我试图使用
python解决Project Euler问题10,但我的程序给出了错误的结果.因为我是一个完整的pyob toob,我在我的(显然是暴力)逻辑中找不到任何错误,我在
java中编写了一个程序(几乎翻译了它),它给出了一个不同的结果,结果证明是正确的.
这是python代码:
from math import *
limit = 2000000
def isPrime(number):
if number == 2: return 1
elif number % 2 == 0: return 0
elif number == 3: return 1
elif number == 5: return 1
elif number == 7: return 1
else:
rootOfNumber = sqrt(number)
tag = 3
while tag < rootOfNumber:
if number % tag != 0:
tag += 2
else:
break ###
if tag >= rootOfNumber: ###EDIT: it should by only tag > rootOfNumber here
return 1 ### Thats what the problem was.
else:
return 0
sum = 2 # 2 is an even prime, something we are not iterating for
for i in range(3, limit, 2):
if isPrime(i) == 1:
sum += i
print(sum)
print('done...')
等效的java代码是:
public class Prob10{
static int limit = 2000000;
static long sum = 2L; // 2 is an even prime, something we are not iterating for
public static void main (String[] args) {
for(int i = 3; i < limit; i+=2) {
if( isPrime(i) )
sum += i;
}
System.out.println(sum);
}
private static boolean isPrime (int number) {
if (number == 2) return true;
else if (number == 3 || number == 5 || number == 7) return true;
else {
double rootOfNumber = Math.sqrt(number);
int tag = 3;
while (tag < rootOfNumber) {
if (number % tag != 0)
tag +=2;
else
break;
}
if (tag > rootOfNumber)
return true;
else
return false;
}
}
}
我想我正在做一些愚蠢的错误或遗漏一些微妙的观点.
附:我知道我的isPrime实现不太好.我不打印输出,因为它可能会破坏其他人的问题.
欢迎任何有关python程序中(坏)样式的评论.
最佳答案 尝试使用代码运行,例如isPrime(49).你应该从那里找出你的问题.你已经取代了>使用> = in if(tag> rootOfNumber)
.作为一些编码风格,你可以用以下代码替换第一行:
if i in (2, 3, 5, 7): return 1
elif number % 2 == 0: return 0
else:
......