python 编程练习——吉尔布雷斯的有趣猜测

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61]
       [1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2] 
         [1, 0, 2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 0, 4] 
           [1, 2, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 4] 
             [1, 2, 0, 0, 0, 0, 2, 2, 2, 2, 0, 0, 2, 2] 
               [1, 2, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 0] 
                 [1, 2, 0, 0, 2, 2, 0, 0, 2, 2, 2, 2] 
                   [1, 2, 0, 2, 0, 2, 0, 2, 0, 0, 0] 
                     [1, 2, 2, 2, 2, 2, 2, 2, 0, 0] 
                       [1, 0, 0, 0, 0, 0, 0, 2, 0] 
                         [1, 0, 0, 0, 0, 0, 2, 2] 
                           [1, 0, 0, 0, 0, 2, 0] 
                             [1, 0, 0, 0, 2, 2] 
                               [1, 0, 0, 2, 0] 
                                 [1, 0, 2, 2] 
                                   [1, 2, 0] 
                                     [1, 2] 
                                       [1] 

@JanetZ
你是用算法实现的吗？贴下代码

import numpy as np
numbers = np.arange(1,65)
length = len(numbers)
n_max = int(np.sqrt(length))
is_prime = np.ones(length,dtype = bool)
is_prime[0] = False
for i in range(2,n_max):
    if i in numbers[is_prime]:
        is_prime[(i**2-1)::i] = False
prime = numbers[is_prime].tolist()
print(prime,end="")
row=[]
result=[]
for j in range(len(prime)):
    for i in range(len(prime)-1):
        row.append(np.abs(prime[i]-prime[i+1]))
    prime = row
    result.append(prime)
    if len(row) < 2 :
        break
    else:
        row = []
r = 4
for i in result:
    print("\n" + (2*r-1) * " " + str(i),end=" ")
    r += 1

C语言实现效果：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int isPrime(int n) 
{ 
    static int d=2; 
    return n<d*d ? (d=2,1): n%d ? (d+=1+(d>2),isPrime(n)) : (d=2,0); 
} 

int main(int argc, char *argv[])
{
    int n=0,index=0;
    scanf("%d",&n);

    int array[n];
    for(int i=2; ;i++)
    {
        if(isPrime(i)){
            array[index]=i;
            index++;
        }
        if(index==n)
        break;
    }

    for(int i=0;i<n;i++)
    printf("%d ",array[i]);

    for(int i=0;i<n;i++){
        printf("\n");
        for(int j=0;j<n-1-i;j++){
            array[j]=abs(array[j+1]-array[j]);
        }
        for(int k=0;k<=i;k++){
            printf(" ");
        }
        for(int j=0;j<n-1-i;j++){
            printf("%d ",array[j]);
        }

    }
    return 0;
}