【Codeforces Round 370 (Div 2) E】【线段树 等比数列 区间合并】Memory and Casinos 赌场区间[l,r] l进r先出的概率

E. Memory and Casinos
time limit per test
4 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi 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... j if he completes a walk such that,

  • He starts on casino i.
  • He never looses in casino i.
  • 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 .
  • 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... 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. pi ≤ pi + 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 ≤ n, q ≤ 100 000),  — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109)  —  is the probability pi 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 |a - b| ≤ 10 - 4.

Example
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
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

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define ff first
#define ss second
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
typedef pair PD;
template inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template inline void gmin(T1 &a, T2 b) { if (b> 1;
	build(lson);
	build(rson);
	a[o] = merge(a[ls], a[rs]);
}
PD V; int P;
void update(int o, int l, int r)
{
	if (l == r)
	{
		a[o] = V;
		return;
	}
	int mid = (l + r) >> 1;
	if (P <= mid)update(lson);
	else update(rson);
	a[o] = merge(a[ls], a[rs]);
}
int L, R;
PD query(int o, int l, int r)
{
	if (l >= L&&r <= R)return a[o];
	int mid = (l + r) >> 1;
	if (R <= mid)return query(lson);
	else if (L > mid)return query(rson);
	else return merge(query(lson), query(rson));
}
int main()
{
	while (~scanf("%d%d", &n, &q))
	{
		build(1, 1, n);
		while (q--)
		{
			int op; scanf("%d", &op);
			if (op == 1)
			{
				double x, y;
				scanf("%d%lf%lf", &P, &x, &y);
				V = { x / y,x / y };
				update(1, 1, n);
			}
			else
			{
				scanf("%d%d", &L, &R);
				printf("%.12f\n", query(1, 1, n).ff);
			}
		}
	}
	return 0;
}
/*
【题意】
有n(1e5)个赌场,我们一开始在1号赌场,
在第i个赌场赌赢的概率为p[i]
如果赌赢了,会进入第i+1个赌场
如果赌输了,会进入第i-1个赌场

有m(1e5)个询问,
问你,对于询问区间[l,r]
如果我们从l位置开始赌博,第一次走出[l,r]区间是因为r赢而不是因为l输的概率是多少。

【类型】
线段树 等比数列 区间合并

【分析】
我们要求的是对于区间[l,r],从l位置开始赌博,第一次走出[l,r]区间是因为r赢而不是因为l输的概率是多少。
我们设其为L[l,r],显然第一次是从[l,r]的l输走出区间的概率是1-L[l,r]

这个问题乍一想很难,但是我们可以从中捕捉到区间合并的影子。
我们考虑已经直到了[l,mid]和[mid+1,r]两个区间内部的属性,并尝试合并。

我们把L[l,mid]设为l1,把L[mid+1,r]设为l2.
那么,我们考虑从mid->mid+1经过了1次,2次,3次,...
其概率公式分别是
l1 * l2
l1 * (1-l2) * probability of [ mid处进入[l,mid],mid处第一次出[l,mid]]

于是我们需要:
对于区间[l,r],从r位置开始赌博,第一次走出[l,r]区间是因为r赢而不是因为l输的概率为R[l,r]。
把R[l,mid]设为r1,把R[mid+1,r]设为r2.
那么之前的概率公式变成了——
l1 * l2
l1 * (1-l2) * r1 * l2
l1 * ((1-l2) * r1)^2 * l2
...
a1=l1*l2
q=((1-l2)*r1)
无穷多项求和=a1/(1-q)

我们还要求R
我们考虑从mid->mid+1经过了0次,1次,2次,3次,...
其概率公式分别是
r2
(1-r2) * r1 * l2
(1-r2) * r1 * (1 - l2) * r1 * l2
(1-r2) * r1 * ((1 - l2) * r1)^2 * l2
一样求和即可

于是对于询问,我们可以用线段树维护区间L和R并合并,最后得到答案。

【时间复杂度&&优化】
O(qlogn)

*/


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