Project Euler 011~015
关于啥是Project Euler 详见 https://projecteuler.net/about
观前声明：
这是个人兴趣使然的对自己入坑pe的记录，仅提供思路和部分代码；各个方面肯定都是有着优化与提升空间的，甚至在许多大佬看来这应该是初学者的浅薄而未经剪枝的丑码，如果能帮到有兴趣的人自是最好，也欢迎能醍醐灌顶的深度讨论。
大佬看到了笑笑就行，还请轻喷。
带着恶意，抬杠者...俺也喷不过您，也不能拿您咋样...毕竟这只是我个人兴趣使然的行为或者说是学习记录分享。 （说是记录，但因为是早先写的所以其实是在各种意义上公开处刑和吐槽自己 并尝试补救优化）
语言是c++，用的VS平台
第11题：
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
手机端看这里：
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的数组中找到4个在相同方向上（上下左右斜）的连续数字，使它们的乘积为最大值
据说明眼人是5min内就能看出来的。
那么.出于节能，还是直接放弃思考的暴搜吧.毕竟规模不大，每个数字检索所有可能的方向全跑一遍也非常快，太简单就不放代码了（其实是自己写的太菜用了几个for还毫无优化）
不过确实 多试试找几个大数字就能试出来了
答案是70600674（至于在哪想依靠肉眼的有缘人就自己发现吧）
第12题：
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是第一个有超过5个因数的三角数，找到第一个有超过500个因数的三角数;
三角数就是n*(n+1)/2没啥好说的；
无非就是搜索—>判断个数的类型的题，但找因子也有几种找法：
对于一个数n，可以按照定义去找，遍历1到n^1/2，如果能整除就说明是因数，令记录的因数加1，最后把结果乘2即可;（注意这里没有算n本身）
或者由质因数分解定理，n=a1^p1*a2^p2*...*an^pn(ai为质数)，那么n的因子个数即为（p1+1）(p2+1)*...*(pn+1)（排列组合问题；注意这里算了n本身）
两种写法分别对应：
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;
}
实际上第二个写法我并没有每次都余质数，但可以发现如果把n余完，那么之后所有n的倍数都不能被余，因此得以避开所有的合数.(实际上是懒得优化) 用第二个写法会比第一个快;
但这两种写法都是在正向考虑因数的个数，实际上还能这么看，遍历所有的1到n，对于每个i，i一定是k*i的因子，所以使用一个数组记录每个数的因子个数fac[n]，在第一个循环中对每个i再用一个循环使所有fac[k*i]++，跑完后1到n的因子个数就记录下来了 只是要提前确定n的上界。
int fac[100000000] = { 0 };//这里将上界设置为100000000.
void getfac(void)
{
for (int n = 1; n <= 100000000; n++)
for (int k = 2; k * n <= 100000000; k++)
fac[k * n]++;
}
在主函数中找第一个fac[n*(n+1)/2]>500的n*(n+1)/2即可
答案：76576500；
第13题：
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
算出这坨100个50位数字之和的前十位。
显然这可以先当字符串再用大整数加法处理;我这里先把这50000个数字按顺序全存进一个字符串s中，然后每次取出其中的50个进行加法(用s.substr(50 * (i - 1), 50))取，(i从1~100))
外部用一个数组sum[55]来记录这个和;令sum[0]为大整数的位数，sum[1]为个位数，sum[2]为十位数，以此类推，因为sum是逆序存储而s是顺序存储，所以加的时候是
sum[50-j] += sd[j] - '0';
加完后进行进位处理
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);//自动配位;
}
主函数大致是这样：
for (i = 1; i <= 100; i++)//对这100个数字串逐个加法;
{
string sd = s.substr(50 * (i - 1), 50);//第i个数组串的第一个字符从50（i-1)开始;
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);//自动配位;
}
}
for (i = sum[0]; i > sum[0] - 10; i--)
cout << sum[i];
答案是5537376230
第14题：
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.
这也算挺出名的一个猜想了，找到1000000以下的能创造出最长链的起始数。
对这一类题自从我意识到可以外部数组记录储存之后都不是很想直视我以前的头铁算法...
但规模不大，头铁也不用很久，几秒就可
这是从原始的定义莫得任何储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;
}
比较机智一点的优化就是外部数组储存每个数字作为起始数字的链长：
A[x]=A[x/2]+1或A[3*x+1]+1
那么递归函数就能这么写：
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;所以还能接着优化;
答案：837799
第15题：
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条路径，那么20×20有几条路径？
水题，高中排列组合题...
将两相邻点之间的横线或竖线视作一步，那么在n×n的格子共要走2n步，所以找出n×n的格子的所有路径个数等价于求将2n个步划分成向右与向下的2等分有多少种划分，即是将2n步每步都标号，每步要么向右要么向下，但属于向右的集合元素个数与向下集合元素个数相同（即为n）;
由排列组合知识可知有 (2n!)/((n!)(n!))个（证明详见推广伯努利分布或多项分布）
（实际上也不难理解，2n!就是全排列，但有些全排列对于选取而言是相同的，所以要除去重复部分，两个n!分别是向右的集合和向下的集合的全排列，选定元素后这两部分的全排列对于实际选取而言是相同的 所以除掉.）
代入n=20，答案为137846528820
又是节能的一天;有缘再见.
