欧拉前线9


11

 

欧拉计划(Project Euler)是另一个有趣的编程挑战网站,可以进行比赛(好玩)。早期的问题起初很温和,但随后的困难却激增至最初的一百多个左右。前几个问题在查找素数,倍数和因数之间具有一些共性,因此可能会有一些有趣的代码微重用的机会可以使用。

因此,编写一个无需先验知识即可解决前9个问题中的任何一个的程序。

  1. 用户通过调用时的参数或运行时的标准输入,由用户选择ASCII(1到9)(包括ASCII)从1到9。(您可以计算所有答案,但只能显示一个。)
  2. 正确答案必须使用ASCII以10为底的新行打印。
  3. 程序应在不到一分钟的时间内执行(建议使用PE)。
  4. 通过“无先验知识”,我的意思是你的代码必须获得答案,而无需外部资源。像这样的程序将被认为是无效的(不过,如果我没有打错,还是正确的):

    print[233168,4613732,6857,906609,232792560,25164150,104743,40824,31875000][input()-1]
    

    对于问题8(涉及1000位数字),您可以从外部文件中读取该数字,只需指定其存储方式(例如,二进制,文本,标头,导入的模块)和/或将其包含在您的答案中(不计入主程序的长度)。

  5. 分数按字节计。

  6. 2周后,字节计数的领导者获得了15个Unicorn Points™奖励。

Answers:


4

Python,505

import f
A,B,C,D=map(range,[22,1000,101,500])
R=reduce
M=int.__mul__
a=x=0
b=1
n=2
p=[]
while b<=4e6:a,b=b,a+b;x+=b*(b%2<1)
while len(p)<=1e4:p+=[n]*all(n%i for i in p);n+=1
q=y=R(M,p[:8])
while any(y%(i+1)for i in A):y+=q
print[
    sum(i for i in B if i%3*(i%5)<1),
    x,
    max(i for i in p if 600851475143%i<1),
    max(a*b for a in B for b in B if`a*b`==`a*b`[::-1]),
    y,
    sum(C)**2-sum(i**2for i in C),
    p[-1],
    max(R(M,map(int,f.s[i:i+5]))for i in B),
    [a*b*(1000-a-b)for a in D for b in D if(a+b)*1e3==5e5+a*b][0]
][input()-1]

空格被添加到最后一行以提高可读性。1000位数字是从名为f.py包含以下行的模块中导入的:

s="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"

3

Java脚本

解决方案:1785个字符

用nodeJS执行代码

关于问题7的注释:算法还可以,但是需要一点时间!如果有人有更有效的解决方案...

z=process.argv[2]
y=console.log
if(z==1){b=0;for(i=1e3;i--;)if(i%3<1||i%5<1)b+=i;y(b)}
if(z==2){d=e=1;f=0;while(e<=4e6)g=d+e,d=e,e=g,f+=e%2<1?e:0;y(f)}
if(z==3){e=Math.sqrt(d=600851475143)|0+1;f=2;while(f<e){g=d/f;if(g==(g|0))h=f,d=g;f+=f==2?1:2}y(h)}
if(z==4){for(a=b=100,c=0;a+b<1998;){d=a++*b;if(d==(""+d).split("").reverse().join("")&&d>c)c=d;if(a>999)a=b+1,b++}y(c)}
if(z==5){for(a=c=1;c;){for(b=20;b>1;)a%b>0?b=0:b--;b==1?c=0:a++}y(a)}
if(z==6){for(a=100,b=Math.pow(a*(a+1)/2,2),d=a+1;d--;)b-=d*d;y(b)}
if(z==7){for(a=3,c=d=0;c<1e4;){for(b=a;b>2;){b=b>3?b-2:2;if(a%b<1)b=0}if(b>0)c++,d=a;a=a+2}y(d)}
if(z==8){a="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";for(e=0,b=996;b--;){d=eval(a.substr(b,5).split("").join("*"));if(d>e)e=d};y(e)}
if(z==9){for(a=b=0;(c=a+b+Math.sqrt(a*a+b*b))!=1e3;)if(++a==500)a=0,b++;y(a,b,c-b-a)}

问题1:39个字符

b=0;for(i=1e3;i--;)if(i%3<1||i%5<1)b+=i

问题2:49个字符

d=e=1;f=0;while(e<=4e6)g=d+e,d=e,e=g,f+=e%2<1?e:0

问题3:85个字符

e=Math.sqrt(d=600851475143)|0+1;f=2;while(f<e){g=d/f;if(g==(g|0))h=f,d=g;f+=f==2?1:2}

问题4:105个字符

for(a=b=100,c=0;a+b<1998;){d=a++*b;if(d==(""+d).split("").reverse().join("")&&d>c)c=d;if(a>999)a=b+1,b++}

问题5:55个字符

for(a=c=1;c;){for(b=20;b>1;)a%b>0?b=0:b--;b==1?c=0:a++}

问题6:51个字符

for(a=100,b=Math.pow(a*(a+1)/2,2),d=a+1;d--;)b-=d*d

问题7:82个字符

for(a=3,c=d=0;c<1e4;){for(b=a;b>2;){b=b>3?b-2:2;if(a%b<1)b=0}if(b>0)c++,d=a;a=a+2}

问题8:1078个字符^ _ ^

a="7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
for(e=0,b=996;b--;){d=eval(a.substr(b,5).split("").join("*"));if(d>e)e=d}

问题9:58个字符

for(a=b=0;a+b+Math.sqrt(a*a+b*b)!=1e3;)if(++a==500)a=0,b++

if(i%3<1||i%5<1)a+=i更短!:)
Michael M.

