Euler-Project(II)

11. Largest product in a grid

Problem Description

In the 20×20 grid below, four numbers along a diagonal line have been marked in bold.

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 the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

Solution

# Simply rotation and comparison in four types of direction
import re
grid_given ='''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'''
# Alist variable is used to slice the given grid
alist = [re.split(' ', everyrow) for everyrow in re.split('\n', grid_given)]
result_list = []
# rows
for row_index in range(len(alist)):
    for columns_index in xrange(len(alist[row_index])-3):
        result_list.append(int(alist[row_index][columns_index])*int(alist[row_index][columns_index+1])*int(alist[row_index][columns_index+2])*int(alist[row_index][columns_index+3]))
# columns
for columns_index in range(len(alist[0])):
    for row_index in xrange(len(alist)-3):
        result_list.append(int(alist[row_index][columns_index])*int(alist[row_index+1][columns_index])*int(alist[row_index+2][columns_index])*int(alist[row_index+3][columns_index]))
# diagonal
for row_index in range(len(alist)-3):
    for columns_index in range(len(alist[row_index])-3):
        # main diagonal
        result_list.append(int(alist[row_index][columns_index])*int(alist[row_index+1][columns_index+1])*int(alist[row_index+2][columns_index+2])*int(alist[row_index+3][columns_index+3]))
        # vice diagonal
        result_list.append(int(alist[row_index+3][columns_index])*int(alist[row_index+2][columns_index+1])*int(alist[row_index+1][columns_index+2])*int(alist[row_index][columns_index+3]))
print max(result_list)

12. Highly divisible triangular number

Problem Description

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?

Solution1

通过观察可以发现，数字N的全部因子可分为大于$\sqrt{N}$和小于$\sqrt{N}$两部分，且这两部分的因子数量相等。故以$\sqrt{N}$为界限，统计前半部分因子数量，当大于250时跳出循环。

# Triangular number
import math
basic = 1
step = 2
while True:
    basic += step
    count = 0
    i = 1
    while i <= int(math.sqrt(basic)):
        if basic % i == 0:
            count += 1
        i += 1
    if count >= 250:
        print basic
        break
    else:
        count = 0
        step += 1

Solution2

每个整数N可以分解为如下模式：

$N = p_{1}^{a1} * p_{2}^{a2} * p_{3}^{a3} * ...$

此处$p{n}$必为素数，$a{n}$是其对应的幂指数。例如$28 = 2^{2} * 7^{1}$。继续推广下去，对于任意正整数N，其因子的数量D(N)可用如下式子表示：

$D(N) = (a_{1}+1) * (a_{2}+1) * (a_{3}+1) * ...$

故首先采用埃氏筛法构造大素数表，随后对循环内每个数字进行分解以获得$a{1}，a{2}，a_{3}$等参数，连乘可得D(N)。对该常规方法进行改进，首先对三角形数进行等差数列求和，结果为$t = n * (n+1) / 2$，此处n和n+1必互素。随后采用如下公式分解D(t)：

$D(t) = D(n/2) * D(n+1)$, if n is even

$D(t) = D(n) * D((n+1)/2)$, if n is odd

实现代码如下：

# Highly divisible triangular number
# First get the possible prime facotrs below 1000
def getPossiblePrimeFactors(upper_bound):
    L = range(2, upper_bound)
    primes_list = findPrime(L)
    return primes_list
def findPrime(L):
    i = 0
    while L[i]**2 < L[-1]:
        # Eratosthenes algorithm
        count = 0
        for j in range(i+1, len(L)):
            if L[j] % L[i] == 0:
                L[j] = 0
                count += 1
        L.sort()
        L = L[count:]
        i += 1
    return L
def countD(n, Dn, count_range, primes_list):
    count = 0
    while count <= count_range:
        n += 1
        n1 = n
        if n1 % 2 == 0:
            n1 /= 2
        Dn1 = 1
        for i in range(1, primes_list[-1]):
            if primes_list[i]**2 > n1:
                '''
                When the prime divisor would be greater than the residual n1, that residual n1 is the last prime factor with an exponent = 1
                '''
                Dn1 = 2*Dn1
                break
            # ause D(n) = (a1+1)*(a2+1)*(a3+1)*...., so initialize the exponent to 1
            exponent = 1
            while n1 % primes_list[i] == 0:
                # resolve n1 to the form of multiplication, count a1, a2, a3, ...
                exponent += 1
                n1 = n1 // primes_list[i]
            if exponent > 1:
                Dn1 = Dn1*exponent
            if n1 == 1: break
        count = Dn*Dn1
        Dn = Dn1
    return n*(n-1) // 2

n = 3
Dn = 2
primes_list = getPossiblePrimeFactors(1000)
final_result = countD(n, Dn, 500, primes_list)
print final_result

