Euler Project 12 欧拉工程12

收藏到CSDN网摘


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, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

翻译:

三角数序列可通过连续自然数求和得到.第7个三角数是1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.前10个三角数是:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

列出前7个三角数的因子:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

可以发现28是第一个有超过5个因子的三角数.

找到第一个有500个因子的三角数.

Euler Project 11 欧拉工程11

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 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 20×20 grid?

计算上面矩阵中横竖斜相邻的4个数的最大乘积.

Euler Project10 欧拉工程10

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

10以内的素数之和为2+3+5+7 = 17
计算2,000,000(两百万)以内的素数之和.
此题的要快速需要用素数筛生成素数列表,然后求和即可.
好在matlab提供了primes函数,因此可以大大省事.下面注释掉的代码使用的是循环+判断.
时间差别是巨大的: 45s对0.08s. %% Find the sum of all the primes below two million.
% ========use loop================
% Elapsed time is 45.486655 seconds.
% ans =
% 142913828922
% ========use primes=============
% Elapsed time is 0.087248 seconds.
% ans =
% 142913828922
function result = euler10()
tic;
result = sum(primes(2000000));
% a = 2:2000000;
% b = isprime(a);
% result = sum(a(b));
toc;
end

Euler Project9 欧拉工程9

寻找和为1000的勾股数之积.

勾股数定义为: 自然数a<b<c且a^2+b^2=c^2.例如3^2+4^2=9+16=25=5^2. 所以3,4,5就是一组勾股数.

存在一组勾股数满足a+b+c=1000.请计算abc之积.

Euler Project8 欧拉工程第八题

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


计算上面100位数中,连续5个数字的最大乘积.
只有100位数字,暴力循环即可.每次取5个数,如果包含0,乘积是0,直接跳过;如果这5个数中max-min大于4,则这5个数不连续.

代码:
%% greatest product of five consecutive digits
% Elapsed time is 0.042594 seconds.
% ans =
% 40824
function result = euler8()
tic;
s = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450';
result = 0;
for i=5:length(s)
temp = s(i-4:i);
if ismember(0,temp) || max(temp)-min(temp)>4 % 0 or not consecutive digits
continue;
end
result = max([result,prod(str2num(temp(:)))]);
end
toc;
end

Euler Project7 欧拉工程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?

寻找第10001个素数. %% find 10001's prime number
% ======use primes==========
% Elapsed time is 0.008245 seconds.
% ans =
% 104743
function result = euler7()
tic;
x = primes(200000);
result = x(10001);
toc;
end

用循环的办法:

% ======use loop===========
% Elapsed time is 5.486655 seconds.
% ans =
% 104743
function result = euler7()
n = 0;
result = 1;
while n<10001
result = result+1;
if isprime(result)
n = n+1;
end
end
end

Euler Project6 欧拉工程6

The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385

The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 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.

计算前100个自然数"和的平方"与"平方的和"的差值.
%% Find the difference between the sum of the squares of 1:100 and the square of the sum.
% Elapsed time is 0.000032 seconds.
% ans =
% 25164150
function result = euler6()
tic;
x = 1:100;
result = sum(x).^2-sum(x.^2);
toc;
end

Euler Project5 欧拉工程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 positive number that is evenly divisible by all of the numbers from 1 to 20?

找到可以整除1-20的最小正整数. 分析过程:

% 整除1-20,其实就是1-20的最小公倍数,计算时可以精简:
% 1-20只需要求[2,3,4,5,6,7,11,13,17,19]的最小公倍数即可
% 其余的
% 1不用计算
% 8 = 2*4
% 9 = 3*3(第二个3来自6=2*3)
% 10 = 2*5
% 12 = 3*4
% 14 = 2*7
% 16 = 2*8
% 18 = 2*9 = 2*3*3(第二个3来自6=2*3)
% 20 = 2*10 = 2*5*2(第二个2来自4=2*2)

==============
代码:
function result = euler5()
tic;
result = prod([2,3,4,5,6,7,11,13,17,19]); % => 8,9(3*3),10,12,14,15,16(2*4*2),18,20
toc;
end
运行时间及结果:
% Elapsed time is 0.000012 seconds.
% ans =
% 232792560

Euler Project4 欧拉工程4

如果一个数字正写和反写结果一样,就称为回文数,请找到可以表示为2个三位数乘积的最大回文数.
%% 3 digital number which remains same from both direction, like 9009
% == from small to big
% Elapsed time is 43.379476 seconds.
% ans =
% 906609
% == from big to small
% Elapsed time is 5.504064 seconds.
% ans =
% 906609
function result = euler4()
tic;
result = 0;
for i=999*999:-1:100*100 % ends with 0, jump over
if mod(i,10)==0
continue;
end
n = num2str(i);
if strcmp(n,n(end:-1:1))
flag = 0;
for j=999:-1:900 % max, so from 999 to 900, try
if i/j<=999 && i==fix(i/j)*j % i/j must be smaller than 999, and i/j must be integer flag = j; break; end end if flag~=0 result = i; break; end end end [j,result/j] toc; end
结果
% Elapsed time is 5.504064 seconds.
% ans =
% 906609

Euler Project3 欧拉工程3

求600851475143的最大质因数
%% 600851475143's biggest prime factor
% Elapsed time is 55.277722 seconds.
% ans =
% 6857
function result = euler3()
tic;
result = 0;
n = 600851475143;
for i=fix(sqrt(n)):-1:3
if isprime(i)==1 && mod(n,i)==0
result = i;
break;
end
end
toc;
end

Euler Project2 欧拉工程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, ...

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

求斐波那契数列中小于4,000,000的值中的偶数项之和.
%% sum of even value elements from fibonacci list smaller than 4,000,000
% Elapsed time is 0.000011 seconds.
% ans =
% 4613732
function result = euler2()
tic;
a = 1;
b = 2;
result = 2;
while (a<4000000 && b<4000000) a = a+b; if mod(a,2)==0 result = result+a; end b = a+b; if mod(b,2)==0 result = result+b; end end toc; end

Euler Project1 欧拉工程1

欧拉工程是一个用编程来解决一连串数学问题的项目。发现它对训练数学及编程挺有作用的。网上也有许多人做上面的题目并发布自己的算法。

上面共有337题，分几个等级，一级比一级难。完成它绝对是时间及精力及耐力及智力的较量。

解决问题的代码贴在这里.主要用matlab,偶尔python,很少c/c++.有时会matlab调python解决,只为快速解决问题. Question 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.
解决
%% sum of number below 1000 and can be divided by 3 or 5
% Elapsed time is 0.002665 seconds.
% ans =
% 233168
function result = euler1()
tic
result = sum(unique([3:3:1000,5:5:999]))
toc;
end