zoukankan      html  css  js  c++  java
  • [暑假集训--数论]poj2262 Goldbach's Conjecture

    In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture: 
    Every even number greater than 4 can be 
    written as the sum of two odd prime numbers.

    For example: 
    8 = 3 + 5. Both 3 and 5 are odd prime numbers. 
    20 = 3 + 17 = 7 + 13. 
    42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

    Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.) 
    Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million. 

    Input

    The input will contain one or more test cases. 
    Each test case consists of one even integer n with 6 <= n < 1000000. 
    Input will be terminated by a value of 0 for n.

    Output

    For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

    Sample Input

    8
    20
    42
    0
    

    Sample Output

    8 = 3 + 5
    20 = 3 + 17
    42 = 5 + 37

    把n>=6分成两个奇质数之和,小的那个要尽量小

    直接暴力

     1 #include<cstdio>
     2 #include<iostream>
     3 #include<cstring>
     4 #define LL long long
     5 using namespace std;
     6 inline LL read()
     7 {
     8     LL x=0,f=1;char ch=getchar();
     9     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    10     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    11     return x*f;
    12 }
    13 int n;
    14 bool mk[1000010];
    15 int p[1000010],len;
    16 inline void getp()
    17 {
    18     for (int i=2;i<=1000000;i++)
    19     {
    20         if (!mk[i])
    21         {
    22             p[++len]=i;
    23             for (int j=2*i;j<=1000000;j+=i)mk[j]=1;
    24         }
    25     }
    26 }
    27 int main()
    28 {
    29     getp();
    30     while (~scanf("%d",&n)&&n)
    31     {
    32         if (n&1||n<6){puts("Goldbach's conjecture is wrong.");continue;}
    33         for (int i=2;p[i]*2<=n;i++)
    34             if (!mk[n-p[i]]){printf("%d = %d + %d
    ",n,p[i],n-p[i]);break;}
    35     }
    36 }
    poj 2262
  • 相关阅读:
    以太坊虚拟机介绍
    以太坊源码学习 – EVM
    Visual Studio Code 常用快捷键
    深入了解以太坊虚拟机第4部分——ABI编码外部方法调用的方式
    深入了解以太坊虚拟机第3部分——动态数据类型的表示方法
    深入了解以太坊虚拟机第2部分——固定长度数据类型的表示方法
    深入了解以太坊虚拟机
    选取文档元素的方法
    flex布局
    什么是功能需求设计文档
  • 原文地址:https://www.cnblogs.com/zhber/p/7285419.html
Copyright © 2011-2022 走看看