Project Euler 1-6

1>求1000以內3或5的倍數的和>>>print sum([i for i in range(1000) if i%3==0 or i%5==0]) 233168

2>求出小於4,000,000的Fibonaci數列中偶數項的和

a=1 b=2 fibonaci=[a] while 1: if b>4000000: break else: fibonaci.append(b) a,b=b,a+b print sum([i for i in fibonaci if i%2==0])

3>求600851475143 的最大素因子

def isprime(number): i=2 while i<number: if number%i==0: return True return False factors=[] number=600851475143 while number>1: i=2 while i<number: if number%i==0: factors.append(i) number=number/i break i=i+1 print factors,number if number>10000: continue else: if isprime(number): break print max([i for i in factors if isprime(i)])

4>迴文數是類似於9009這樣得數,用二位數相乘得出的最大的迴文數是9009=91*99,問三位數得出的最大的迴文數是多少

def ispa(number): s_number=str(number) left=s_number[:len(s_number)/2+len(s_number)%2] right=s_number[len(s_number)/2:][::-1] #print left,right if left==right: return True else: return False def checkFactor(number): #檢查其因子是否爲2個三位數 for i in range(100,1000): if number%i==0: factor=number/i s_factor=str(factor) if len(s_factor)==3: return number,i,factor break for i in range(999*999,100*100,-1): if ispa(i): r=checkFactor(i) if r: print r break

5>求可以被1-20整除的最小偶數

def beishu(a,b): begin=max(a,b) while 1: if begin%a==0 and begin%b==0: return begin else: begin+=1 a,b=1,2 while b<21: a,b=beishu(a,b),b+1 print a,b

6>計算1-100的平方和和的平方之差

sumsquare=sum([i for i in range(101)])**2 squaresum=sum([i*i for i in range(101)]) squaresum-sumsquare