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：

  1 #include <algorithm>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <string>
  5 #include <queue>
  6 #include <map>
  7 #include <set>
  8 #include <stack>
  9 #include <ctime>
 10 #include <cmath>
 11 #include <iostream>
 12 #include <assert.h>
 13 #pragma comment(linker, "/STACK:102400000,102400000")
 14 #define max(a, b) ((a) > (b) ? (a) : (b))
 15 #define min(a, b) ((a) < (b) ? (a) : (b))
 16 #define mp std :: make_pair
 17 #define st first
 18 #define nd second
 19 #define keyn (root->ch[1]->ch[0])
 20 #define lson (u << 1)
 21 #define rson (u << 1 | 1)
 22 #define pii std :: pair<int, int>
 23 #define pll pair<ll, ll>
 24 #define pb push_back
 25 #define type(x) __typeof(x.begin())
 26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
 27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
 28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
 29 #define dbg(x) std::cout << x << std::endl
 30 #define dbg2(x, y) std::cout << x << " " << y << std::endl
 31 #define clr(x, i) memset(x, (i), sizeof(x))
 32 #define maximize(x, y) x = max((x), (y))
 33 #define minimize(x, y) x = min((x), (y))
 34 using namespace std;
 35 typedef long long ll;
 36 const int int_inf = 0x3f3f3f3f;
 37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
 38 const int INT_INF = (int)((1ll << 31) - 1);
 39 const double double_inf = 1e30;
 40 const double eps = 1e-14;
 41 typedef unsigned long long ul;
 42 typedef unsigned int ui;
 43 inline int readint() {
 44     int x;
 45     scanf("%d", &x);
 46     return x;
 47 }
 48 inline int readstr(char *s) {
 49     scanf("%s", s);
 50     return strlen(s);
 51 }
 52
 53 class cmpt {
 54 public:
 55     bool operator () (const int &x, const int &y) const {
 56         return x > y;
 57     }
 58 };
 59
 60 int Rand(int x, int o) {
 61     //if o set, return [1, x], else return [0, x - 1]
 62     if (!x) return 0;
 63     int tem = (int)((double)rand() / RAND_MAX * x) % x;
 64     return o ? tem + 1 : tem;
 65 }
 66 ll ll_rand(ll x, int o) {
 67     if (!x) return 0;
 68     ll tem = (ll)((double)rand() / RAND_MAX * x) % x;
 69     return o ? tem + 1 : tem;
 70 }
 71
 72 void data_gen() {
 73     srand(time(0));
 74     freopen("in.txt", "w", stdout);
 75     int kases = 1;
 76     //printf("%d\n", kases);
 77     while (kases--) {
 78         ll sz = 1000;
 79         printf("%d %d\n", sz, sz);
 80         FOR(i, 1, sz) {
 81             int x = Rand(1e2, 1);
 82             int y = Rand(1e9, 1);
 83             if (x > y) swap(x, y);
 84             printf("%d %d\n", x, y);
 85         }
 86         FOR(i, 1, sz) {
 87             int o = Rand(2, 0);
 88             if (o) {
 89                 printf("1 ");
 90                 int pos = Rand(1000, 1);
 91                 int x = Rand(1e9, 1), y = Rand(1e9, 1);
 92                 if (x > y) swap(x, y);
 93                 printf("%d %d %d\n", pos, x, y);
 94             } else {
 95                 printf("2 ");
 96                 int x = Rand(1000, 1), y = Rand(1e3, 1);
 97                 if (x > y) swap(x, y);
 98                 printf("%d %d\n", x, y);
 99             }
100         }
101     }
102 }
103
104 const int maxn = 1e5 + 10;
105 struct Seg {
106     double l1, l2, r1, r2;
107 }seg[maxn << 2];
108 int n, q;
109 pii a[maxn];
110
111 void push_up(int u) {
112     seg[u].l2 = seg[lson].l2 * seg[rson].l2 / (1 - seg[lson].r1 * seg[rson].l1);
113     seg[u].l1 = seg[lson].l1 + seg[lson].l2 * seg[lson].r2 * seg[rson].l1 / (1 - seg[lson].r1 * seg[rson].l1);
114     seg[u].r1 = seg[rson].r1 + seg[rson].r2 * seg[rson].l2 * seg[lson].r1 / (1 - seg[lson].r1 * seg[rson].l1);
115     seg[u].r2 = seg[lson].r2 * seg[rson].r2 / (1 - seg[lson].r1 * seg[rson].l1);
116 }
117
118 double query1(int u, int l, int r, int L, int R);
119 double query3(int u, int l, int r, int L, int R);
