题意描述:
给一个偶数,让你分成两个素数(因为素数一定是奇数所以就不考虑奇数了)对于素数的要求是在分解的所有素数对中这两个素数差值最大(实际上只要从最小的素数找这句话就可以不用管)没找到就输出Goldbach’s conjecture is wrong.;
解题思路:为方便描述假设公式为s=n+m从最小的素数开始判断,要是找到了n,就判断m是不是素数如果是就直接输出n,m,不是的话就继续往下找下一个n(相对于后面的来说目前是最小的),同样判断m是否为素数,接着进行上述判断,如果一直没有找到(找到的最小素数也比s大了)就输出Goldbach’s conjecture is wrong.
原文:
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
AC代码:
#include<stdio.h>
#include<math.h>
int prime(int s)
{
int i,j,k=0,m;
m=(int)sqrt(s);
for(i=2; i<=m;i++)
{
if(s%i==0)
{
k=1;
break;
}
}
if(k==1)
return 0;
else
return 1;
}
int main()
{
int n,i,t;
while(scanf("%d",&n),n!=0)
{
t=0;
for(i=3; i<=n/2; i++)
{
if(prime(i)==1&&prime(n-i)==1)
{
printf("%d = %d + %d\n",n,i,n-i);
t=1;
break;
}
}
if(t==0)
printf("Goldbach's conjecture is wrong.\n");
}
return 0;
}