POJ 2262 Goldbach's Conjecture哥德巴赫猜想
http://poj.org/problem?id=2262
Goldbach's Conjecture
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 33301 | Accepted: 12782 |
Description
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
For example:
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.
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.
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
数论题。主要是判断一个数是否为质数，代码如下：
#include"stdio.h"
int a[1000002];
int b[41540];
int main()
{
int N;
int i,j;
for(i=2;i<1002;i++)
{
if(!a[i])
{
for(j=i*i;j<1000002;j+=i)
a[j]=1;
}
}
b[0]=2;
for(j=1,i=3;i<500002;i+=2)
{
if(!a[i])
{
b[j++]=i;
}
}
while(scanf("%d",&N)!=EOF && N)
{
for(i=0;b[i]<N;i++)
if(!a[N-b[i]])
break;
printf("%d = %d + %d\n",N,b[i],N-b[i]);
}
return 0;
}
| yinjili | 2262 | Accepted | 4244K | 141MS | C++ | 427B |
耗时 141 MS，有点慢，尝试改进代码：
#include"stdio.h"
int a[1000002];
int main()
{
int N;
int i,j;
for(i=2;i<1002;i++)
{
if(!a[i])
{
for(j=i*i;j<1000002;j+=i)
a[j]=1;
}
}
while(scanf("%d",&N)!=EOF && N)
{
for(i=2;i<N;i++)
if(!a[i] && !a[N-i])
break;
printf("%d = %d + %d\n",N,i,N-i);
}
return 0;
}
| 2262 | Accepted | 4076K | 125MS | C++ | 325B |
耗时 125 MS，还是有点慢，尝试改进代码：
#include"stdio.h"
char a[1000002];
int main()
{
int N;
int i,j,t;
for(i=0x03;i<0x3ea;i+=0x02)
{
if(!a[i])
{
for(j=i*i;j<0xf4242;j+=i)
a[j]=0x01;
}
}
while(scanf("%d",&N)!=EOF && N)
{
t=N/2;
for(i=0x03;i<=t;i+=0x02)
if(!a[i] && !a[N-i])
break;
printf("%d = %d + %d\n",N,i,N-i);
}
return 0;
}
| yinjili | 2262 | Accepted | 1140K | 94MS | C++ | 354B |
| 71 | 11749546(3) | yinjili | 1140K | 94MS | C++ | 354B |
耗时 94 MS。在C++中排名 71 。
尝试改善代码，提高性能，不仅仅满足AC !