实际上题目要求的
也就是说是筛的时候每次枚举质因子后
当前枚举的质数是次大质因子的
也就是内的所有质数
预处理出质数个数,筛一下就完了
不会的看这个
#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define re register
#define cs const
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
cs int N=1000005;
ll f1[N],f2[N],l,r,ans,n;
int pr[N],tot,lim;
inline void init(ll k){
if(k<=1)return;
lim=sqrt(k),tot=0;
for(int i=1;i<=lim;i++)f1[i]=i-1,f2[i]=k/i-1;
for(int p=2;p<=lim;p++){
if(f1[p]==f1[p-1])continue;
pr[++tot]=p;
for(int i=1;i<=lim/p;i++)f2[i]-=(f2[i*p]-f1[p-1]);
for(int i=lim/p+1;1ll*i*p*p<=k&&i<=lim;i++)f2[i]-=(f1[k/i/p]-f1[p-1]);
for(int i=lim;i>=1ll*p*p;i--)f1[i]-=(f1[i/p]-f1[p-1]);
}
}
inline ll F(ll x){
return (x<=lim)?f1[x]:f2[n/x];
}
inline void dfs(int pos,ll res){
if(res<pr[pos])return;
for(int i=pos;i<=tot;i++){
if(1ll*pr[i]*pr[i]>res)return;
for(ll now=pr[i],xs=1;now<=res;now*=pr[i],xs++){
if(now*pr[i]<=res)dfs(i+1,res/now),ans+=pr[i]*(F(res/now)-i+1);
}
}
}
inline ll calc(ll x){
n=x;
init(x);
ans=0;
dfs(1,x);
return ans;
}
signed main(){
scanf("%lld%lld",&l,&r);
cout<<calc(r)-calc(l-1)<<'
';
}