Codeforces Round #341 (Div. 2) C. Wet Shark and Flowers

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There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i and i + 1 are neighbours for all i from 1 to n - 1. Sharks n and 1 are neighbours too.

Each shark will grow some number of flowers s_i. For i-th shark value s_i is random integer equiprobably chosen in range from l_i to r_i. Wet Shark has it's favourite prime number p, and he really likes it! If for any pair of neighbouring sharks i and j the product s_i·s_j is divisible by p, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.

At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.

Input

The first line of the input contains two space-separated integers n and p (3 ≤ n ≤ 100 000, 2 ≤ p ≤ 10⁹) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that p is prime.

The i-th of the following n lines contains information about i-th shark — two space-separated integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10⁹), the range of flowers shark i can produce. Remember that s_i is chosen equiprobably among all integers from l_i to r_i, inclusive.

Output

Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10^- 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $\text{[math]}$ .

Sample test(s)

input

3 21 2420 421420420 420421

output

4500.0

input

3 51 42 311 14

output

0.0

Note

A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.

Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (s₀, s₁, s₂) each shark grows:

(1, 420, 420420): note that s₀·s₁ = 420, s₁·s₂ = 176576400, and s₂·s₀ = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.
(1, 420, 420421): now, the product s₂·s₀ is not divisible by 2. Therefore, sharks s₀ and s₂ will receive 1000 dollars, while shark s₁will receive 2000. The total is 4000.
(1, 421, 420420): total is 4000
(1, 421, 420421): total is 0.
(2, 420, 420420): total is 6000.
(2, 420, 420421): total is 6000.
(2, 421, 420420): total is 6000.
(2, 421, 420421): total is 4000.

The expected value is $\text{[math]}$ .

In the second sample, no combination of quantities will garner the sharks any money.

题意是给定n个范围，在每个范围中选取一个数，相邻的如第i和i -1以及i和i+1这两个范围中如果有这两个数相乘是给定的一个数p的倍数的话，那么这组选取加2000，最后要求取总价值与选取个数的除数。

这个题抛开题意，不要想选取的问题，因为最后的选取一定是都加起来的，所以我们只需要将每个加2000的概率算出来乘以2000就好。这里需要注意边界的处理和最后的输出的范围，代码如下：

#include <bits/stdc++.h>using namespace std;int n;long long p;double num[100010];int main(void){    cin >> n >> p;    long long x, y;    for(int i = 0; i < n; i ++){        scanf("%I64d %I64d", &x, &y);        num[i] = (y / p - (x - 1)/ p) * 1.0  / (y + 1 - x);    }    long double ans = 0;    for(int i = 0; i < n; i ++){        int j = (i + 1) % n;        ans += 1 - (1 - num[i]) * (1 - num[j]);    }    cout << fixed << setprecision(15) << ans * 2000.0 << endl;}

查看原文：http://chilumanxi.org/2016/01/31/codeforces-round-341-div-2-c-wet-shark-and-flowers/

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