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Project Euler:Problem 58 Spiral primes
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed. 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18  5  4  3 12 29 40 19  6  1  2 11 ...
分类:其他好文   时间:2015-07-08 22:37:59    阅读次数:218
Project Euler:Problem 56 Powerful digit sum
A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each num...
分类:其他好文   时间:2015-07-08 09:39:56    阅读次数:101
Project Euler :Problem 54 Poker hands
In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way: High Card: Highest value card.One Pair: Two cards of the same value.Two Pairs: Tw...
分类:其他好文   时间:2015-07-07 17:09:16    阅读次数:131
Project Euler:Problem 55 Lychrel numbers
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. Not all numbers produce palindromes so quickly. For example, 349 + 943 = 1292, 1292 + 2921 = 4213 4213 + 3124 = 7337 That is...
分类:其他好文   时间:2015-07-07 16:56:48    阅读次数:113
Project Euler:Problem 52 Permuted multiples
It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order. Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, con...
分类:其他好文   时间:2015-07-02 10:22:55    阅读次数:165
Project Euler:Problem 50 Consecutive prime sum
The prime 41, can be written as the sum of six consecutive primes: 41 = 2 + 3 + 5 + 7 + 11 + 13 This is the longest sum of consecutive primes that adds to a prime below one-hundred. The longest s...
分类:其他好文   时间:2015-06-07 17:35:13    阅读次数:129
Project Euler:Problem 48 Self powers
The series, 11 + 22 + 33 + ... + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000. #include #include using namespace std; string powe(int n) { stri...
分类:其他好文   时间:2015-06-07 12:39:28    阅读次数:128
Project Euler:Problem 46 Goldbach's other conjecture
It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square. 9 = 7 + 2×12 15 = 7 + 2×22 21 = 3 + 2×32 25 = 7 + 2×32 27 = 19 + 2...
分类:数据库   时间:2015-06-07 11:12:04    阅读次数:148
Project Euler:Problem 47 Distinct primes factors
The first two consecutive numbers to have two distinct prime factors are: 14 = 2 × 7 15 = 3 × 5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2² × 7 × 23 ...
分类:其他好文   时间:2015-06-07 11:08:41    阅读次数:106
Project Euler:Problem 42 Coded triangle numbers
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... By converting each letter in a word to a...
分类:其他好文   时间:2015-06-05 12:29:01    阅读次数:128
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