Source：http://codeforces.com/problemset/problem/621/C
Description
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 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 neighbouring sharks 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
Examples
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 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.(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.
2.(1, 420, 420421): now, the product s2·s0 is not divisible by 2. Therefore, sharks s0 and s2 will receive 1000 dollars, 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.
题目大意不难读懂，关键得知道这题求的是期望值。
思路： 此题要求l~r区间内是p的倍数的数的个数，一般思路会想到用前缀和，但此处数据范围为1e9，无法开这么大的数组，故我们需另寻他法。注意到这里的p保证为素数，故可以想到通过(r/p - (l-1)/p)来求得区间内符合条件数的个数。然后求出相邻两组符合条件的对数，再根据相关公式算出每组期望，最后相加即可。
ACODE:
/7777777
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<string>
#include<queue>
#include<utility>
#include<vector>
#include<map>
#include<set>
#define lson l , m , rt << 1
#define rson m+1 , r , rt << 1 | 1
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const double pi = 3.1415926535;
const double eps = 1e-6;
const int MXN = 1e5 + 7;
const int MXM = 1e3 + 7;
const int MX = 1e9 + 7;
const int maxbit = 18;
const double val = pi/180.0;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e4 + 7;
ll p,n;
double ans = 0;
struct node
{
ll t;//区间内符合条件数的个数
ll l;//区间长度
}a[MXN];
int main()
{
scanf("%lld%lld",&n,&p);
ll x,y;
for(int i = 1;i <= n;++i)
{
scanf("%lld %lld",&x,&y);
//类似前缀和的求法 故为(x-1)!!!
a[i].t = (y)/p - (x-1)/p;
a[i].l = y - x + 1;
//printf("%d %d\n",a[i].t,a[i].l);
}
for(int i = 1;i < n;++i)
{
//两种需要考虑的情况数（注意避免重复）
ll r1 = a[i].t * a[i+1].l;
ll r2 = a[i+1].t * (a[i].l - a[i].t);
//依次求得每组的期望并累加
ans += (double)(r1 + r2) * 2000.0 / (a[i].l * a[i+1].l);
//printf("%.2lf\n",ans);
}
//单独求最后一组
ll r1 = a[n].t * a[1].l;
ll r2 = a[1].t * (a[n].l - a[n].t);
ans += (double)(r1 + r2) * 2000.0 / (a[n].l * a[1].l);
//此题输出%lf即可保证精度
printf("%lf",ans);
return 0;
}