题目
Problem 008: Largest product in a series
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
解答
本题要求找出 1000 位整数中连续 13 个数字的最大乘积。使用 Haskell 解答如下:
import Data.List
import Data.Char
main = do
a <- readFile "008.txt"
print . maximum . map (product . take 13) . tails $ map digitToInt $ filter isDigit a
在上述程序中:
- 首先使用 readFile 函数将 008.txt 文件的内容(1000 位整数)读入到字符串 a 中。
- 最后一行从右往左分析,filter isDigit a 过滤掉 \n 等字符,只留下数字字符。
- map digitToInt 将数字字符转换为(只有一位的)整数,得到的是单个数字的列表:[7,3,1,6,7,...,4,5,0]。
- tails 函数返回列表中所有尾部元素组成的列表的列表:[ [7,3,1,6,7,...,4,5,0], [3,1,6,7,...,4,5,0], [1,6,7,...,4,5,0], [6,7,...,4,5,0], [7,...,4,5,0], ..., [4,5,0], [5,0], [0], [] ]。
- map (product . take 13) 对列表中的每个元素(也是列表)进行以下操作:take 13 从列表开头取出 13 个元素(如果不够 13 个则有多少个就取多少个),product 将这些元素全部乘起来。结果是:[ 5000940, 0, 0, 0, 0, ..., 0, 0, 0, 1 ]。
- maximum 函数找出这些乘积中最大的。
- print 函数输出结果。大功告成。
这个程序的运行时间是 0.006 秒。
用scala写了下
val begin:Long = System.currentTimeMillis()
val content = Source.fromFile("file/text").filter(_.isDigit).map(_.toInt - '0').toList
val lists =
for (i <- 0 to content.size - 13) yield content.drop(i).take(13)
println(lists.maxBy(_.reduce(_*_)))
val end:Long = System.currentTimeMillis()
println(end-begin)
读取文件转换为List就花了80毫秒 一共110毫秒的样子...话说haskell挺快啊..我用java的FileChannel试了下读取保存的1019B的文件都花了10毫秒左右的样子..
I/O依旧是10MS
处理起来 只需要17MS左右的样子....
话说图灵的评论不支持markdown嘛...
long begin = System.currentTimeMillis();
byte[] bytes = Files.readAllBytes(Paths.get("file/text"));
List<Integer> list=new ArrayList<>();
for(int i=0;i<bytes.length;i++){
if(bytes[i]-'0'>=0&&bytes[i]-'0'<=9) list.add(bytes[i]-'0');
}
int max=-1;
List<Integer> maxList=new ArrayList<>();
List<Integer> temp=new ArrayList<>();
for(int i=0;i<=list.size()-13;i++){
int value=1;
for(int j=i;j<i+13;j++){
temp.add(list.get(j));
value*=list.get(j);
}
if(value > max) {
max = value;
maxList.clear();
maxList.addAll(temp);
}
temp.clear();
}
System.out.println(maxList);
long end = System.currentTimeMillis();
System.out.println(end - begin);
Array.prototype.tails = function(){var z = [], len=this.length; for(var i=len;i>=0;i--){z.push(this.slice(i,len));} return z; }
function product13(arr){var p=1, len=arr.length, max=len>13?13:len; for(var i=0;i<max;i++){p=p*arr[i];}; return p; }
Math.max.apply(this,str.replace(/\s/g,"").split("").tails().map(product13));
procedure TForm1.btn1Click(Sender: TObject);
var
nBegin, nEnd, nInterval, nFrequency: Int64;
nProduct: Integer;
oFile: TextFile;
sNumbers: string;
bIsSupport: Boolean;
begin
QueryPerformanceCounter(nBegin);;
AssignFile(oFile, '..\..\008.txt');
Reset(oFile);
try
while not Eof(oFile) do
begin
Readln(oFile, sNumbers);
end;
finally
CloseFile(oFile);
end;
//去掉空格及换行符号
sNumbers := StringReplace(sNumbers, ' ', '', [rfReplaceAll]);
sNumbers := StringReplace(sNumbers, #13#10, '', [rfReplaceAll]);
sNumbers := StringReplace(sNumbers, #13, '', [rfReplaceAll]);
Assert(Length(sNumbers) = 1000);
nProduct := GetMaxAdjacentProdouct(13, sNumbers);
QueryPerformanceCounter(nEnd);
bIsSupport := QueryPerformanceFrequency(nFrequency);
nInterval := nEnd - nBegin;
nFrequency := IfThen(bIsSupport, nFrequency, 1);
ShowMessage(Format('乘积为 %d , 用时 %f ms.', [nProduct, (nInterval * 1000 / nFrequency)]));
end;
function TForm1.GetMaxAdjacentProdouct(AdjacentNum: Integer; const ASrcNumbers: string): Integer;
var
nIndex, nCommonProduct, nRemain, I: Integer;
bIsFirst: Boolean;
begin
Result := 0;
nIndex := 1;
bIsFirst := True;
nCommonProduct := -1;
while nIndex <= Length(ASrcNumbers) - AdjacentNum + 1 do
begin
nRemain := IfThen(bIsFirst, StrToInt(ASrcNumbers[nIndex]),
StrToInt(ASrcNumbers[nIndex + AdjacentNum]));
if nRemain = 0 then
begin
Inc(nIndex, AdjacentNum);
bIsFirst := True;
Continue;
end;
if bIsFirst then
begin
nCommonProduct := 1;
for I := nIndex + 1 to nIndex + AdjacentNum - 1 do
begin
nCommonProduct := nCommonProduct * StrToInt(ASrcNumbers[I]);
if nCommonProduct = 0 then
Break;
end;
end;
Result := Max(Result, nRemain * nCommonProduct);
Inc(nIndex);
bIsFirst := not bIsFirst;
end;
end;
python解欧拉计划008-相邻13个数字最大乘积.py
http://pe-cn.github.io/8/
2016年4月9日 07:01:13 codegay
我开了一个github项目来解欧拉计划,欢迎大家贡献自己的原创代码:
https://github.com/FGFW/projecteuler
"""
n=''.join("""73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450""".split())
import operator
import functools
result=max([functools.reduce(operator.mul,[int(r) for r in n[k:k+13]]) for k,v in enumerate(n)])
print(result)
#23514624000
#[Finished in 0.1s]