素數篩法

標籤：

線性篩更快。

1.埃氏篩法

1 int m=sqrt(n+0.5);2 memset(vis,0,sizeof(vis));3 for(int i=2;i<=m;i++)4     if(!vis[i])5     for(int j=i*i;j<=n;j+=i)6     vis[j]=1;

 

Write a program to read in a list of integers and determine whether or not each number is prime. A number, n, is prime if its only divisors are 1 and n. For this problem, the numbers 1 and 2 are not considered primes. 

Input

Each input line contains a single integer. The list of integers is terminated with a number<= 0. You may assume that the input contains at most 250 numbers and each number is less than or equal to 16000. 

Output

The output should consists of one line for every number, where each line first lists the problem number, followed by a colon and space, followed by "yes" or "no". 

Sample Input

1 2 3 4 5 17 0 

Sample Output

1: no 2: no 3: yes 4: no 5: yes 6: yes

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include<cmath> 6 using namespace std; 7 int n; 8 int vis[16005]; 9 void prime()10 {11     int m=sqrt(16000+0.5);12     memset(vis,0,sizeof(vis));13     for(int i=2;i<=m;i++)14         if(!vis[i])15         for(int j=i*i;j<=16000;j+=i)16         vis[j]=1;17     vis[1]=vis[2]=1;18 }19 int main()20 {21     prime();22     //printf("%d 123213",vis[4]);23     int cnt=0;24     while(~scanf("%d",&n))25  {26     if(n<=0)27         break;28     printf("%d: ",++cnt);29     if(vis[n]==0)30         printf("yes\n");31     if(vis[n]==1)32         printf("no\n");33  }34  return 0;35 }36 //137 //238 //339 //440 //541 //1742 //0

2.線性篩

 1     memset(vis,false,sizeof(vis)); 2     tot=0; 3     for(int i=2;i<=maxn;i++) 4     { 5         if(!vis[i]) 6           prime1[tot++]=i; 7         for(int j=0;j<tot;j++) 8         { 9             if(i*prime1[j]>maxn)10             break;11             vis[i*prime1[j]]=true;12             if(i%prime1[j]==0)13                 break;

Description

Goldbach‘s conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:

Every even integer, greater than 2, can be expressed as the sum of two primes [1].

Now your task is to check whether this conjecture holds for integers up to 107.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case starts with a line containing an integer n (4 ≤ n ≤ 107, n is even).

Output

For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of (a, b) where

1)      Both a and b are prime

2)      a + b = n

3)      a ≤ b

Sample Input

2

6

4

Sample Output

Case 1: 1

Case 2: 1

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include<cmath> 6 using namespace std; 7 #define maxn 10000000 8 bool vis[10000002]; 9 int prime1[1000020];10 int tot;11 void prime()12 {13     memset(vis,false,sizeof(vis));14     tot=0;15     for(int i=2;i<=maxn;i++)16     {17         if(!vis[i])18           prime1[tot++]=i;19         for(int j=0;j<tot;j++)20         {21             if(i*prime1[j]>maxn)22             break;23             vis[i*prime1[j]]=true;24             if(i%prime1[j]==0)25                 break;26         }27     }28 }29 int main()30 {31 32     int t,n;33     int ha=0;34     scanf("%d",&t);35     prime();36     while(t--)37     {38         int cnt=0;39         scanf("%d",&n);40         printf("Case %d: ",++ha);41         for(int i=0;prime1[i]<=n/2;i++)42             if(!vis[n-prime1[i]])43                 cnt++;44             printf("%d\n",cnt);45 46     }47  return 0;48 }

 

素數篩法