C. Wet Shark and Flowers
time limit per test
memory limit per test
input
output
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
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
At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.
Input
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
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
10 - 6.
a, and the answer of the jury is b. The checker program will consider your answer correct, if
.
Examples
input
3 21 2 420 421 420420 420421
output
4500.0
input
3 51 4 2 3 11 14
output
0.0
[l, r]之间p的倍数的个数为cnt = r /p - l / p; if l % p == 0, cnt++;对于两个相邻的sharks, 他们产生的flowers之积为p的倍数的情况数为node[i].cnt * node[j].len + node[j].cnt * (node[i].len - node[i].cnt)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;
struct Node{
double len;
int cnt;
}node[maxn];
int main(){
// freopen("in.txt", "r", stdin);
int n, p, l, r;
scanf("%d%d", &n, &p);
for(int i = 0; i < n; i++){
scanf("%d%d", &l, &r);
node[i].cnt = r / p - l / p;
if(l % p == 0)
node[i].cnt++;
node[i].len= r - l + 1;
}
double pp = 0;
for(int i = 0; i < n; i++){
int j = (i + 1) % n;
pp += node[i].cnt / node[i].len;
pp += node[j].cnt * (node[i].len - node[i].cnt) / node[i].len / node[j].len;
}
printf("%.12lf\n", pp * 2000);
return 0;
}
