poj 2262 筛法求素数(巧妙利用数组下标!)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 41582 | Accepted: 15923 |
Description
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
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
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
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 1000000
int n,a[N];
bool pr[N];
void pdd()
{
int i,j;
for(i=;i<;i++) a[i]=;
a[]=; a[]=;
for(i=;i<;i++)
{
if(a[i]==)
{
for(j=i*;j<;j+=i) a[j]=;
}
}
}
int main()
{
int num;
int i,flag;
pdd();
while(scanf("%d",&num)!=EOF&&num!=)
{
for(flag=,i=;i<num;i++)
{
if(a[i]==&&a[num-i]==)
{
printf("%d = %d + %d\n",num,i,num-i);
flag=;
break;
}
}
if(flag==)
printf("Goldbach's conjecture is wrong.\n");
}
return ;
}
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