No.907 Continuous Kadomatu
Problem Statement
Suppose a sequence of "real numbers" $a_1,\ldots,a_N$ of length $N$ is constructed by the following method. Find the probability that the constructed sequence is a Kadomatsu sequence.
- Choose a "real number" between $A_i$ and $B_i$, inclusive, uniformly at random, and let it be $a_i$.
A sequence $a_1,\ldots,a_N$ is called a Kadomatsu sequence if $a_1 \le a_2 \ge a_3 \le a_4 \ge a_5 \le \cdots $. That is, the following holds for all $1 \le i \le N-1$.
- If $i$ is odd, then $a_i \le a_{i+1}$.
- If $i$ is even, then $a_i \ge a_{i+1}$.
Note that the constructed sequence is not an integer sequence but a "real sequence".
Input
$N$ $A_1\ B_1$ $\vdots$ $A_N\ B_N$
$3 \le N \le 200$
$1 \le A_i < B_i \le 10^9$
All inputs are integers.
Output
The answer can be expressed as an irreducible fraction $p/q$, so output $p\,q^{-1} \bmod{10^9+7}$. That is, output the smallest non-negative integer $x$ satisfying $xq \equiv p \pmod{10^9+7}$. In this problem, this value always exists.
Sample
Sample 1
Input
3 1 3 1 3 1 3
Output
333333336
The answer is $1/3$. This can be rephrased as the probability that the second value is the largest when three real numbers are randomly generated, which is $1/3$ by symmetry. Note that the constructed sequence is not an integer sequence but a "real sequence".
Sample 2
Input
3 1 2 2 3 1 2
Output
1
The answer is $1/1$.
Sample 3
Input
3 2 3 1 2 2 3
Output
0
The answer is $0/1$. The only possible Kadomatsu sequence is $2,2,2$, but the probability of it appearing is $0$.
Sample 4
Input
5 1 4 2 3 1 3 2 4 1 4
Output
715740746
The answer is $269/1080$.
Sample 5
Input
9 766526787 974319581 941243410 942612052 427056305 964028266 313356130 664522966 480128631 531723103 1626524 615851774 293907611 496577233 1955784 467846274 83687962 802560435
Output
175278186
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