LightOJ-1197 Help Hanzo

Posted on | Edited on | In | Views:

题意

求 $a\sim b$ 间素数个数 $(1 \le a \le b < 2^{31}, b - a \le 1e5)$

题解

$b-a$ 这个区间比较小，所以可以用区间素数筛选的办法解决这个题目。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define inf 0x3f3f3f3f
#define mod (int)1e9+7
#define for0(i, n) for (int i = 0; i < n; i++)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forn(i, n) for (int i = n; i >= 0; i--)
#define mem0(a) memset(a, 0, sizeof(a))
#define mem1(a) memset(a, 1 , sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define meminf(a) memset(a, inf, sizeof(a))
using namespace std;

const int N=100005;
bool a[N+1];
vector<int> prime;
void getPrime(){
    memset(a,true,sizeof(a));
    for(int i=2;i<=N;i++){
        if(a[i]) prime.push_back(i);
        for(int j=0;j<(int)prime.size()&&i*prime[j]<=N;j++){
            a[i*prime[j]]=false;
            if(!(i%prime[j])) break;
        }
    }
}
int main(){
  ll t,a,b;
  bool isprime[100005];
  getPrime();
  cin>>t;
  for1(k,t){
    int ans=0;
    cin>>a>>b;
    mem1(isprime);
    for(int i=0;prime[i]*prime[i]<=b&&i<(int)prime.size();i++){
      ll l=a/prime[i];
      if (l*prime[i]<a) l++;
      if (l<2) l=2;
      for (; l*prime[i]<=b; l++) isprime[l*prime[i]-a]=0;
    }
    if (a==1) isprime[0]=0;
    for0(i,b-a+1)
    if (isprime[i])ans++;
    printf("Case %d: %d\n",k,ans );
  }
  return 0;
}