Project Euler|欧拉计划 1~10 (python 3.6)
一，Project Euler. Problem 1： Multiples of 3 and 5
Multiples of 3 and 5
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
Project Euler第一题，问题描述
如果我们列出所有低于10的自然数，它们是3或5的倍数，则得到3、5、6和9。这些倍数的总和为23。
找出1000以下3或5的所有倍数的总和。
解题思路
直接遍历1000以内能被3或5整除的数，然后累加这些数即可。
sum = 0
for i in range(1,1000):
if (i % 3 == 0 or i % 5 == 0):
print(i)
sum += i
print(sum)
答案为：233168
二，Project Euler.Problem 2:Even Fibonacci numbers
Even Fibonacci numbers
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
Project Euler第二题，问题描述
斐波那契数列中的每个新项都是通过将前两个项相加而生成的。 从1和2开始，前10个项将是：
1，2，3，5，8，13，21，34，55，89，...
通过考虑斐波那契数列中值不超过400万的项，找到偶值项的总和。
解题思路
找到斐波那契值不超过4000000的项，将其保存，然后求其偶数项之和。
fib = [1,2]
new_fib = fib[-1]+fib[-2]
while new_fib < 4000000:
fib.append(new_fib)
new_fib = fib[-1]+fib[-2]
sum = 0
for i in fib:
if i%2 == 0:
sum += i
else:
sum += 0
print(fib)
print(sum)答案为：4613732
三，Project Euler. Problem 3:Largest prime factor
Largest prime factor
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
Project Euler第三题，问题描述
13195的质因数是5、7、13和29。
600851475143的最大质数是多少？
解题思路
遍历600851475143的所有质因数，然后求其最大。
num = int(input("请输入一个整数"))
temp = []
while num != 1:
for i in range(2,int(num+1)):
if num % i == 0:
temp.append(i)
num = num/i
break
print (temp)
max_num = max(temp)
print(max_num)
答案为：6857
四，Project Euler. Problem 4：Largest palindrome product
Largest palindrome product
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
Project Euler第四题，问题描述
回文数（palindromic number）指的是不管从哪边读都是一样的数字，且能够被分解为两个数的乘积。
由两个数字组成的最大回文数（palindromic number）是9009，且9009=91×99.
查找由两个3位数字的乘积组成的最大回文数（palindromic number）。
解题思路
三位数的乘积在10000和998001之间，介于五位数六位数之间，由于要求解的是最大回文数，所以只考虑六位数中的最大回文数，那么要找的这个数具有以下特点：左边第一个数和右边第一个数相等，左边第二个数和右边第二个数相等，中间两个数相等，如abccba这种类型。先计算出100到1000之间的所有的三位数的成绩，并保存。然后找出乘积为六位数的的数，再在六位数的成绩中找出最大回文数。
a = []
b = 100
for i in range(100,1000):
for j in range(100,1000):
c = i * j
a.append(c)
p =[]
for m in a:
if m >=100001:
p.append(m)
print(p)
t = []
for j in p:
j = str(j)
if j[0] == j[-1] and j[1] == j[-2] and j[2] == j[-3]:
t.append(j)
print(t)
print(max(t))答案为：906609
五，Project Euler. Problem 5：Smallest multiple
Smallest multiple
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
Project Euler第五题，问题描述
2520是可以除以1到10的每个数字而没有任何余数的最小数字。
能被1到20的所有数整除的最小正数是多少？
解题思路
找出1~20的最小公倍数
yushu = 1
beishu = 2
while yushu > 0:
new_num = beishu * 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19
yushu = new_num % 20 + new_num % 18 + new_num % 16+ new_num % 15 + new_num % 14 + new_num % 12+ new_num % 10 + new_num % 9 + new_num % 8+ new_num % 7 + new_num % 6+ new_num % 5 + new_num % 4+ new_num % 3 + new_num % 2
beishu += 1
print(new_num)答案为：232792560
六，Project Euler. Problem 6：Sum square difference
Sum square difference
The sum of the squares of the first ten natural numbers is,
12+22+...+102=385The square of the sum of the first ten natural numbers is,
(1+2+...+10)2=552=3025Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025−385=2640.
Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
Project Euler第六题，问题描述
前十个自然数的平方和是
12 + 22 + ... + 102 = 385
前十个自然数之和的平方是，
（1 + 2 + ... + 10）2 = 552 = 3025
因此，前十个自然数的平方和与和的平方之差为
3025−385 = 2640
求出前一百个自然数的平方和与和的平方之差。
解题思路
先求1~100的平方和，再求1~100和的平方，最后求差。
import math
s_sum = 0
sum_s = 0
for i in range(1,101):
s_sum += i**2
sum_s =math.pow(sum(range(1,101)),2)
print(int(sum_s - s_sum))
答案为：25164150
七，Project Euler. Problem 7：10001st prime
10001st prime
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
Project Euler第七题，问题描述
通过列出前六个质数：2、3、5、7、11和13，我们可以看到第6个质数是13。
第10 001个素数是多少？
解题思路
定义一个函数，判断一个数是否为质数，然后再定义一个函数，找出第10001个。
import math
def is_prime(x):
for i in range(2, int(math.sqrt(x) + 1)):
if x % i == 0:
return False
return True
def find_prime(n):
count = 1
prime = []
for i in range( 3 , 999999999 , 2 ):
if count == n:
print(count)
return (print(prime[-1]))
if is_prime(i):
count += 1
prime.append(i)
if __name__ == '__main__':
find_prime(10001)
答案为：104743
八，Project Euler. Problem 8：Largest product in a series
Largest product in a series
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
Project Euler第八题，问题描述
在1000个数字中，4个相连数字的最大乘积是9*9*8*9=5832，其位置如上标红处所示。
那么请找出13个相连数字的最大乘积。
解题思路
将1000个数字以字符串的形式读取到内存中，然后逐个遍历，并计算相邻12个数的乘积。保存最大的乘积即可。
with open('1000.txt','r') as f :
data = f.read()
print(data)
# print(len(data))
s = []
for i in range(988):
succcc = 1
for j in range(13):
p = i + j
b = int(data[p])
succcc = succcc * b
s.append(succcc)
print(max(s))
答案为：23514624000
九，Project Euler. Problem 9：Special Pythagorean triplet
Special Pythagorean triplet
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
Project Euler第九题，问题描述
一组毕达哥拉斯数是由三个数组成，且满足勾股定理，比如3，4，5
由唯一的一组毕达哥拉斯数的和为1000，求这组数的乘积
解题思路
三个数的和为1000 ，根据勾股定理算出三个数并球其积。
for a in range(1,1000):
for b in range(1,1000):
c = 1000 - a - b
if a**2 + b**2 == c**2 and a < b < c:
print(a, b, c)
print(a*b*c)
答案为：31875000（三个数分别为：200 375 425）
十，Project Euler. Problem 10：Summation of primes
Summation of primes
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
Project Euler第十题，问题描述
小于10的所有质数的和为2+3+5+7=17
那么小于2000000的所有质数的和为多少？
解题思路
先定义一个函数判断是否为质数，然后在定义一个函数找出2000000以内的所有质数，然后再求和。
import math
def is_prime(num):
for i in range(2, int(math.sqrt(num) + 1)):
if num % i == 0:
return False
return True
def find_prime(x):
prime = []
for j in range(3, x+1, 2):
if is_prime(j):
prime.append(j)
return prime
if __name__ == '__main__':
a = find_prime(2000000)
print(a)
sum = 2
for i in a:
sum += i
print(sum)
答案为：142913828922