简短的问题3:e=Math.sqrt(d=600851475143)|0+1;f=2;while(f<e){g=d/f;if(g==g|0)h=f,d=g;f+=f==2?1:2}
Florent 2014年

简短的问题4:for(a=b=100,c=0;a+b<1998;){d=a++*b;if(d==(""+d).split("").reverse().join("")&&d>c)c=d;if(a>999)a=b+1,b++}
Florent

@Florent:问题3 g==g|0不起作用g|0必须在括号之间
guy777 2014年

1
@Florent:解决方案在变量中h,而不是脚本返回的值
guy777 2014年

1

R 684个字符

f=function(N){r=rowSums;p=function(x,y)r(!outer(x,y,`%%`));S=sum;x=c(1,1);n=600851475143;a=900:999;b=20;c=1:100;m=2:sqrt(n);M=m[!n%%m];A=a%o%a;d=3;P=2;z=1:1e3;Z=expand.grid(z,z);Y=cbind(Z,sqrt(r(Z^2)));W=gsub("\n","",xpathApply(htmlParse("http://projecteuler.net/problem=8"),"//p",xmlValue)[[2]]);switch(N,S(which(p(1:999,c(3,5))>0)),{while(tail(x,1)<4e6)x=c(x,S(tail(x,2)));S(x[!x%%2])},max(M[p(M,M)<2]),max(A[sapply(strsplit(c(A,""),""),function(x)all(x==rev(x)))],na.rm=T),{while(any(b%%1:20>0))b=b+20;b},S(c)^2-S(c^2),{while(P<=1e4){d=d+2;if(sum(!d%%2:d)<2)P=P+1};d},max(sapply(5:nchar(W),function(i)prod(as.integer(strsplit(substr(W,i-4,i),"")[[1]])))),prod(Y[r(Y)==1000,][1,]))}

缩进:

f=function(N){
    r=rowSums
    p=function(x,y)r(!outer(x,y,`%%`))
    S=sum
    x=c(1,1)
    n=600851475143
    a=900:999
    b=20
    c=1:100
    m=2:sqrt(n)
    M=m[!n%%m]
    A=a%o%a
    d=3
    P=2
    z=1:1e3
    Z=expand.grid(z,z)
    Y=cbind(Z,sqrt(r(Z^2)))
    W=gsub("\n","",xpathApply(htmlParse("http://projecteuler.net/problem=8"),"//p",xmlValue)[[2]])
    switch(N,S(which(p(1:999,c(3,5))>0)),
             {while(tail(x,1)<4e6)x=c(x,S(tail(x,2)));S(x[!x%%2])},
             max(M[p(M,M)<2]),
             max(A[sapply(strsplit(c(A,""),""),function(x)all(x==rev(x)))],na.rm=T),
             {while(any(b%%1:20>0))b=b+20;b},
             S(c)^2-S(c^2),
             {while(P<=1e4){d=d+2;if(sum(!d%%2:d)<2)P=P+1};d},
             max(sapply(5:nchar(W),function(i)prod(as.integer(strsplit(substr(W,i-4,i),"")[[1]])))),
             prod(Y[r(Y)==1000,][1,]))
    }

用法:

> f(1)
[1] 233168
> f(2)
[1] 4613732
> f(3)
[1] 6857
> f(4)
[1] 906609
> f(5)
[1] 232792560
> f(6)
[1] 25164150
> f(7)
[1] 104743
> f(8)
[1] 40824
> f(9)
[1] 31875000

分别:

1:48个字符 sum(which(rowSums(!outer(1:999,c(3,5),`%%`))>0))

2:64个字符 x=c(1,1);while(tail(x,1)<4e6)x=c(x,sum(tail(x,2)));sum(x[!x%%2])

3:73个字符 n=600851475143;m=2:sqrt(n);M=m[!n%%m];max(M[rowSums(!outer(M,M,`%%`))<2])

4:88个字符 a=900:999;b=a%o%a;max(b[sapply(strsplit(c(b,""),""),function(x)all(x==rev(x)))],na.rm=T)

5:34个字符 a=20;while(any(a%%1:20>0))a=a+20;a

6:25个字符 a=1:100;sum(a)^2-sum(a^2)

7:54个字符 d=3;P=2;while(P<=1e4){d=d+2;if(sum(!d%%2:d)<2)P=P+1};d

8:181个字符

第一行从项目欧拉网站上读取数字,第二行实际执行计算。

W=gsub("\n","",xpathApply(htmlParse("http://projecteuler.net/problem=8"),"//p",xmlValue)[[2]])
max(sapply(5:nchar(W),function(i)prod(as.integer(strsplit(substr(W,i-4,i),"")[[1]]))))

9:87个字符 z=1:1e3;Z=expand.grid(z,z);Y=cbind(Z,sqrt(rowSums(Z^2)));prod(Y[rowSums(Y)==1000,][1,])


到目前为止,f(7)需要花费2分钟来计算,因此它实际上并不符合执行时间的限制。我会尽力而为。(f(5)在我的计算机上需要48s)
plannapus

1

Ĵ 245 236 232

load'n'
echo".>(<:".1!:1]1){<;._1'!+/I.+./0=5 3|,:~i.1e3!+/}:(],4&*@:{:+_2&{)^:(4e6>{:)^:_]0 2!{:q:600851475143!>./(#~(-:|.)@":"0),/*/~i.1e3!*./>:i.20!(([:*:+/)-[:+/*:)i.101!p:1e4!>./5*/\"."0 n!x:*/{.(#~1e3=+/"1)(+.,|)"0,j./~i.500'

n是一个包含以下内容的文件:

n=:'7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450'

0

TI-BASIC(工作中)

对于您的TI-83或TI-84计算器

主程序,15个字节

Input X:OpenLib(1):X:ExecLib:Disp Ans

库1(4个字节确实计入总字节数:

L1(Ans

然后,我们得到列表1:

work in progress
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