[暑假集训--数论]poj2262 Goldbach's Conjecture

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

 

把n>=6分成两个奇质数之和，小的那个要尽量小

直接暴力

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define LL long long
 5 using namespace std;
 6 inline LL read()
 7 {
 8     LL x=0,f=1;char ch=getchar();
 9     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
10     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
11     return x*f;
12 }
13 int n;
14 bool mk[1000010];
15 int p[1000010],len;
16 inline void getp()
17 {
18     for (int i=2;i<=1000000;i++)
19     {
20         if (!mk[i])
21         {
22             p[++len]=i;
23             for (int j=2*i;j<=1000000;j+=i)mk[j]=1;
24         }
25     }
26 }
27 int main()
28 {
29     getp();
30     while (~scanf("%d",&n)&&n)
31     {
32         if (n&1||n<6){puts("Goldbach's conjecture is wrong.");continue;}
33         for (int i=2;p[i]*2<=n;i++)
34             if (!mk[n-p[i]]){printf("%d = %d + %d\n",n,p[i],n-p[i]);break;}
35     }
36 }

poj 2262

 

转载于:https://www.cnblogs.com/zhber/p/7285419.html