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 << ) - );
 const double double_inf = 1e30;
 const double eps = 1e-;
 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 ;
     int tem = (int)((double)rand() / RAND_MAX * x) % x;
     return o ? tem +  : tem;
 }
 ll ll_rand(ll x, int o) {
     if (!x) return ;
     ll tem = (ll)((double)rand() / RAND_MAX * x) % x;
     return o ? tem +  : tem;
 }

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

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

 void push_up(int u) {
     seg[u].l2 = seg[lson].l2 * seg[rson].l2 / ( - seg[lson].r1 * seg[rson].l1);
     seg[u].l1 = seg[lson].l1 + seg[lson].l2 * seg[lson].r2 * seg[rson].l1 / ( - seg[lson].r1 * seg[rson].l1);
     seg[u].r1 = seg[rson].r1 + seg[rson].r2 * seg[rson].l2 * seg[lson].r1 / ( - seg[lson].r1 * seg[rson].l1);
     seg[u].r2 = seg[lson].r2 * seg[rson].r2 / ( - 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) >> ;
     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 / (. - 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) >> ;
     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 / (. - 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) >> ;
     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 / (. - 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) >> ;
     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 / (. - R1 * L1);
 }

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

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

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