    if (l == L && R == r) return seg[u].l1;
    int mid = (l + r) >> 1;
    if (R <= mid) return query1(lson, l, mid, L, R);
    else if (L >= mid) return query1(rson, mid, r, L, R);
    double tem = query1(lson, l, mid, L, mid);
    double L1 = query1(rson, mid, r, mid, R);
    double L2 = query(lson, l, mid, L, mid);
    double R2 = query4(lson, l, mid, L, mid);
    double R1 = query3(lson, l, mid, L, mid);
    return tem + L2 * L1 * R2 / (1. - R1 * L1);
}

void build(int u, int l, int r) {
    if (r - l < 2) {
        double p = (double)a[l].first / a[l].nd;
        seg[u].l1 = 1 - p;
        seg[u].l2 = p;
        seg[u].r1 = p;
        seg[u].r2 = 1 - p;
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid), build(rson, mid, r);
    push_up(u);
}

void update(int u, int l, int r, int L, int R, int lhs, int rhs) {
    if (l == L && r == R) {
        double p = (double)lhs / rhs;
        seg[u].l1 = 1 - p;
        seg[u].l2 = p;
        seg[u].r1 = p;
        seg[u].r2 = 1 - p;
        return;
    }
    int mid = (l + r) >> 1;
    if (R <= mid) update(lson, l, mid, L, R, lhs, rhs);
    else update(rson, mid, r, L, R, lhs, rhs);
    push_up(u);
}

double __get(int x, int y) {
    return query(1, 1, n + 1, x, y + 1);
}

void __set(int x, int y, int z) {
    update(1, 1, n + 1, x, x + 1, y, z);
}

int main() {
    //data_gen(); return 0;
    //C(); return 0;
    int debug = 0;
    if (debug) freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while (~scanf("%d%d", &n, &q)) {
        FOR(i, 1, n) scanf("%d%d", &a[i].first, &a[i].nd);
        build(1, 1, n + 1);
        FOR(i, 1, q) {
            int op, x, y, z;
            scanf("%d%d%d", &op, &x, &y);
            if (op == 1) {
                z = readint();
                __set(x, y, z);
            } else {
                double ans = __get(x, y);
                printf("%.10f
", ans);
            }
        }
    }
    return 0;
}