Time Limit: 1000MS Memory Limit: 65536KByte 64 IO Format:%I64d & %I64u
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the
following conjecture:
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 Input8
20
42
0
Sample Output
8 = 3 + 5
20 = 3 + 17
42 = 5 + 37
#include<stdio.h>
#include<math.h>
int judge_prime(int x)
{
if(!(x % 2))
return 0;//是大于2的偶数必定不是素数，直接返回0；
for(int i = 3; i * i <= x; i+= 2)//奇数，能除尽非本身的奇数，这样的奇数一定不是素数
{
if(!(x % i))
return 0;
}
return 1;
}
int main()
{
int num, ans, flag, i, j;
while(scanf("%d",&num), num)
{
flag = 0;
for(i = 3; i * 2 <= num; i+= 2)//奇素数
{
if(judge_prime(i) && judge_prime(num - i))
{
printf("%d = %d + %d\n", num, i, num - i);
flag = 1;
break;
}
}
if(!flag)
printf("Goldbach's conjecture is wrong.\n");
}
}
Goldbach's Conjecture
For any even number n greater than or equal to 4, there exists at least one pair of prime
numbers p1 and p2 such that
n = p1 + p2
This conjecture has not been proved nor refused yet. No one is sure whether this conjecture
actually holds. However, one can find such
a pair of prime numbers, if any, for a given even number. The problem here is to write a
program that reports the number of all the pairs
of prime numbers satisfying the condition in the conjecture for a given even number.
A sequence of even numbers is given as input. There can be many such numbers.
Corresponding to each number, the program should
output the number of pairs mentioned above. Notice that we are interested in the number
of essentially different pairs and therefore you
should not count (p1, p2) and (p2, p1) separately as two different pairs.
An integer is given in each input line. You may assume that each integer is even, and
is greater than or equal to 4 and less than 215.
The end of the input is indicated by a number 0.
Each output line should contain an integer number. No other characters should
appear in the output.
6
10
12
0
1
2
1
#include<stdio.h>
#include<math.h>
#define Max 32769
int prime[Max] = {0}, plen;
void getprime()
{
int temp, j, i, isprime;
plen = 1;
prime[0] = 2;
for(i = 3; i < Max; i+= 2)
{
isprime = 1;
for(j = 0; j < plen; j++)
{
if(i % prime[j] == 0)
{
isprime = 0;
break;
}
}
if(isprime)
prime[plen++] = i;
}
}
int main()
{
int i, j, num, ans;
getprime();
while(scanf("%d",&num), num)
{
ans = 0;
i = 0;
j = plen - 1;
for(; i <= j;)
{
if(prime[i] + prime[j] == num)
{
i++;
j--;
ans++;
}
else if(prime[i] + prime[j] > num)
--j;
else
i++;
}
printf("%d\n",ans);
}
}