POJ2262_Goldbach's Conjecture【素数判断】【水题】
Goldbach's Conjecture
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 38024
Accepted: 14624
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
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
Source
Ulm Local 1998
题目大意：给你一个数n，拆分成两个奇素数相加的形式，另这两个素数的距离最大
思路：从三开始枚举奇数，判断数i和n-i是否都为素数，若为素数则输出结果。
#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)
{
Prime[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;
}