120 double query4(int u, int l, int r, int L, int R);
121
122 double query(int u, int l, int r, int L, int R) {
123     if (l == L && R == r) return seg[u].l2;
124     int mid = (l + r) >> 1;
125     if (R <= mid) return query(lson, l, mid, L, R);
126     else if (L >= mid) return query(rson, mid, r, L, R);
127     double lhs = query(lson, l, mid, L, mid), rhs = query(rson, mid, r, mid, R);
128     double L1 = query1(rson, mid, r, mid, R), R1 = query3(lson, l, mid, L, mid);
129     return lhs * rhs / (1. - L1 * R1);
130 }
131
132 double query3(int u, int l, int r, int L, int R) {
133     if (l == L && r == R) return seg[u].r1;
134     int mid = (l + r) >> 1;
135     if (R <= mid) return query3(lson, l, mid, L, R);
136     else if (L >= mid) return query3(rson, mid, r, L, R);
137     double tem = query3(rson, mid, r, mid, R);
138     double R2 = query4(rson, mid, r, mid, R);
139     double R1 = query3(lson, l, mid, L, mid);
140     double L2 = query(rson, mid, r, mid, R);
141     double L1 = query1(rson, mid, r, mid, R);
142     return tem + R2 * R1 * L2 / (1. - L1 * R1);
143 }
144
145 double query4(int u, int l, int r, int L, int R) {
146     if (l == L && r == R) return seg[u].r2;
147     int mid = (l + r) >> 1;
148     if (R <= mid) return query4(lson, l, mid, L, R);
149     else if (L >= mid) return query4(rson, mid, r, L, R);
150     double lhs = query4(lson, l, mid, L, mid) * query4(rson, mid, r, mid, R);
151     double rhs = query3(lson, l, mid, L, mid) * query3(rson, mid, r, mid, R);
152     return lhs / (1. - rhs);
153 }
154
155 double query1(int u, int l, int r, int L, int R) {
156     if (l == L && R == r) return seg[u].l1;
157     int mid = (l + r) >> 1;
158     if (R <= mid) return query1(lson, l, mid, L, R);
159     else if (L >= mid) return query1(rson, mid, r, L, R);
160     double tem = query1(lson, l, mid, L, mid);
161     double L1 = query1(rson, mid, r, mid, R);
162     double L2 = query(lson, l, mid, L, mid);
163     double R2 = query4(lson, l, mid, L, mid);
164     double R1 = query3(lson, l, mid, L, mid);
165     return tem + L2 * L1 * R2 / (1. - R1 * L1);
166 }
167
168 void build(int u, int l, int r) {
169     if (r - l < 2) {
170         double p = (double)a[l].first / a[l].nd;
171         seg[u].l1 = 1 - p;
172         seg[u].l2 = p;
173         seg[u].r1 = p;
174         seg[u].r2 = 1 - p;
175         return;
176     }
177     int mid = (l + r) >> 1;
178     build(lson, l, mid), build(rson, mid, r);
179     push_up(u);
180 }
181
182 void update(int u, int l, int r, int L, int R, int lhs, int rhs) {
183     if (l == L && r == R) {
184         double p = (double)lhs / rhs;
185         seg[u].l1 = 1 - p;
186         seg[u].l2 = p;
187         seg[u].r1 = p;
188         seg[u].r2 = 1 - p;
189         return;
190     }
191     int mid = (l + r) >> 1;
192     if (R <= mid) update(lson, l, mid, L, R, lhs, rhs);
193     else update(rson, mid, r, L, R, lhs, rhs);
194     push_up(u);
195 }
196
197 double __get(int x, int y) {
198     return query(1, 1, n + 1, x, y + 1);
199 }
200
201 void __set(int x, int y, int z) {
202     update(1, 1, n + 1, x, x + 1, y, z);
203 }
204
205 int main() {
206     //data_gen(); return 0;
207     //C(); return 0;
208     int debug = 0;
209     if (debug) freopen("in.txt", "r", stdin);
210     //freopen("out.txt", "w", stdout);
211     while (~scanf("%d%d", &n, &q)) {
212         FOR(i, 1, n) scanf("%d%d", &a[i].first, &a[i].nd);
213         build(1, 1, n + 1);
214         FOR(i, 1, q) {
215             int op, x, y, z;
216             scanf("%d%d%d", &op, &x, &y);
217             if (op == 1) {
218                 z = readint();
219                 __set(x, y, z);
220             } else {
221                 double ans = __get(x, y);
222                 printf("%.10f\n", ans);
223             }
224         }
225     }
226     return 0;
227 }