qq_21466587的博客

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.Find the sum of all the primes below two million.----------------------------------------------------------------------------------fr

def d(i):    k=0                      #判断一个数（小于28123）是否写成两个过剩数的和，不能的话返回为真    for m in range(0,len(M)):        for n in range(0,len(M)):             #M为小于28123的所有的过剩数            if i==M[m]+M[n]

def d(i):    k=0                      #判断一个数（小于28123）是否写成两个过剩数的和，不能的话返回为真    for m in range(0,len(M)):        for n in range(0,len(M)):             #M为小于28123的所有的过剩数            if i==M[m]+M[n]

n! means n × (n − 1) × ... × 3 × 2 × 1For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.Find the sum of the digi

sum=1x=1for i in range(1,501):    for j in range(0,4):        x=x+2*i        sum=sum+xprint (sum)===========================思路12，3，4，5，6，7，8，9、10，11，12，13，14，15，16，17，18，19，20，21

注意set的使用===================print (len(set(a**b for a in range(2, 101) for b in range(2, 101))))

逆康拓展开式import mathL=[0,1,2,3,4,5,6,7,8,9]M=[]m=999999for i in range(9,-1,-1):    x=(m/(math.factorial(i)))    print x    y=(m%(math.factorial(i)))    m=y    M.append(L[x])    L.po

def f(n):    if n==1 or n==2:        return 1    else:        a=D[n-1]+D[n-2]        D[n]=a        return an=1D={1:1,2:1}while len(str(f(n)))    n=n+1    print f(n)print n===

def cycle_length(n):    i = 1    if n % 2 == 0: return cycle_length(n / 2)    if n % 5 == 0: return cycle_length(n / 5)    while True:        if (10**i - 1) % n == 0: return i        else:

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:1, 3, 6, 10, 15, 21, 2

You are given the following information, but you may prefer to do some research for yourself.1 Jan 1900 was a Monday.Thirty days has September,April, June and November.All the rest have thirty

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.37 42 4 68 5 9 3That is, 3 + 7 + 4 + 9 = 23.Find th

A Pythagorean triplet is a set of three natural numbers, a b c, for which,a2 + b2 =c2For example, 32 + 42 = 9 + 16 = 25 = 52.There exists exactly one Pythagorean triplet for which a + b +

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 0849 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.37107287533902102798797998220837590246510135740250463769376774900097126481248969700780504170182605387432498

The following iterative sequence is defined for the set of positive integers:n → n/2 (n is even)n → 3n + 1 (n is odd)Using the rule above and starting with 13, we generate the following sequen

def f(n):    value={'A':1,'B':2,'C':3,'D':4,'E':5,'F':6,'G':7,'H':8,'I':9,'J':10,'K':11,'L':12,'M':13,'N':14,'O':15,'P':16,'Q':17,'R':18,'S':19,'T':20,'U':21,'V':22,'W':23,'X':24,'Y':25,'Z':26,} 

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.How many such routes are there through a 2

215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.What is the sum of the digits of the number 21000?---------------------------------------------------------------------------a=str

import mathfrom time import clocktime1=clock()def IsPrime(x):      if x        return False      for i in range(2,int(math.sqrt(x))+1):          if x%i==0:              return

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.If all the numbers from 1 to 1000 (one thousand) inclusive we

def d(n):    result=[a for a in range(1,n) if n%a==0]    y=sum(result)    return yL=[]for i in range(2,10001):    m=d(i)    if d(m)==i and m!=i:        L.append(i)    i=i+1print su