poj2262 简单素数判定
来源:互联网 发布:分辨率300的软件 编辑:程序博客网 时间:2024/06/07 05:48
Goldbach's Conjecture
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 45296 Accepted: 17255
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
For example:
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.
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.
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
820420
Sample Output
8 = 3 + 520 = 3 + 1742 = 5 + 37
题意:输入一个数,将其写成俩个素数相加的形式。
很简单的一道题,只要将第一次出现的符合条件的情况输出即可。
代码:
#include<iostream>#include<cmath>#include<cstdio>using namespace std;bool isodd(int x){ int y=sqrt(x+0.5); for(int i=2;i<=y;i++) { if(x%i==0) { return false; } } return true;}int main(){ int n; while(scanf("%d",&n)&&n!=0) { for(int i=3;i<=n-3;i++) { if(isodd(i)&&isodd(n-i)) { printf("%d = %d + %d\n",n,i,n-i); break; } } } return 0;}
0 0
- poj2262 简单素数判定
- POJ2262 素数判定,万能的素数筛选
- POJ2262 素数筛法
- POJ2262 素数筛
- poj2262 - 素数判断
- poj2262 筛选法求素数
- poj2262裸的素数筛
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- 素数判定
- mysql view 更新问题
- P1006 传纸条
- 蓝桥杯省赛12
- 最大连续子序列求和详解
- 利用VS2013在win7 64位机器上搭建xgboost 0.6+Anaconda3 环境
- poj2262 简单素数判定
- Spring MVC学习(六)-------注解式控制器详解1
- ueditor二次加载(getEditor)渲染失败(加载失败)的原因解决方案
- mooc_排序
- POJ1651Multiplication Puzzle(区间dp)
- ZOJ 3705 Applications(模拟题)
- 搜索结果处理——变色
- softmax回归
- HDU1247:Hat’s Words(字典树)