題干：

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

題目大意：

? ? ?給定一個大于4的數num，求兩個奇素數使得num = p1 + p2.

解題報告：

? ? 當時寫的代碼太丑陋了23333.

其實這一性質可以當一個結論去記住。。。給定一個大于4的偶數num，求兩個奇素數使得num = p1 + p2.

不難分析出給出的num一定是個偶數（因為奇+奇=偶）。另外我們只需要從小到大遍歷其中一個，然后看另一個是否在里面就可以了，不需要雙重循環。因為我們通過其中一個，和num，就可以推斷出另一個是多少，所以不需要在進行一輪循環，做了很多無用的操作。

AC代碼：

#include<cstdio> #include<cstring> #include<iostream> #define MAX 1000001//求MAX范圍內的素數 using namespace std; long long su[MAX],cnt; bool isprime[MAX]; void prime() { cnt=1; memset(isprime,1,sizeof(isprime)); isprime[0]=isprime[1]=0;for(long long i=2;i<=MAX;i++) { if(isprime[i]) {su[cnt++]=i; } for(long long j=1;j<cnt&&su[j]*i<MAX;j++) { isprime[su[j]*i]=0;} } } int main() {prime();int n;int i,flag; while(scanf("%d",&n) && n ) {flag = 0;for( i= 3; i+i<=n; i+=2) {if(isprime[ i ] == 1 && isprime[n-i] == 1) {flag = 1;break;}}if(flag == 1) {printf("%d = %d + %d\n",n,i,n-i);}else {printf("Goldbach's conjecture is wrong.\n");}}return 0 ;}

但是為什么這樣會WA？

#include<cstdio> #include<cstring> #include<iostream> #define MAX 1000001 using namespace std; long long su[MAX],cnt; bool isprime[MAX]; void prime() { cnt=1; memset(isprime,1,sizeof(isprime)); isprime[0]=isprime[1]=0;for(long long i=2;i<=MAX;i++) { if(isprime[i]) {su[cnt++]=i; } for(long long j=1;j<cnt&&su[j]*i<MAX;j++) { isprime[su[j]*i]=0;} } } int main() {prime();int n;int i,flag; while(cin>>n) {flag = 0;for( i= 2; su[i]<=n; i++) {if(isprime[ n - su[i] ] == 1) {flag = 1;break;}}if(flag == 1) {printf("%d = %lld + %lld\n",n,su[i],n-su[i] );}else {printf("Goldbach's conjecture is wrong.\n");}}return 0 ;}

?

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