Race Week flowers solution to a problem (number theory)

Topic Link

Subject to the effect

Individual flowers n (cyclic) to give the number of flowers each section may, if the product of the number of adjacent two flowers is a multiple of p, each of the two obtained 1000,

Demand expectations for all wages. If the answer is a / b is written as a * inv [b]

Topic ideas

1: How to judge [L, R] middle how many multiples of p, but not very easy to want to think about or know is actually R / p- (L-1) / p

2: The subject is not necessary to obtain a final probability, and so with the least common multiple operating directly inverse to solving

3: n is 1 and 2 will have to judge Patent, Laid attention judgment can not a [n + 1] = 0, as this is the inverse operation, directly reduce the cycle judged Laid

Code

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=1e6+5;
int n,p,l,r; 
ll a[maxn],ans;
int extgcd(int a,int b,int &x,int &y){//扩欧 
	int d=a;
	if(b!=0){
		extgcd(b,a%b,y,x);
		y-=(a/b)*x;
	}else{
		x=1,y=0;
	}
	return d;
} 
int main(){
    scanf("%d %d",&n,&p);
  	for(int i=1;i<=n;i++){
  		scanf("%d %d",&l,&r);	
		int x,y;
		extgcd(r-l+1,mod,x,y);
		x=(mod+x%mod)%mod;
		a[i]=(((long long)(r/p-(l-1)/p))*x)%mod;
	}
	a[n+1]=a[1];//选择n和1 
	if(n==1&&n==2){//特判必须这么特判不能给a[n+1]赋值为0 
		n--;
	}	
	for(int i=1;i<=n;i++){
		ans=((mod+1-(mod+1-a[i])*(mod+1-a[i+1])%mod)%mod+ans)%mod;//这个我觉得最好就是先写出平常的式子，然后再加上mod 
	}
	ans=ans*2000%mod;
	printf("%lld",ans);
    return 0;
}