【数学】Prime-Factor Prime

Prime-Factor Prime

题目描述

A positive integer is called a "prime-factor prime" when the number of its prime factors is prime. For example, 12 is a prime-factor prime because the number of prime factors of 12=2×2×3 is 3, which is prime. On the other hand, 210 is not a prime-factor prime because the number of prime factors of 210=2×3×5×7 is 4, which is a composite number.

In this problem, you are given an integer interval [l,r]. Your task is to write a program which counts the number of prime-factor prime numbers in the interval, i.e. the number of prime-factor prime numbers between l and r, inclusive.

输入

The input consists of a single test case formatted as follows.

l r
A line contains two integers l and r (1≤l≤r≤109), which presents an integer interval [l,r]. You can assume that 0≤r−l<1,000,000.

输出

Print the number of prime-factor prime numbers in [l,r].

样例输入

1 9

样例输出

4

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【题解】

区间素数筛即可解决。

【队友代码】

 1 //
 2 //  IniAully
 3 //
 4 //#pragma GCC optimize("Ofast,no-stack-protector")
 5 //#pragma GCC optimize("O3")
 6 #pragma GCC optimize(2)
 7 #include <bits/stdc++.h>
 8 #define inf 0x3f3f3f3f
 9 #define linf 0x3f3f3f3f3f3f3f3fll
10 #define pi acos(-1.0)
11 #define nl "\n"
12 #define pii pair<ll,ll>
13 #define ms(a,b) memset(a,b,sizeof(a))
14 #define FAST_IO ios::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL)
15 using namespace std;
16 typedef long long ll;
17 const ll mod = 1e9+7;
18 ll qpow(ll x, ll y){ll s=1;while(y){if(y&1)s=s*x%mod;x=x*x%mod;y>>=1;}return s;}
19 //ll qpow(ll a, ll b){ll s=1;while(b>0){if(b%2==1)s=s*a;a=a*a;b=b>>1;}return s;}
20 inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
21   
22 const int maxn = 1e5+5;
23 ll vis[maxn], prime[maxn], cnt;
24 void isprime()
25 {
26     memset(vis,0,sizeof(vis));
27     for(ll i=2;i<maxn;i++)
28     {
29         if(!vis[i])prime[cnt++]=i;
30         for(ll j=0;j<cnt && i*prime[j]<maxn;j++)
31         {
32             vis[i*prime[j]]=1;
33             if(i%prime[j]==0)break;
34         }
35     }
36     vis[1] = 1;
37     vis[0] = 1;
38 }
39   
40 const int N = 1e6+5;
41 ll bas[N];
42 int amo[N];
43   
44 int main()
45 {
46     int l, r;
47     isprime() ;
48   
49     scanf("%d%d",&l,&r);
50     for(int i=0;i<=r-l+5;i++) bas[i] = 1;
51     for(int i=0;i<cnt;i++)
52     {
53         int u = (l-1)/prime[i] + 1;
54         int v = r/prime[i];
55         for(int j=u;j<=v;j++)
56         {
57             int _ = j*prime[i], __=_;
58             while(_%prime[i]==0){
59                 amo[__-l+1]++;
60                 _ /= prime[i];
61                 bas[__-l+1] *= prime[i];
62             }
63         }
64     }
65     //cout<<1<<nl;
66     int ans = 0;
67     for(int i=l;i<=r;i++)
68     {
69         if(i<=3) continue;
70         //cout<<i<<" : "<<amo[i-l+1]<<nl;
71         if(bas[i-l+1] != i)amo[i-l+1] ++;
72         if(!vis[amo[i-l+1]]) ans ++;
73     }
74     cout<<ans<<nl;
75   
76     return 0;
77 }

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