關(guān)于啥是Project Euler 詳見 https://projecteuler.net/about?

觀前聲明：? ??

1. 這是個人興趣使然的對自己入坑pe的記錄，僅提供思路和部分代碼；各個方面肯定都是有著優(yōu)化與提升空間的，甚至在許多大佬看來這應該是初學者的淺薄而未經(jīng)剪枝的丑碼，如果能幫到有興趣的人自是最好，也歡迎能醍醐灌頂?shù)纳疃扔懻摗?/p>

2. 大佬看到了笑笑就行，還請輕噴。

3. 帶著惡意，抬杠者...俺也噴不過您，也不能拿您咋樣...畢竟這只是我個人興趣使然的行為或者說是學習記錄分享。?（說是記錄，但因為是早先寫的所以其實是在各種意義上公開處刑和吐槽自己 并嘗試補救優(yōu)化）

4. 語言是c++，用的VS平臺

第11題：

Largest product in a grid

Problem 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

手機端看這里：

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?

在20×20的數(shù)組中找到4個在相同方向上（上下左右斜）的連續(xù)數(shù)字，使它們的乘積為最大值

據(jù)說明眼人是5min內(nèi)就能看出來的。

那么.出于節(jié)能，還是直接放棄思考的暴搜吧.畢竟規(guī)模不大，每個數(shù)字檢索所有可能的方向全跑一遍也非常快，太簡單就不放代碼了（其實是自己寫的太菜用了幾個for還毫無優(yōu)化）

不過確實 多試試找?guī)讉€大數(shù)字就能試出來了

答案是70600674（至于在哪想依靠肉眼的有緣人就自己發(fā)現(xiàn)吧）

第12題：

Highly divisible triangular number

Problem 12

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?

28是第一個有超過5個因數(shù)的三角數(shù)，找到第一個有超過500個因數(shù)的三角數(shù);

三角數(shù)就是n*(n+1)/2沒啥好說的；

無非就是搜索—>判斷個數(shù)的類型的題，但找因子也有幾種找法：

對于一個數(shù)n，可以按照定義去找，遍歷1到n^1/2，如果能整除就說明是因數(shù)，令記錄的因數(shù)加1，最后把結(jié)果乘2即可;（注意這里沒有算n本身）

或者由質(zhì)因數(shù)分解定理，n=a1^p1*a2^p2*...*an^pn(ai為質(zhì)數(shù))，那么n的因子個數(shù)即為（p1+1）(p2+1)*...*(pn+1)（排列組合問題；注意這里算了n本身）

兩種寫法分別對應：

int facnum(int n)

{

int i, count = 0;

for (i = 1; i <= sqrt(n); i++)

if (n % i == 0)count++;

return 2 * count;

}

和

int facnum(int n)

{

int i = 1, s = 1, t;

while (n > 1)

{

i++, t = 0;

while (n % i == 0)

{

n /= i;

t++;

}

s *= (t + 1);

}

return s-1;

}

實際上第二個寫法我并沒有每次都余質(zhì)數(shù)，但可以發(fā)現(xiàn)如果把n余完，那么之后所有n的倍數(shù)都不能被余，因此得以避開所有的合數(shù).(實際上是懶得優(yōu)化) 用第二個寫法會比第一個快;

但這兩種寫法都是在正向考慮因數(shù)的個數(shù)，實際上還能這么看，遍歷所有的1到n，對于每個i，i一定是k*i的因子，所以使用一個數(shù)組記錄每個數(shù)的因子個數(shù)fac[n]，在第一個循環(huán)中對每個i再用一個循環(huán)使所有fac[k*i]++，跑完后1到n的因子個數(shù)就記錄下來了 只是要提前確定n的上界。

int fac[100000000] = { 0 };//這里將上界設置為100000000.

void getfac(void)

{

for (int n = 1; n <= 100000000; n++)

for (int k = 2; k * n <= 100000000; k++)

fac[k * n]++;

