ProjectEuler Problem 27: Quadratic primes
Mar 24, 2014
Problem 27: Quadratic primes #
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive integer values 0 <= n <= 39. However, when n = 40, 40² + 40 + 41 = 40(40+1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values 0 <= n <= 79. The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| <= 1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
分析 #
二次质数。欧拉发现了著名的二次方程：n² + n + 41。当n=0~39时，这个方程能够连续产生40个质数。但是，当n=40时，402 + 40 + 41 = 40(40 + 1) + 41可以被41整除，当然，还有当n = 41, 41² + 41 + 41 可以被41整除。
发现了更奇妙的方程： n² - 79n + 1601，当n=1~79时，它能够连续产生80个质数。方程系数的积为 –79 * 1601 = -126479.
考虑二次方程的形式：
n² + an + b, |a| < 1000 且 |b| <= 1000
|n| 是n 的绝对值。
找到系数a和b，当n从0开始连续递增时，可以产生最多的质数。 给出a*b的值。
方法1 穷举法 #
穷举遍历。a=-1000~1000，b=-1000~1000，对每组a,b, 从n=0开始往上递增，判断f = nn+an+b是否是质数，如果是，则n++,cnt++,如果不是则结束，此时a,b能产生的最多质数就是cnt个。
C++ #
#include <stdio.h>
/************************************************************************
* 判断n是否是质数
************************************************************************/
bool isPrime(int n)
{
if(n <= 1 ) return false;
if(n == 2 || n == 3) return true;
if(n % 2 == 0) return false;
// 关于质数分布的规律：大于等于5的质数一定和6的倍数相邻
// 即大于等于5的质数一定是 6x-1 和 6x+1，6x+2 ~ 6x+4 一定不是质数
// 即大于等于5的质数一定满足 n%6==1 or n%6==5
// https://blog.csdn.net/songyunli1111/article/details/78690447
if(n % 6 != 1 && n % 6 != 5 ) return false;
// 从 [5, sqrt(n)] 遍历寻找是否是其因数
for(int i=5; i * i <= n; i++){
if( n % i == 0 ) return false;
}
return true;
}
/************************************************************************
* 返回 n*n + a*n + b，n 从 0 开始，能连续产生的质数的个数
************************************************************************/
int f(int a, int b)
{
int cnt = 0;
for (int i=0;;i++){
if ( isPrime(i*i+a*i+b) ) {
cnt ++;
}else break;
}
return cnt;
}
int main()
{
int max_a,max_b,max_primes=0;
for (int a = -1000; a < 1000; a++){
for(int b = -1000; b < 1000; b++){
if( f(a,b) > max_primes){
max_a = a;
max_b = b;
max_primes = f(a,b);
}
}
}
printf("a=%d, b=%d, a*b=%d, 连续质数个数为：%d\n", max_a, max_b, max_a * max_b, max_primes);
return 0;
}
Golang #
package main
import "fmt"
func isPrime(n int64) bool {
if n <= 1 {
return false
}
if n == 2 || n == 3 {
return true
}
if n%2 == 0 {
return false
}
// 关于质数分布的规律：大于等于5的质数一定和6的倍数相邻
// 即大于等于5的质数一定是 6x-1 和 6x+1，6x+2 ~ 6x+4 一定不是质数
// 即大于等于5的质数一定满足 n%6==1 or n%6==5
// https://blog.csdn.net/songyunli1111/article/details/78690447
if n%6 != 1 && n%6 != 5 {
return false
}
// 从 [5, sqrt(n)] 遍历寻找是否是其因数
for i := int64(5); i*i <= n; i++ {
if n%i == 0 {
return false
}
}
return true
}
/************************************************************************
* 返回 n*n + a*n + b，n 从 0 开始，能连续产生的质数的个数
************************************************************************/
func f(a, b int) int {
cnt := 0
for i := 0; ; i++ {
if isPrime(int64(i*i + a*i + b)) {
cnt++
} else {
break
}
}
return cnt
}
func main() {
maxA, maxB, maxPrimes := 0, 0, 0
for a := -1000; a < 1000; a++ {
for b := -1000; b < 1000; b++ {
if f(a, b) > maxPrimes {
maxA = a
maxB = b
maxPrimes = f(a, b)
}
}
}
fmt.Printf("a=%d, b=%d, a*b=%d, 连续质数个数为：%d\n", maxA, maxB, maxA*maxB, maxPrimes)
}
Python #
from math import sqrt
def isPrime(n):
if n <= 1 : return False
if n == 2 or n == 3 : return True
if n % 2 == 0 : return False
# 关于质数分布的规律：大于等于5的质数一定和6的倍数相邻
# 即大于等于5的质数一定是 6x-1 和 6x+1，6x+2 ~ 6x+4 一定不是质数
# 即大于等于5的质数一定满足 n%6==1 or n%6==5
# https://blog.csdn.net/songyunli1111/article/details/78690447
if n % 6 != 1 and n % 6 != 5 : return False
# 从 [5, sqrt(n)] 遍历寻找是否是其因数
for i in range(5, int(sqrt(n)) + 1):
if n % i == 0 : return False
return True
def f(a,b):
'''return the count of consecutive primes n*n + a*n + b can produce'''
i, cnt = 0, 0
while True:
if isPrime(i*i+a*i+b):
cnt += 1
i += 1
else:
break
return cnt
#just one line
#print max((f(a,b),a,b,a*b) for a in range(-1000,1000) for b in range(-1000,1000)) #(71, -61, 971, -59231)
#or several lines in detail
max_a, max_b, max_primes, max_product = 0, 0, 0, 0
for a in range(-1000,1000):
for b in range(-1000,1000):
if f(a,b) > max_primes:
max_a, max_b, max_product, max_primes = a, b, a * b, f(a,b)
print(max_a, max_b, max_product, max_primes)
方法2 不编程，纯数学计算 #
来自论坛第一页用户 hk 的答案
hk的原答案如下：
I did not use a computer to solve this problem.
The formula n^2-79n+1601 is nothing more than (n-40)^2+n-40+41
so that all the forty primes of n^2+n+41 are met twice that's why 80 primes are found, but only 40 different ones.
So what I did was:
take (n-p)^2+n-p+41, working out this formula gives:
n^2-(2p-1)n+p^2-p+41.
Now |2p-1|<1000 and |p^2-p+41|<1000.
The second condition gives -30<=p<=31
The value p=31 gives the most primes.
So the numbers are -(2*31-1)=-61 and 31^2-31+41=971.
See also: http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
答案 #
-59231