[暑假集训--数论]poj2262 Goldbach's 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
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
把n>=6分成两个奇质数之和,小的那个要尽量小
直接暴力
#include<cstdio>
#include<iostream>
#include<cstring>
#define LL long long
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n;
bool mk[];
int p[],len;
inline void getp()
{
for (int i=;i<=;i++)
{
if (!mk[i])
{
p[++len]=i;
for (int j=*i;j<=;j+=i)mk[j]=;
}
}
}
int main()
{
getp();
while (~scanf("%d",&n)&&n)
{
if (n&||n<){puts("Goldbach's conjecture is wrong.");continue;}
for (int i=;p[i]*<=n;i++)
if (!mk[n-p[i]]){printf("%d = %d + %d\n",n,p[i],n-p[i]);break;}
}
}
poj 2262
[暑假集训--数论]poj2262 Goldbach's Conjecture的更多相关文章
- [暑假集训--数论]poj2909 Goldbach's Conjecture
For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 ...
- poj2262 Goldbach's Conjecture
poj2262 Goldbach's Conjecture 用欧拉筛把素数筛出来,再枚举一下. #include<iostream> #include<cstdio> #inc ...
- [POJ2262] Goldbach’s Conjecture
Goldbach's Conjecture Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 48161 Accepted: ...
- [暑假集训--数论]hdu2136 Largest prime factor
Everybody knows any number can be combined by the prime number. Now, your task is telling me what po ...
- [暑假集训--数论]hdu1019 Least Common Multiple
The least common multiple (LCM) of a set of positive integers is the smallest positive integer which ...
- [暑假集训--数论]poj2115 C Looooops
A Compiler Mystery: We are given a C-language style for loop of type for (variable = A; variable != ...
- [暑假集训--数论]poj1365 Prime Land
Everybody in the Prime Land is using a prime base number system. In this system, each positive integ ...
- [暑假集训--数论]poj2034 Anti-prime Sequences
Given a sequence of consecutive integers n,n+1,n+2,...,m, an anti-prime sequence is a rearrangement ...
- [暑假集训--数论]poj1595 Prime Cuts
A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In ...
随机推荐
- oc字符串截取 数组字典运用
#define NSLog(FORMAT, ...) printf("%s\n", [[NSString stringWithFormat:FORMAT, ##__VA_ARGS_ ...
- C#冒泡排序程序
考虑到很多面试可能会考察冒泡排序的用法,所以特地花时间厘清了一下思路.下面说一下我的思路:冒泡排序核心就是比较方法,冒泡排序的比较方法顾名思义就是像气泡一样,最大(或者最小)的数往上冒.普通比较几个数 ...
- XML格式与实体类的转换
背景 本人头一回写博客,请大家多多关照.通过读取XML文件获取用户管理权限,其中涉及三部分: 1.XML文件的生成: 2.XML文件的读取: 3.XML文件的保存: 如何做 第一步:自己先将XML文件 ...
- IOS ViewTable
// // ViewController.swift // UITableView // // Created by lanou on 16/11/7. // Copyright (c) 20 ...
- cf492E. Vanya and Field(扩展欧几里得)
题意 $n \times n$的网格,有$m$个苹果树,选择一个点出发,每次增加一个偏移量$(dx, dy)$,最大化经过的苹果树的数量 Sol 上面那个互素一开始没看见,然后就GG了 很显然,若$n ...
- inotifywait实时监控文件目录
一.inotify简介 inotify 是一种强大的.细粒度的.异步文件系统监控机制,它满足各种各样的文件监控需要,可以监控文件系统的访问属性.读写属性.权限属性.创建删除.移动等操作,也可以监控文件 ...
- 【JAVA】apachehttpclient设置http1.0短链接
HttpPost httpPost = new HttpPost(url);httpPost.setEntity(new StringEntity(text, ContentType.create(& ...
- 动态规划:HDU-1398-Square Coins(母函数模板)
解题心得: 1.其实此题有两种做法,动态规划,母函数.个人更喜欢使用动态规划来做,也可以直接套母函数的模板 Square Coins Time Limit: 2000/1000 MS (Java/Ot ...
- IQueryable与IEnumerable区别
前者可以延迟加载,即执行完后不马上执行数据库语句,用到再加载.
- Android开发——网易云音乐使用的开源组件集合
前言 网易云音乐Android版从第一版使用到现在,全新的 Material Design 界面,更加清新.简洁.同样也是音乐播放器开发者,我们确实需要思考,相同的功能,会如何选择.感谢开源,让我们有 ...