10311 - Goldbach and Euler
来源:互联网 发布:淘宝清仓活动入口 编辑:程序博客网 时间:2024/05/16 10:21
Problem L
Goldbach and Euler
Input: standard input
Output: standard output
Time Limit: 40 seconds
Memory Limit: 40 MB
“That every number which is resolvable into two prime numbers can be resolved into as many prime numbers as you like, can be illustrated and confirmed by an observation which you have formerly communicated to me, namely that every even number is a sum of two primes, and since (n-2) is also a sum of two prime numbers,n must be a sum of three, and also four prime numbers, and so on. If, however,n is an odd number, then it is certainly a sum of three prime numbers, since(n-1) is a sum of two prime numbers, and can therefore be resolved into as many prime numbers as you like. However, that every number is a sum of two primes, I consider a theorem which is quite true, although I cannot demonstrate it.”
-- Euler to Goldbach, 1742
The above conjecture about all numbers being the sum of two primes (where 1 counts as a prime) is not always true, but it is more true for even numbers. Your task is to test the conjecture for specified integers, considering that prime numbers are the numbers which are positive and divisible by exactly two positive integers. Your program must be very efficient.
Input
The input file contains 100000 lines of input. Each line contains a single integern (0<n<=100000000). Input is terminated by end of file.
Output
For each line of input produce one line of output. This line should be of one of the following types:
n is not the sum of two primes!n is the sum of p1 and p2.
For the second case, always make sure that (p2-p1) is positive and minimized.
Sample Input
11
12
Sample Output
11 is not the sum of two primes!12 is the sum of 5 and 7.
这道题坑爹的两个素数不能相同,害我wa了好久。。
#include<cstdio>#include<iostream>#include<cmath>using namespace std;int isprime(int a){ if(a==1) return 0; for(int i=2;i<=sqrt(a);i++) if(a%i==0) return 0; return 1;}int main(void){ int case1,n,left,right,flag; flag = 0; while(scanf("%d",&case1)!=EOF){ n = case1; if(case1==1){ printf("%d is not the sum of two primes!\n",n); goto end;} if(case1%2==0){ flag =0; for ( int i = case1 / 2; i >= 2; --i ) if ( isprime ( i ) && isprime ( case1 - i ) && i != ( case1 - i ) ) { left = i, right =case1 - i, flag = 1; break; } if(flag==1) printf("%d is the sum of %d and %d.\n",n,left,right); else{ printf("%d is not the sum of two primes!\n",n);} goto end; } case1 = case1 - 2;//奇数 if(isprime(case1)==1) printf("%d is the sum of 2 and %d.\n",n,case1); else printf("%d is not the sum of two primes!\n",n);end:;}}
- 10311 - Goldbach and Euler
- UVa 10311-Goldbach and Euler
- UVa 10311 - Goldbach and Euler
- UVa10311 - Goldbach and Euler
- Project Euler:Problem 46 Goldbach's other conjecture
- Goldbach
- Euler
- 【Project Euler】【Problem 1】Multiples of 3 and 5
- Project Euler 题解 #45 Triangular, pentagonal, and hexagonal
- [Project Euler]Problem 1:Multiples of 3 and 5
- Project Euler #1: Multiples of 3 and 5
- Codeforces 508D Tanya and Password 欧拉通路Euler
- Project Euler:Problem 45 Triangular, pentagonal, and hexagonal
- 1 Multiples of 3 and 5 - Project Euler
- Project Euler 001 Multiples of 3 and 5
- Project Euler Problem 1: Multiples of 3 and 5
- 长沙 Goldbach
- ZOJ3856 Goldbach
- 在线全部免费技术视频、在线查看API文档,
- uva 1330 - City Game(扫描法)
- 关于 "ubuntu-12.04.3-desktop-i386.iso" U盘制作后停留在syslinux版权声明界面问题解决
- round 172Div1 B
- SAMBA
- 10311 - Goldbach and Euler
- 阿里巴巴笔试题
- ActivityForResult
- 项目管理的6点经验学习
- OCP-1Z0-053-V12.02-584题
- final关键字修饰变量是引用不能变
- come chiamerai il tuo bebe scegli tra oltre nome 意大利小孩名字
- JS前端优化
- 通过EPROCESS获取进程名