理解上述代码过程中，值得注意的是countD()函数的返回结果为n*(n-1) / 2而非n*(n+1) / 2，从而代码每次迭代计算的D(t)由传入while循环之前的n值表示，与上面提到的性质对应。

13. Large sum

Problem Description

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

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

Solution

all_numbers = given_50_digit_numbers
sum_result = 0
for item in all_numbers.split('\n'):
    if item == '':
        continue
    else:
        sum_result += int(item)
print str(sum_result)[0:10]

14. Longest Collatz sequence

Problem Description

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 sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Solution1

按照题意直接进行暴力遍历，可以采用递归和迭代的方法。采用靠近计算机底层的位运算代替普通运算可以适当提高效率：n & 1 代替 n % 2，n >> 1代替 n / 2。

# Iteration
# Longest Collatz Sequence
# Which starting number, under one million, produces the longest chain
def getCollatzSquence(n):
    count = 0
    while n != 1:
        if n & 1 == 0:
            n = n >> 1
        else:
            n = 3*n+1
        count += 1
    return count
if __name__ == "__main__":
    maxlength = 1
    flag = 0
    for i in range(100, 1000000):
        count_number = getCollatzSquence(i)
        if count_number > maxlength:
            maxlength = count_number
            flag = i
    print (maxlength, flag)


# Recursion
def countCollatzChain(n):
    if n == 1:
        return 1
    if n & 1 == 0:
        return 1 + countCollatzChain(n >> 1)
    else:
        return 1 + countCollatzChain(3 * n + 1)

if __name__ == "__main__":
    longest_chain = 0
    result = -1
    for number in range(100, 10**6-1):
        if countCollatzChain(number) > longest_chain:
            longest_chain = countCollatzChain(number)
            result = number
    print result

二者效率对比如下，可以看出迭代调用所需时间明显小于递归调用。

14_sufficiency.PNG-10.6kB \1.PNG)

Solution2

无论采用递归或是迭代的方法进行暴力遍历，我们都可以明确看出有部分数值经过重复计算，导致运算时间过长。于是进行以下优化：

引入字典存储已经计算过的考拉兹链长度；
根据考拉兹猜想提出的运算规则，我们可以得到Collatz(n) = Collatz(n/2) + 1。因此对于所有整数k，Collatz(2k) > Collatz(k)必定成立，所以我们不必计算小于LIMIT/2的所有k值，本例中即无需计算低于500000的k的考拉兹链长度；
若n是奇数，则3*n+1必定为偶数，从而n经过考拉兹变换后最终得到(3*n+1) / 2。故当n是奇数时，采用以下公式简化运算过程：Collatz(n) = Collatz((3*n+1) / 2) + 2。

代码复现如下，但效率无法达到相关题解中提及的1.5s。

# Find the longest Collatz sequence chain
# Which starting number, under one million, produces the longest chain
def countCollatzChain(n, values):
    if n in values.keys():
        return values[n]
    
    if n & 1 == 0:
        values[n] = 1 + countCollatzChain(n >> 1, values)
    else:
        values[n] = 2 + countCollatzChain((3 * n + 1) >> 1, values)
    return values[n] 

if __name__ == "__main__":
    longest_chain = 0
    final_result = -1
    values = {1: 1}
    for number in range(500000, 10**6 - 1):
        if countCollatzChain(number, values) > longest_chain:
            longest_chain = countCollatzChain(number, values)
            final_result = number
    
    print final_result

15. Lattice paths

Problem Description

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.

4.png-9.7kB \2.PNG)

How many such routes are there through a 20×20 grid?

