ProjectEuler题目日记2.5-2.7

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ProjectEuler题目日记2.5-2.7

Problem1 Multiples of 3 or 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.

如果我们将10以下的能够被3或5整除的自然数拿出来，有3 5 6 9他们的和是23.找到1000以下的数字的和。

编写代码即可

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n = 1000
sum = 0
for i in range(1, n):
    if i % 3 == 0 or i % 5 == 0:
        #print(i)
        sum += i
print(sum)

Problem 2 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.

寻找斐波那契数列中小于四百万的项中偶数项的和

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fib = [0,1]
sum = 0
while fib[-1] < 4000000:
    fib.append(fib[-1] + fib[-2])
    #print(fib[-1])
    if fib[-1] % 2 == 0:
        sum += fib[-1]
print(sum)

Problem 3 Largest Prime Factor

The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor of the number 600851475143?

寻找600851475143的最大的因子

本质为求解最大质因数问题，不断寻找能够整除的质数即可

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import math as m
n = 600851475143
k = 2
if n%2 == 0:
    n /= 2
k = 3
maxfactor = m.sqrt(n)
while(n>1 and k<=maxfactor):
    if n%k == 0:
        n /= k
    else:
        k += 2
print(k)

Problem 4 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.

最大回文数，由两位数的乘积组成的最大的回文数是9009，找到由三位数的乘积组成的最大回文数是多少？

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def is_huiwen(n):
    s = str(n)
    return s == s[::-1]

min_n = 100
max_n = 999
max_huiwen  = 0

for i in range(max_n, min_n - 1, -1):
    if i * i <= max_huiwen:
        break
    for j in range(i, min_n - 1, -1):
        temp = i * j
        if temp <= max_huiwen:
            break
        if is_huiwen(temp):
            max_huiwen = temp

print(max_huiwen)

Problem 5 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?

2520是能被1-10都整除的最小的数字，那么能被1-20都整除的最小的数字是多少？

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import math

def lcm(a, b):
    return a * b // math.gcd(a, b)

# 计算 1 到 20 的最小公倍数
lcm_result = 1
for i in range(1, 21):
    lcm_result = lcm(lcm_result, i)

print(lcm_result)

Problem 6 Sum Square Difference

The sum of the squares of the first ten natural numbers is,

$$
1^2+2^2+…+10^2=385.
$$

The square of the sum of the first ten natural numbers is,

$$
(1+2+…+10)^2=55^2=3025.
$$

Hence 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.

求1-100平方的和与和的平方的差值

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## Problem 6
sum_of_squares = 0
square_of_sum = 0

for i in range(1,101):
    sum_of_squares += i**2
    square_of_sum += i

square_of_sum = square_of_sum**2

print(square_of_sum - sum_of_squares)

## 可以使用数列进行优化

Problem 7 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 10001st prime number?

找到第10001个质数

先写一个常规的质数筛，然后用标记变量寻找第10001个质数，因为会多找一个所以-2。

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## Problem 7
### Prime Checker
def is_prime(n):
    if n == 2:
        return True
    elif n % 2 == 0:
        return False
    else:
        for i in range(3, int(n**0.5) + 1, 2):
            if n % i == 0:
                return False
    return True
n = 3
i = 2
### Prime Finder
while i <= 10001:
    if is_prime(n):
        i += 1
    n += 2
print(n-2)

Problem 8 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
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

在给定的 1000 位数字 中，找到 相邻的 13 个数字，使它们的乘积最大，并求出这个最大乘积的值。

暴力窗口求解即可：

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num_str = "73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450"

def multiple(x):
    product = 1
    for i in x:
        product *= i
    return product

windows_length = 13
max_product = -999

for i in range(len(num_str) - windows_length + 1):
    current_window = list(map(int, num_str[i:i+windows_length]))
    product = multiple(current_window)
    if product > max_product:
        max_product = product
print(max_product)

‍

Problem 9 Special Pythagorean Triplet

A Pythagorean triplet is a set of three natural numbers, $a<b<c$, for which,$a^2+b^2=c^2$.

For example, $3^2+4^2=9+16=25=5^2$.

There exists exactly one Pythagorean triplet for which a+b+c=1000. Find the product abc.

存在一组递增的abc，存在上方的平方等式，求 当abc的和为1000的时候，abc的乘积。

易从单位圆推得勾股数组通式：

$a = m^2-n^2\ b = 2mn \ c = m^2 + n^2$

暴搜即可：

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m = 3
while m < 1000:
    for n in range(1, m):
        a = m**2 - n**2
        b = 2 * m * n
        c = m**2 + n**2
        if a + b + c == 1000:
            if a**2 + b**2 == c**2:
                print(f"The Pythagorean triplet is: a = {a}, b = {b}, c = {c}")
                print(f"The product abc is: {a * b * c}")
                break
    m += 1

Problem 10 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.

找到200W以下质数的和。

同样暴搜：

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# Problem 10
## Prime Checker
def is_prime(n):
    if n == 2:
        return True
    elif n % 2 == 0:
        return False
    else:
        for i in range(3, int(n**0.5) + 1, 2):
            if n % i == 0:
                return False
    return True
sum = 0
for i in range(2, 2000000):
    if is_prime(i):
        sum += i
print(sum)

‍