Goldbach's Conjecture
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
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
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
Sample Input
8
20
42
0
Sample Output
8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
bool isprime ( int k )
{
int t = sqrt ( k + 0.5 ) ;
for ( int i = ; i <= t ; i ++ )
if ( k % i == )
return false ;
return true ;
}
int main()
{
// freopen ("a.txt" , "r" , stdin );
int n ;
while ( scanf ("%d", &n) , n )
{
int i ;
int t = n / ;
for ( i = ; i <= t ; i += )
if ( isprime ( i ) && isprime ( n - i ) )
break ;
printf ( "%d = %d + %d\n" , n , i , n - i ) ;
}
return ;
}
n = isprime(i) + isprime(n - i)
Goldbach's Conjecture的更多相关文章
- Poj 2262 / OpenJudge 2262 Goldbach's Conjecture
1.Link: http://poj.org/problem?id=2262 http://bailian.openjudge.cn/practice/2262 2.Content: Goldbach ...
- poj 2262 Goldbach's Conjecture(素数筛选法)
http://poj.org/problem?id=2262 Goldbach's Conjecture Time Limit: 1000MS Memory Limit: 65536K Total ...
- HDOJ 1397 Goldbach's Conjecture(快速筛选素数法)
Problem Description Goldbach's Conjecture: For any even number n greater than or equal to 4, there e ...
- Goldbach's Conjecture(哥德巴赫猜想)
Goldbach's Conjecture Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Ot ...
- UVa 543 - Goldbach's Conjecture
题目大意:给一个偶数,判断是否是两个素数的和. 先用sieve方法生成一个素数表,然后再进行判断即可. #include <cstdio> #include <vector> ...
- 【LightOJ1259】Goldbach`s Conjecture(数论)
[LightOJ1259]Goldbach`s Conjecture(数论) 题面 Vjudge T组询问,每组询问是一个偶数n 验证哥德巴赫猜想 回答n=a+b 且a,b(a<=b)是质数的方 ...
- POJ 2262 Goldbach's Conjecture (打表)
题目链接: https://cn.vjudge.net/problem/POJ-2262 题目描述: In 1742, Christian Goldbach, a German amateur mat ...
- 题目1440:Goldbach's Conjecture(哥达巴赫猜想)
题目链接:http://ac.jobdu.com/problem.php?pid=1440 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: ...
- Goldbach`s Conjecture(素筛水题)题解
Goldbach`s Conjecture Goldbach's conjecture is one of the oldest unsolved problems in number theory ...
随机推荐
- 3.SQLAlchemy文档-SQLAlchemy Core(中文版)
这里的文描述了关于SQLAlchemy的的SQL渲染引擎的相关内容,包括数据库API的集成,事务的集成和数据架构描述服务.与以领域为中心的ORM使用模式相反,SQL表达式语言提供了一个数据构架为中心的 ...
- C# GC 垃圾回收机制
今天来谈谈C# 的GC ,也就是垃圾回收机制,非常的受教,总结如下 首先:谈谈托管,什么叫托管,我的理解就是托付C# 运行环境帮我们去管理,在这个运行环境中可以帮助我们开辟内存和释放内存,开辟内存一般 ...
- javascript 事件传播与事件冒泡,W3C事件模型
说实话笔者在才工作的时候就听说了什么"事件冒泡",弄了很久才弄个大概,当时理解意思是子级dom元素和父级dom元素都绑定了相同类型的事件,这时如果子级事件触发了父级也会触发,然后这 ...
- Bootstrap系列 -- 33. 等分按钮
等分按钮也常被称为是自适应分组按钮,其实现方法也非常的简单,只需要在按钮组“btn-group”上追加一个“btn-group-justified”类名. <div class="bt ...
- [BZOJ1070][SCOI2007]修车(最小费用最大流)
题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1070 分析: 把每个工人拆成N个点.记为A[i,j]表示第i个工人修倒数第j辆车. 每 ...
- Boostrap(1)
1.简介 Bootstrap 是一个用于快速开发 Web 应用程序和网站的前端框架.Bootstrap 是基于 HTML.CSS.JAVASCRIPT 的,可以认为bootstrap就是一个样式库. ...
- margin的理解
1.盒子模型 在进行网页设计的时候,我们使用的是盒子模型,其内容如下: 整个网页就是大盒子套小盒子,小盒子又套更小的盒子来实现的.但是在做网页设计时总是搞不清margin和padding的使用方式,在 ...
- python 逐行读取文件的三种方法
方法一: 复制代码代码如下: f = open("foo.txt") # 返回一个文件对象 line = f.readline() ...
- beta汇总
第一天:http://www.cnblogs.com/hxh969012806/p/5034085.html 第二天:http://www.cnblogs.com/zyk150910/p/503783 ...
- The web application [/codeMarket] registered the JBDC driver[.........] but failed to unregister it when the web application was stopped. To prevent
如果你报错了上面的这个严重,那么你的tomcat版本一定是在6.0.25之上的 原因:tomcat 6.025以后在sever.xml中引入了内存泄露侦测,对于垃圾回收不能处理的对像,它就会做日志. ...