description

analysis

• 头尾相差不到 $10^6$，自然而然想到暴力筛 $10^6$内的质数

• 然后 $[p,10^6]$内的质数暴力判断，拿并查集把模数相同的连一下维护就好了

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 1000005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define O3 __attribute__((optimize("-O3")))

using namespace std;

ll prime[MAXN],fa[MAXN<<1],f[MAXN<<1];
bool bz[MAXN<<1];
ll n,m,p,tot,ans;

O3 inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
O3 inline void init()
{
	memset(bz,1,sizeof(bz));
	fo(i,2,MAXN)
	{
		if (bz[i])prime[++tot]=i;
		for (reg j=1;j<=tot && i*prime[j]<=MAXN;++j)
		{
			bz[i*prime[j]]=0;
			if (i%prime[j]==0)break;
		}
	}
}
O3 inline ll getfa(ll x)
{
	return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
O3 inline void merge(ll x,ll y)
{
	fa[getfa(x)]=getfa(y);
}
O3 int main()
{
	//freopen("T4.in","r",stdin);
	n=read(),m=read(),p=read(),init();
	fo(i,n%MAXN,n%MAXN+(m-n+1))fa[i]=i;
	fo(j,1,tot)if (prime[j]>=p)
	{
		ll last=0;
		fo(i,n,m)if (i%prime[j]==0)
		{
			if (last)merge(n%MAXN+(last-n+1),n%MAXN+(i-n+1));
			last=i;
		}
	}
	memset(bz,1,sizeof(bz));
	fo(i,n%MAXN,n%MAXN+(m-n+1))fa[i]=getfa(fa[i]);
	fo(i,n%MAXN+1,n%MAXN+(m-n+1))ans+=bz[fa[i]],bz[fa[i]]=0;
	printf("%lld\n",ans);
	return 0;
}