#容斥，搜索，线性筛#CF83D Numbers

洛谷

CF83D

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分析

题意就是\(\sum_{i=l}^r[k|i]*[mn[\frac{i}{k}]\geq k]\)

首先线性筛每个数的最小质因数，如果\(\frac{r}{k}\)较小直接暴力

否则\(k\)一定比较小，那么直接容斥解决即可

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代码

#include <cstdio>
#include <cctype>
#include <algorithm>
#define rr register
using namespace std;
const int N=5000000;
int prime[N+101],v[N+101],cnt,Tot,ans,nn;
inline signed iut(){
	rr int ans=0; rr char c=getchar();
	while (!isdigit(c)) c=getchar();
	while (isdigit(c)) ans=(ans<<3)+(ans<<1)+(c^48),c=getchar();
	return ans;
}
inline void pro(){
	for (rr int i=2;i<=N;++i){
		if (!v[i]) v[i]=prime[++cnt]=i;
		for (rr int j=1;j<=cnt&&prime[j]<=N/i;++j){
			v[i*prime[j]]=prime[j];
			if (i%prime[j]==0) break;
		}
	}
}
inline bool Is_Prime(int k){
	if (k<=N) return v[k]==k;
	for (rr int i=1;i<=cnt;++i){
		if (prime[i]>k/prime[i]) break;
		if (k%prime[i]==0) return 0;
	}
	return 1;
}
inline signed brute(int n,int k){
	ans=1;
	for (rr int i=2;i<=n;++i) ans+=v[i]>=k;
	return ans;
}
inline void dfs(int dep,int now,int op){
	if (dep>Tot){
		ans+=op*(nn/now);
		return;
	}
	dfs(dep+1,now,op);
	if (now<=nn/prime[dep])
	    dfs(dep+1,now*prime[dep],-op);
}
inline signed Pro_DFS(int n,int k){
	Tot=lower_bound(prime+1,prime+1+cnt,k)-prime-1,
	ans=0,nn=n,dfs(1,1,1); return ans;
}
inline signed answ(int n,int k){
	if (n<k||!Is_Prime(k)) return 0;
	if (n/k<=N) return brute(n/k,k);
	    else return Pro_DFS(n/k,k);
}
signed main(){
	rr int l=iut(),r=iut(),k=iut(); pro();
	return !printf("%d",answ(r,k)-answ(l-1,k));
}