CodeForces 621CWet Shark and Flowers

Posted 2020-06-13

题

There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i andi + 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 si. For i-th shark value si is random integer equiprobably chosen in range from li to ri. Wet Shark has it‘s favourite prime number p, and he really likes it! If for any pair of neighbouringsharks i and j the product si·sj 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 li and ri(1 ≤ li ≤ ri ≤ 109), the range of flowers shark i can produce. Remember that si is chosen equiprobably among all integers from li to ri, 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 .

Sample test(s)

input

3 2
1 2
420 421
420420 420421

output

4500.0

input

3 5
1 4
2 3
11 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 to421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (s0, s1, s2) each shark grows:

1. (1, 420, 420420): note that s0·s1 = 420, s1·s2 = 176576400, and s2·s0 = 420420. For each pair, 1000dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000dollars.
2. (1, 420, 420421): now, the product s2·s0 is not divisible by 2. Therefore, sharks s0 and s2 will receive 1000dollars, while shark s1 will receive 2000. The total is 4000.
3. (1, 421, 420420): total is 4000
4. (1, 421, 420421): total is 0.
5. (2, 420, 420420): total is 6000.
6. (2, 420, 420421): total is 6000.
7. (2, 421, 420420): total is 6000.
8. (2, 421, 420421): total is 4000.

The expected value is .

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

题意

n个人围成环，如果两个相邻人的数字相乘后可以整除p，那么这两个人都得到1000元，现在给我们每个人的数字的范围，让我们求所有人得到的钱的期望。

分析

只要两个相邻的人数字中有一个可以整除那就两个人都能得到1000，我们求出每个人不能整除p的概率pp[i]=1-(r/p-l/p+(l%p==0))/(r-l+1)，因为这里是整除，所以r/p就是1到r有几个p的倍数，当l是p的倍数时，那就再加1，因为你减去的是1到l里p的倍数的个数，总个数是r-l+1。

然后两个邻居可以得到的概率就是只有两个人都不能整除时，就不能得到钱，即1-pp[i]*pp[i%n+1]，概率累加起来再乘上2000就好了。

代码

#include <stdio.h>
long long n,p,l,r;
double pp[100005],ans;
int main()
{
    scanf("%lld%lld",&n,&p);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&l,&r);
        pp[i]=1-(double)(r/p-l/p+(l%p==0))/(r-l+1);
    }
    for(int i=1;i<=n;i++)
        ans+=1-pp[i]*pp[i%n+1];
    printf("%lf",ans*2000);
    return 0;
}