Goldbach's Conjecture
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 35054
Accepted: 13441
Description
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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.
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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.
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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
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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
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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
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8
20
42
0
Sample Output
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8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
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#include
#include
#define MAX 1000001
int prime[MAX]={1};
void ProductData()
{
int i, j, k;
for(i = 2; i*i <= MAX; i++)
{
// 排除偶数的情况
if( i!=2 && i%2==0) continue;
// 排除不必要的奇数判断 奇数相乘一定是两个奇数
for(k = 3; k*k <= i; k += 2)
if(i%k == 0) break;
if(k*k <= i) continue;
// 筛选法 除去 i 的倍数
if(prime[i] == 1)
for(j = 2*i; j <= MAX; j+=i)
prime[j] = 0;
}
}
int main()
{
int n, i;
ProductData();
while(scanf("%d",&n)!=EOF && n!= 0)
{
if(n%2 != 0 || n < 6)
return 0;
for(i = 3; i*i <= n; i+=2)
{
if(prime[i] && prime[n-i])
{
printf("%d = %d + %d\n",n, i, n-i);
break;
}
}
if(i*i > n) printf("Goldbach's conjecture is wrong.\n");
}
return 0;
}
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