有志者事竟成...

No. Title
1 Multiples of 3 and 5
2 Even Fibonacci numbers
3 Largest prime factor
4 Largest palindrome product
5 Smallest multiple
6 Sum square difference
7 10001st prime
8 Largest product in a series
9 Special Pythagorean triplet
10 Summation of primes

1. 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.

找出小于1000的能被3或5整除的数的和。

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

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, \cdots

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

找出小于四百万的斐波那契数列中的偶数和。

sum,a,b,c=2,1,2,3
    b=c
    if c>4000000:
        break
    if c%2==0:
         sum+=c
print(sum)

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 ?

找出600851475143最大的素因数。

def find(n=600851475143):
    i=2
    while i*i<n:
        while n%i==0:
            n/=i
        i+=2 if i>2 else 1
    return n
print(find())

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.

找出最大的由两个三位数相乘的得到的回文数。

def is_palindrome(num):
    text=str(num)
    i=0
    j=len(text)-1
    while i<j:
        if text[i]!=text[j]:
            return False
        i+=1
        j-=1
    return True
ans=0
for i in range(100,1000):
    for j in range(100,1000):
        if is_palindrome(i*j) and ans<i*j:
            ans=i*j
print(ans)

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?

找出最小的正整数,使得其可以被1-20整除。

def get(i):
    j=2
    while j<=20:
        if i%j!=0:
            return 0
        j+=1
    return 1       
i=1 
while i:
    if get(i)==1:
        print(i)
        break
    i+=1

6. Sum square difference

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

1^2+2^2+3^2+\cdots+10^2=385

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

(1+2+3+\cdots+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.

计算1-100的和的平方与1-100的平方和之差。

n=100
ans=n*(n+1)/2;
ans*=ans
ans-=n*(n+1)*(2*n+1)/6
print(ans)

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?

找出第10001个素数。

from numpy import *
def isPrime(n):
    if n==1:
        return False
    if n<4:
        return True
    if n%2==0:
        return False
    if n<9:
        return True
    if n%3==0:
        return False
    r=floor(sqrt(n))
    f=5
    while f<=r:
        if n%f==0:
            return False
            break
        if n%(f+2)==0:
            return False
            break
        f+=6
    return True
inf=10001
cnt=1
num=1
while cnt!=inf:
    num+=2
    if isPrime(num):
        cnt+=1
print(num)

8. Largest product in a series

The four adjacent digits in the 1000-digit number that have the greatest product are 9 \times 9 \times 8 \times 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?

找出上述数字块中连续13个数字里乘积的最大值。

digits='73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450'
ans=0
for i in range(12,len(digits)):
    product=1
    for j in range(13):
        product*=int(digits[i-j])
    if ans<product:
        ans=product
print(ans)

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.

找出满足 a^2+b^2=c^2 条件且 a+b+c=1000 的三正整数乘积。

for a in range(0,1001):
    for b in range(a+1,1001):
        c=1000-a-b
        if b<c and a*a+b*b==c*c:
            print(a*b*c)

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.

找出小于2000000的所有素数之和。

size=2000000
is_prime=[True for i in range(size)]

for i in range(2,size):
    if is_prime[i]:
        j=2
        while i*j<size:
            is_prime[i*j]=False
            j+=1
print sum(filter(lambda x:is_prime[x],range(2,size)))

ProjectEuler 欧拉计划 题解

《鲁滨逊漂流记》题解
上一篇 «
《整形溢出》修订版修改内容
» 下一篇