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Goldbach's Conjecture
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
For example: 8 = 3 + 5. Both 3 and 5 are odd prime numbers.
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.) 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 |
解题思路：
和刚才上面有一道题很像，但是这道题的数据变大了 1000000,第一次用了Ｏ（ｎ＾２） 果断ＴＬＥ了，其实不用这么ＳＢ的算法，只要在这个筛选出来的素数表中把
ｎ－ｉ和ｉ的项找出来，分别判断两者是否同时为１就可以了，如果两者同时为１，那么，就可以认为找到了，这组我们需要的解，如果两者不同时为１，那么就continue.如果ｉ的变化范围超过了n/2都没有找到合适的解，那么就可以认为无解了．
代码：
# include<cstdio>
# include<iostream>
using namespace std;
# define MAX 1000000
int a[MAX];
int b[MAX];
void dabiao()
{
for ( int i = 2;i <= MAX;i++ )
a[i] = 1;
for ( int i = 2;i <= MAX;i++ )
{
if ( a[i] == 1 )
{
for ( int j = 2*i;j <= MAX;j+=i )
{
if ( j%i == 0 )
a[j] = 0;
}
}
}
}
int main(void)
{
dabiao();
int n;
while ( cin>>n )
{
int x = 0;
int y = 0;
if ( n == 0 )
break;
int i;
for ( i = 3;i <= n/2;i++ )
{
if ( a[i] == 1 && a[n-i] == 1 )
{
printf("%d = %d + %d\n", n, i, n-i);
break;
}
}
if ( i > n/2 )
{
printf("Goldbach's conjecture is wrong.\n");
}
}
// }
return 0;
}