Goldbach‘s Conjecture
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 48161 | Accepted: 18300 |
Description
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
Source
提交地址 : POJ2262
题解 :
怎么又是水题...
先筛法出素数;
然后暴力判断;
Code:
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 bool isprime[1000010]; //1 -> 合数, 0 -> 质数 6 7 int main() 8 { 9 isprime[0] = isprime[1] = 1; 10 for (register int i = 2 ; i <= 1000001 ; i ++) 11 { 12 if (isprime[i]) continue; 13 for (register int j = i ; j <= 1000001/i ; j ++) 14 isprime[i*j] = 1; 15 } 16 17 int n; 18 while (scanf("%d", &n) != EOF) 19 { 20 if (n == 0) return 0; 21 for (register int i = 0 ; i <= n ; i ++) 22 { 23 if (!isprime[i] and !isprime[n-i]) 24 { 25 printf("%d = %d + %d\n", n, i, n-i); 26 break; 27 } 28 } 29 } 30 31 }
原文地址:https://www.cnblogs.com/zZh-Brim/p/9152918.html