2 huan1043269994 huan1043269994 于 2016.03.03 22:13 提问

求助一到素数题/uva,543
uva

Problem A : Goldbach's Conjecture
From: UVA, 543Submit

Time Limit: 3000 MS
In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:
Every number greater than 2 can be written as the sum of three prime numbers.
Goldbach cwas considering 1 as a primer number, a convention that is no longer followed. Later on, Euler re-expressed the conjecture as:
Every even number greater than or equal to 4 can be expressed as the sum of two 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 as expressed by Euler for all even numbers less than a million.

Input The input file will contain one or more test cases.
Each test case consists of one even integer n with .
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

820420
Sample Output

8 = 3 + 520 = 3 + 1742 = 5 + 37

Miguel A. Revilla
1999-01-11
新手写代码:
#include
#include
#include
#define N 1000
using namespace std;
int main()
{
int a[N];
int n;
memset(a,0,sizeof(a));
int flag1=0,k=1;
a[0]=3;
for(int i=5;i<N;i++)
{
for(int j=2;j*j<=i;j++)
{
if(i%j==0)
{
flag1=1;
break;
}
}
if(!flag1)
{
a[k++]=i;
}
flag1=0;
}
while((scanf("%d",&n))!=EOF&&(n!=0))
{
int flag2=0;
for(int i=0;a[i]<=n;i++)
{
for(int j=i+1;j<=k;j++)
{
if((n-a[i])==a[j])
{
flag2=1;
printf("%d = %d + %d\n",n,a[i],a[j]);
break;
}
}
if(flag2)
break;
}
if(!flag2)
printf("Goldbach's conjecture is wrong.");
}
return 0;
}

2个回答

u013596119
u013596119   Rxr 2016.03.05 00:04

你的算法不能改了,没办法实现,因为你的算法是先获得一个全是素数的数组,976696 = 53 + 976643,这种test case已经溢出了

u013596119
u013596119   Rxr 2016.03.05 00:14

这是我ac的代码,上面说错了,你的可以试试看把数组换成list,说不定也可以,不能说没办法实现

 #include <iostream>
using namespace std;
bool isPrime(int m){
    for(int i=2;i*i<=m;i++){
        if(m%i==0){
            return false;
        }
    }
    return true;
}
int main(){
    int n;
    cin>>n;
    while(n!=0){
        bool flag=false;
        for(int i=3;i<=n/2;i++){
            if(isPrime(i)){
                if(isPrime(n-i)){
                    cout<<n<<" = "<<i<<" + "<<n-i<<endl;
                    flag=true;
                    break;
                }
            }
        }
        if(!flag){
            cout<<"Goldbach's conjecture is wrong."<<endl;
        }
        cin>>n;
    }
}
Csdn user default icon
上传中...
上传图片
插入图片
准确详细的回答,更有利于被提问者采纳,从而获得C币。复制、灌水、广告等回答会被删除,是时候展现真正的技术了!