• Codeforces Round #270 A. Design Tutorial: Learn from Math【数论/埃氏筛法】


    time limit per test
     1 second
    memory limit per test
     256 megabytes
    input
     standard input
    output
     standard output

    One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.

    For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.

    You are given an integer n no less than 12, express it as a sum of two composite numbers.

    Input

    The only line contains an integer n (12 ≤ n ≤ 106).

    Output

    Output two composite integers x and y (1 < x, y < n) such that x + y = n. If there are multiple solutions, you can output any of them.

    Sample test(s)
    input
    12
    
    output
    4 8
    
    input
    15
    
    output
    6 9
    
    input
    23
    
    output
    8 15
    
    input
    1000000
    
    output
    500000 500000
    
    Note

    In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.

    In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.

    【注意】:埃氏筛法筛素数时,注意开头i*2非i*i(数字大爆int)

    【代码】:

    #include<bits/stdc++.h>
    using namespace std;
    int prime[1001000];
    void get_prime()
    {
        prime[1]=prime[0]=1;//0,1非素数
        for(int i=2;i<=1000010;i++)
        {
            if(prime[i]==0)
                for(int j=i*2;j<=1000010;j+=i)
                   prime[j]=1;
        }
    }
    int main()
    {
        int n;
        get_prime();
        cin>>n;
        for(int i=2;i<=n;i++)
            if( prime[i]&& prime[n-i] )
            {
                printf("%d %d
    ",i,n-i);
                return 0;
            }
        return 0;
    }
    筛法
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  • 原文地址:https://www.cnblogs.com/Roni-i/p/7954057.html
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