Solution

本题目为简单格子路径问题，可以采用迭代、递归与组合数三种方法求解。

Recursive Solution

将题目所给信息转换为一般问题，即求从(0, 0)点运动到(m, n)点的所有路径数量，该数量等于(0, 0)点到点(m-1, n)和(0, 0)点到点(m, n-1)的路径数量之和。以此类推，当m或n等于0时，(0, 0)点到达该点只存在一直向右或向下两条道路，此时递归算法返回1。值得注意的是，该方法存在重复计算问题，故可引入大容量数组存储可能需要的计算结果。

Iterative Solution

递归法较易编写，但需要消耗较多计算资源，故考虑结合动态规划进行迭代求解。如果说递归法是”执果索因”，动态规划就是”由因导果”。首先建立20x20数组，由于第一行和第一列所有元素到达点(0, 0)只有一条路径，故数组中对应位置全部设置为1。随后从第二行第二列开始按照grid[i][j] = grid[i][j-1] + grid[i-1][j]进行数组赋值，目标位置grid[m][n]即为待求结果。代码如下：

# Lattice paths problem
# Simple dynamic programming
# Solution1 
def getAllPaths(m, n):
    dp = [[0] * n for _ in range(m)]
    for i in range(m):
        dp[i][0] = 1
    for j in range(n):
        dp[0][j] = 1
    for k in range(1, m):
        for l in range(1, n):
            dp[k][l] = dp[k-1][l] + dp[k][l-1]
    print dp
    return dp[m-1][n-1]
if __name__ == "__main__":
    # Get all possible paths of 20 x 20 grids
    final_result = getAllPaths(21, 21)
    print final_result

Combinatorial Solution

以上两种方法时间复杂度均为O(mn)，我们可以使用组合数学以提高效率。

首先分析一般问题的本质，即从点(0, 0)到点(m, n)共需要走m+n步，其中需要向下走m步，向右走n步。于是引出简单无顺序组合问题即$\binom{m+n}{m}$。然而本题给出m=n=20，从而得到如下公式：

15_formula.PNG-20.8kB \3.PNG)

至此我们得到复杂度为O(n)的算法，实现如下：

# Solution2
def getAllPaths2(n):
    result = 1
    for i in range(1, n+1):
        result = result * (n+i) / i
    return result

if __name__ == "__main__":
    # Get all possible paths of 20 x 20 grids
    final_result2 = getAllPaths2(20)
    print final_result2

Reference

方格问题升级之路（详细讨论格子路径问题）：https://blog.csdn.net/cookieZZ/article/details/70306757

格子路径问题+组合数学：https://www.cnblogs.com/yhm138/p/13610626.html#102-lattice-paths-without-restrictions-无限制格子路径

16. Power digit sum

Problem Description

$2^{15} = 32768$ and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number $2^{1000}$?

Solution

# Power digit sum
def getDigitSum(input_number):
    digitsum = 0
    while input_number != 0:
        digitsum += input_number % 10
        input_number /= 10
    return digitsum
if __name__ == "__main__":
    final_result = getDigitSum(2**1000)
    print final_result

17. Number letter counts

Problem Description

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 were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of “and” when writing out numbers is in compliance with British usage.

Solution

最简单的方法为建立三个字典，分别存储个位数字1-9，十位数字1-9（即10-90）以及11-19，然后判断输入数字的位数并进行相关处理。该方法需要讨论的情况较多，例如三位数字便需要讨论100，1X0，10X，11X，1XX五种情况，代码如下：

# Count the letters of numbers 1 to 1000
def countLetters(input_number, length, dict1, dict2, dict3):
    if length == 1:
        return dict1[input_number]
    elif length == 2:
        if input_number / 10 == 1:
            return dict2[input_number]
        elif input_number % 10 == 0:
            return dict3[input_number/10]
        else:
            return dict3[input_number/10] + dict1[input_number%10]
    else:
        if input_number % 100 == 0:
            return dict1[input_number/100] + 7
        elif input_number % 10 == 0:
            return dict1[input_number/100] + dict3[input_number%100/10] + 10
        elif input_number / 10 % 10 == 1:
            return dict1[input_number/100] + 10 + dict2[input_number%100]
        elif input_number / 10 % 10 == 0:
            return dict1[input_number/100] + 10 + dict1[input_number%10]
        else:
            return dict1[input_number/100] + dict3[input_number%100/10] + dict1[input_number%10] + 10

if __name__ == "__main__":
    # digits
    dict1 = {1:3, 2:3, 3:5, 4:4, 5:4, 6:3, 7:5, 8:5, 9:4}
    # 10-tens
    dict2 = {10:3, 11:6, 12:6, 13:8, 14:8, 15:7, 16:7, 17:9, 18:8, 19:8}
    # tens
    dict3 = {1:3, 2:6, 3:6, 4:5, 5:5, 6:5, 7:7, 8:6, 9:6}
    total_sum = 0
    for item in xrange(1, 1000):
        print item
        item_length = len(str(item))
        # print item_length
        total_sum += countLetters(item, item_length, dict1, dict2, dict3)
    
