Goldbach's Conjecture
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In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Each test case consists of one even integer n with 6 ≤ n < 1000000.
Input will be terminated by a value of 0 for n.
Source: University of Ulm Local Contest 1998
my code :
// 赛选找出 小于n的所有素数~~~~~~~~~~~~~~~~~~~
#include <iostream>
using namespace std;
const int mr=1000000;
int p[100000]={0};
bool y[1000000]={0};
int r=0;
void getp()
{
memset(y,1,sizeof(y));
for(int i=2;i*i<mr;i++)
if(y[i])for(int j=i*i;j<mr;j+=i)y[j]=0;
for(int i=2;i<mr;i++)
if(y[i])p[++r]=i;
}
int main(){
int n;
scanf("%d",&n);
getp();
while(n){
int k=0;
for(int i=0;p[i]<n;i++){
if(y[n-p[i]]){
printf("%d = %d + %d/n",n,p[i],n-p[i]);
k++;
break;
}
}
scanf("%d",&n);
}
return 0;
}
Every even number greater than 4 can beFor example:
written as the sum of two odd prime numbers.
- 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.
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.
Input Specification
The input file 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 Specification
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: University of Ulm Local Contest 1998
my code :
// 赛选找出 小于n的所有素数~~~~~~~~~~~~~~~~~~~
#include <iostream>
using namespace std;
const int mr=1000000;
int p[100000]={0};
bool y[1000000]={0};
int r=0;
void getp()
{
memset(y,1,sizeof(y));
for(int i=2;i*i<mr;i++)
if(y[i])for(int j=i*i;j<mr;j+=i)y[j]=0;
for(int i=2;i<mr;i++)
if(y[i])p[++r]=i;
}
int main(){
int n;
scanf("%d",&n);
getp();
while(n){
int k=0;
for(int i=0;p[i]<n;i++){
if(y[n-p[i]]){
printf("%d = %d + %d/n",n,p[i],n-p[i]);
k++;
break;
}
}
scanf("%d",&n);
}
return 0;
}
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