1 /**
2 给定一定范围求其内的素数
3
4 注意：
5 **/
6
7 #include <iostream>
8 #include <math.h>
9 #include <cstring>
10 using namespace std;
11 #define maxn 1000000
12 long long prime[500000];
13 long long cprime[500000];
14 long long isprime[maxn+100];
15 long long vis[maxn+100];
16 long long q;
17 void getprime(){
18 //memset(isprime,0,sizeof(isprime));
19 q =-1;
20 long long i,j;
21 isprime[0] = isprime[1] =1;
22 for(i=2;i<maxn;i++){
23 if(!isprime[i])
24 prime[++q] = i;
25 for(j=0;j<=q&&prime[j]*i<maxn;j++){
26 isprime[prime[j]*i] =1;
27 if(i%prime[j]==0)
28 break;
29 }
30 }
31 }
32
33 long long qt;
34 void getprime1(long long a, long long b){
35 qt =-1;
36 if(b<maxn){
37 for(long long i=a;i<=b;i++){
38 if(!isprime[i])
39 cprime[++qt] = i;
40 }
41 }else{
42 memset(vis,0,sizeof(vis));
43 long long i,k;
44 for(i=0;i<=b-a;i++){
45 vis[i] =1;
46 }
47 for(i=0;prime[i]*prime[i]<=b&&i<=q;i++){
48 k = a/prime[i];
49 if(k*prime[i]<a) k++;
50 if(k<=1) k++;
51 while(k*prime[i]<=b){
52 vis[k*prime[i]-a] =0;
53 k++;
54 }
55 }
56 for(i=0;i<=b-a;i++){
57 if(vis[i]==1)
58 cprime[++qt] = i+a;
59 }
60 }
61
62 }
63 int main()
64 {
65 getprime();
66 //for(int i=0;i<10;i++)
67 // cout<<prime[i]<<endl;
68 long long a,b;
69 while(cin>>a>>b){
70 long long maxnum =-1;
71 long long minnum =0x3f3f3f3f;
72 getprime1(a,b);
73 long long max1,max2,min1,min2;
74 // for(int i=0;i<=qt;i++)
75 // cout<<cprime[i]<<endl;
76 for(long long i=0;i<qt;i++){
77 long long temp = cprime[i+1]-cprime[i];
78 if(temp>maxnum){
79 maxnum =temp;
80 max1 = cprime[i];
81 max2 = cprime[i+1];
82 }
83 if(temp<minnum){
84 minnum =temp;
85 min1 = cprime[i];
86 min2 = cprime[i+1];
87 }
88 }
89 if((qt+1)<=1)
90 cout<<"There are no adjacent primes."<<endl;
91 else
92 cout<<min1<<","<<min2<<" are closest, "<<max1<<","<<max2<<" are most distant."<<endl;
93 }
94 return 0;
95 }
poj 2689 大范围内素数筛选
时间： 2024-11-06 15:35:37