Poj2262_goldbach's conjecture "prime judgment" "Water problem"

Goldbach ' s conjectureTime limit:1000ms Memory Limit:65536ktotal submissions:38024Accepted:14624description


In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the Followin G conjecture:
Every even number greater than 4 can be
Written as the sum of the 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 are right. (Oh wait, I had the proof of course, but it was too long to write it on the "this page.")
Anyway, your task is now-Verify Goldbach ' s conjecture for all even numbers less than a million.
Input


The input would contain one or more test cases.
Each test case consists of one even an integer n with 6 <= N < 1000000.
Input would 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 is odd primes. Numbers and operators should is separated by exactly one blank like in the sample output below. If there is more than a pair of odd primes adding up to N, choose the pair where the difference b-a is maximized. If There is no such pair, the 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

ULM Local 1998

The main idea: to give you a number n, split into two odd primes in the form of the addition of the two primes the maximum distance

Idea: Enumerate odd numbers from three, judge whether the number I and n-i are prime, if the prime is the output result.

#include <stdio.h>int prime[1000010];void IsPrime () {for    (int i = 2; I <= 1000000; i++)        prime[i] = 1;    for (int i = 2; I <= 1000000, i++)    {        if (Prime[i])        {for           (int j = i+i; J <= 1000000; j+=i)           {               Pr IME[J] = 0;}}}    int main () {    int n;    IsPrime ();    while (~SCANF ("%d", &n) && N)    {for        (int i = 3; I <=n/2; i+=2)        {            if (Prime[i] && Prime[n-i])            {                printf ("%d =%d +%d\n", n,i,n-i);                Break    ;    }}} return 0;}

Poj2262_goldbach's conjecture "prime judgment" "Water problem"