Project Euler 1~10
Problem 1
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.
Answer:
|
233168 |
代码:
print sum(a for a in range(1000) if a%3==0 or a%5==0)
Problem 2
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, ...
Find the sum of all the even-valued terms in the sequence which do not exceed four million.
Answer:
|
4613732 |
代码:
def firstThink():
list=[1,2]
for i in range(2,4000001):
r=list[i-1]+list[i-2]
if r<4000000:
list.append(r)
else:
break
print sum(i for i in list if i %2==0)
def secondThink():
a=1
b=1
c=a+b
sum=0
while c<=4000000:
sum=sum+c
a=b+c
b=c+a
c=a+b
print sum
def thirdThink():
'E(n)=4*E(n-1)+E(n-2)'
a=2
b=8
sum=2
while b<=4000000:
h=4*b+a
a=b
b=h
sum=sum+a
print sum
firstThink()
secondThink()
thirdThink()
Problem 3
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
Answer:
|
6857 |
代码:
import math
def primer(n):
m=int(math.sqrt(n))
for i in range(2,m):
if n%i ==0:
return 0
return 1
b=600851475143
a=int(math.sqrt(b))
maximum=0
while(a):
if b%a==0 and primer(a)==1:
maximum=a
break
a=a-1
print maximum
Problem 4
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.
Answer:
|
906609 |
一句话代码:
print max(m*n for m in range(100,1000) for n in range(m,1000) if str(m*n)==str(m*n)[::-1])
代码:
def ispalmindron2(num):
num=str(num)
return num==num[::-1]
def func(length):
print max(m*n for m in range(10**(length-1),10**length) for n in range(m,10**length) if ispalmindron2(m*n))
import time
start=time.time()
if __name__=='__main__':
func(3)
print time.time()-start
Problem 5
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 number that is evenly divisible by all of the numbers from 1 to 20?
Answer:
|
232792560 |
代码:
def func(num):
result=1
list=range(1,num+1)
for i in range(1,num):
for j in range(i+1,num):
if list[j]%list[i]==0:
list[j]/=list[i]
result=result*list[i]
print result
func(20)
Problem 6
The sum of the squares of the first ten natural numbers is,
The square of the sum of the first ten natural numbers is,
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.
Answer:
|
25164150 |
print sum(range(101))**2-sum(i*i for i in range(101))
Problem 7
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?
Answer:
|
104743 |
import math
def isprime(num):
for i in range(2,int(math.sqrt(num)+1)):
if num%i==0:
return False
return True
def func1(num):
flag=1
flagnum=3
while flag<num:
if isprime(flagnum):
flag=flag+1
## print '%d---%d' %(flag,flagnum)
flagnum=flagnum+2
print '*'*20
print flagnum-2
if __name__=='__main__':
func1(10001)
Problem 8
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Answer:
|
40824 |
代码:
str='73167176531330624919225119674426574742355349194934\
96983520312774506326239578318016984801869478851843\
85861560789112949495459501737958331952853208805511\
12540698747158523863050715693290963295227443043557\
66896648950445244523161731856403098711121722383113\
62229893423380308135336276614282806444486645238749\
30358907296290491560440772390713810515859307960866\
70172427121883998797908792274921901699720888093776\
65727333001053367881220235421809751254540594752243\
52584907711670556013604839586446706324415722155397\
53697817977846174064955149290862569321978468622482\
83972241375657056057490261407972968652414535100474\
82166370484403199890008895243450658541227588666881\
16427171479924442928230863465674813919123162824586\
17866458359124566529476545682848912883142607690042\
24219022671055626321111109370544217506941658960408\
07198403850962455444362981230987879927244284909188\
84580156166097919133875499200524063689912560717606\
05886116467109405077541002256983155200055935729725\
71636269561882670428252483600823257530420752963450'
def func1():
max=0
for i in range(996):
t=1
for x in str[i:i+5]:
t*=int(x)
if t>max:
max=t
print max
func1()
Problem 9
A Pythagorean triplet is a set of three natural numbers, a b c, for which,
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
Answer:
|
31875000 |
代码:
print[a*b*(1000-a-b) for a in range(1,1000) for b in range(a,1000) if a*a+b*b==(1000-a-b)**2]
Problem 10
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
筛选法代码:
-
#!/usr/bin/env python
-
import math
-
-
def sieve_method(num):
-
prime_data = [x for x in xrange(num + 1)]
-
for i in xrange(2,int(math.sqrt(num) + 1)):
-
if prime_data[i]:
-
start = i**2
-
step = i
-
prime_data[start::step]=((num-start)/step + 1) * [0]
-
print sum(prime_data) -1#1 is not prime number
-
-
if __name__ == "__main__":
-
sieve_method(2000000)
2009年6月04日 18:08
没想到一上programmer就找到一个玩过project euler的朋友。
2009年6月05日 01:15
@MasterLuo: 玩过一点,不多,用来练习Python,呵呵……
2023年4月23日 19:31
crediblebh
2023年10月07日 19:04
I ordered this coat from America Jackets and I must say I am amazed with the high quality and fit. You guys can also try it.