}

在主函數(shù)中找第一個fac[n*(n+1)/2]>500的n*(n+1)/2即可

答案：76576500；

第13題：

Large sum

Problem 13

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

算出這坨100個50位數(shù)字之和的前十位。

顯然這可以先當字符串再用大整數(shù)加法處理;我這里先把這50000個數(shù)字按順序全存進一個字符串s中，然后每次取出其中的50個進行加法(用s.substr(50 * (i - 1), 50))取，(i從1~100))

外部用一個數(shù)組sum[55]來記錄這個和;令sum[0]為大整數(shù)的位數(shù)，sum[1]為個位數(shù)，sum[2]為十位數(shù)，以此類推，因為sum是逆序存儲而s是順序存儲，所以加的時候是

sum[50-j] += sd[j] - '0';

加完后進行進位處理

for (j = 1; j <= sum[0]; j++)

{

if (sum[j] < 10)continue;

sum[j + 1] += sum[j] / 10;

sum[j] %= 10;

sum[0] += (sum[0] == j);//自動配位;

}

主函數(shù)大致是這樣：

for (i = 1; i <= 100; i++)//對這100個數(shù)字串逐個加法;

{

string sd = s.substr(50 * (i - 1), 50);//第i個數(shù)組串的第一個字符從50（i-1)開始;

for (j = 49; j >= 0; j--)

sum[50-j] += sd[j] - '0';

for (j = 1; j <= sum[0]; j++)

{

if (sum[j] < 10)continue;

sum[j + 1] += sum[j] / 10;

sum[j] %= 10;

sum[0] += (sum[0] == j);//自動配位;

}

}

for (i = sum[0]; i > sum[0] - 10; i--)

cout << sum[i];

答案是5537376230

第14題：

Longest Collatz sequence

Problem 14

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.

這也算挺出名的一個猜想了，找到1000000以下的能創(chuàng)造出最長鏈的起始數(shù)。

對這一類題自從我意識到可以外部數(shù)組記錄儲存之后都不是很想直視我以前的頭鐵算法...

但規(guī)模不大，頭鐵也不用很久，幾秒就可

這是從原始的定義莫得任何儲存和感情的直接暴搜：

long long change(long long n)

{

if (n == 1)return 1;

if (n % 2 == 0)return change(n / 2) + 1;

return change(3 * n + 1) + 1;

}

int main()

{

long long i, maxn = 0, maxterm = 0;

for (i = 1; i <= 1000000; i++)

{

long long temp = change(i);

if (temp > maxterm)?

{

maxterm = temp;

maxn = i;

}

}

cout << maxn << " " << maxterm;

return 0;

}

比較機智一點的優(yōu)化就是外部數(shù)組儲存每個數(shù)字作為起始數(shù)字的鏈長：

A[x]=A[x/2]+1或A[3*x+1]+1

那么遞歸函數(shù)就能這么寫：

long long dfs(long long x)

{

if (x == 1)return 1;

if (A[x])return A[x];

A[x] = (x % 2 == 0) ? dfs(x / 2) + 1 : dfs(3 * x + 1) + 1;

return A[x];

}

注意到dfs(3x+1)+1就是dfs((3x+1)/2)+2;所以還能接著優(yōu)化;

答案：837799

第15題：

Lattice paths

Problem 15

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 20×20 grid?

2×2的格子中左上到右下有6條路徑，那么20×20有幾條路徑？

水題，高中排列組合題...

將兩相鄰點之間的橫線或豎線視作一步，那么在n×n的格子共要走2n步，所以找出n×n的格子的所有路徑個數(shù)等價于求將2n個步劃分成向右與向下的2等分有多少種劃分，即是將2n步每步都標號，每步要么向右要么向下，但屬于向右的集合元素個數(shù)與向下集合元素個數(shù)相同（即為n）;

由排列組合知識可知有? (2n!)/((n!)(n!))個（證明詳見推廣伯努利分布或多項分布）

（實際上也不難理解，2n!就是全排列，但有些全排列對于選取而言是相同的，所以要除去重復部分，兩個n!分別是向右的集合和向下的集合的全排列，選定元素后這兩部分的全排列對于實際選取而言是相同的 所以除掉.）

代入n=20，答案為137846528820

又是節(jié)能的一天;有緣再見.