    total_sum += 11
    print total_sum

由于三位数字与两位数字相比仅增加了对百位数字的讨论，本质为增加X hundred and这几个字符。故可以1-99为基础进行适当求和，从而省略了三位数字包含字符数量的判断，节约了运算时间，关键代码如下：

for item in xrange(1, 100):
        item_length = len(str(item))
        total_sum += countLetters(item, item_length, dict1, dict2, dict3)
    block_sum = total_sum
    for hundredstype in xrange(1, 10):
        total_sum += dict1[hundredstype] + 7 + (dict1[hundredstype] + 10) * 99 + block_sum
    total_sum += 11
    print total_sum

18. Maximum path sum I

Problem Description

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.

18_problem_1.PNG-2.4kB \4.PNG)

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

18_problem_2.PNG-29.6kB \5.PNG)

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

Solution

根据题目说明采用动态规划进行逐级递归。

首先将所给的字符串类型转换为二维数组，随后进行分析，核心思想为将上一行的数字更新为其本身与下一行相邻两数字中较大数字之和：以倒数第二行元素为例，63可更新为63+max(04, 62)即125，66可更新为66+max(62, 98)即164…以此类推，更新结束后二维数组第一个元素即为所求最长路径。代码如下：

# Find the maximum total from top to bottom of the triangle below:
def transformToList(grid_string):
    # First transform the string to list
    tmp_list = [item.split(' ') for item in grid_string.split('\n')]
    # print tmp_list
    list_length = len(tmp_list)
    for i in range(list_length):
        for j in range(i+1):
            tmp_list[i][j] = int(tmp_list[i][j])
    return tmp_list
def dynamicProgramming(grid_list):
    list_length = len(grid_list)
    # Start from the last but one line
    for i in range(list_length-2, -1, -1):
        for j in range(i+1):
            grid_list[i][j] = grid_list[i][j] + max(grid_list[i+1][j], grid_list[i+1][j+1])
    return grid_list[0][0]
if __name__ == "__main__":
    grid = '''75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23'''
    result_list = transformToList(grid)
    final_result = dynamicProgramming(result_list)
    print final_result

19. Counting Sundays

Problem Description

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-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Solution1

# How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
# First we need to find the sum of the days from Febrary to December every year from 1901 to 2000
def countSundaysPerYear(input_year, first_day):
    # 31 days per month
    month_list1 = [1, 3, 5, 7, 8, 10, 12]
    # 30 days per month
    month_list2 = [4, 6, 9, 11]
    count = 0
    month = 1
    # input the location of the first day in this year as variable first_day
    while month <= 12:
        if first_day % 7 == 0:
            count += 1
        if month in month_list1:
            first_day += 31
        elif month in month_list2:
            first_day += 30
        else:
            if (input_year % 4 == 0 and input_year % 100 != 0) or (input_year % 400 == 0):
                first_day += 29
            else:
                first_day += 28
        month += 1
    # print count
    return (first_day % 7, count)
if __name__ == "__main__":
    start_day = 2
    final_count = 0
    for year in xrange(1901, 2001):
        # print (year, start_day)
        # print "-------"
        result_tuple = countSundaysPerYear(year, start_day)
        start_day = result_tuple[0]
        final_count += result_tuple[1]
    print final_count

Solution2

1
2
3

import calendar
result_list = [calendar.weekday(year,month,1) for year in range(1901, 2001) for month in range(1, 13)]
print result_list.count(6)

python提供calendar模块实现日历功能，提供对日期的操作函数，常用函数说明如下：https://www.cnblogs.com/liuxiaowei/p/7263888.html。

20. Factorial digit sum

Problem Description

n! means n × (n − 1) × … × 3 × 2 × 1

For 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 digits in the number 100!

Solution

# Count the sum of digits of 100!
import math 
def countDigitSum(input_number):
    sum_number = 0
    while input_number != 0:
        sum_number += input_number % 10
        input_number /= 10
    return sum_number
if __name__ == "__main__":
    final_result = countDigitSum(math.factorial(100))
    print final_result