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[交流] Euler Project Q17. 歐拉工程第十七題已有3人參與

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.

如果數(shù)字1-5寫(xiě)成英文單詞形式 one, two, three, four, five, 那么一共使用了3 + 3 + 5 + 4 + 4 = 19個(gè)字母.

如果1-1000的數(shù)字被寫(xiě)成單詞形式,一共使用多少字母?

注意:不要計(jì)算空格,連字符.例如343 (three hundred and forty-two)包含23個(gè)字母,115 (one hundred and fifteen) 包含20個(gè)字母. 數(shù)字轉(zhuǎn)單詞時(shí)"and"用法使用英國(guó)格式.

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matlab/VB/python/c++/Java寫(xiě)程序請(qǐng)發(fā)QQ郵件:790404545@qq.com

1樓 2011-05-28 03:49:14

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微塵、夢(mèng)想(金幣+5): 謝謝參與！ 2011-05-29 16:31:59

完成了Python版

CODE:

#/bin/env python

ones = ["", "one", "two", "three", "four", "five",

        "six", "seven", "eight", "nine", "ten",

        "eleven", "twelve", "thirteen", "fourteen",

        "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]

tens = ["", "twenty", "thirty", "forty", "fifty",

        "sixty", "seventy", "eighty", "ninety"]

words = ["", " thousand", " million", " billion"]

def num2words(n):

    "Convert an integer to english words"

    def part(num, s):

        "Convert a part of numbers"

        if num < 20:

            return s + ones[num]

        if num < 100:

            return s + tens[num / 10 - 1] + " " +  ones[num % 10]

        return ones[num / 100] + " hundred " + s + part(num % 100, "")

    # divided into several part

    # each part contain three numbers

    t = 1000

    # results list

    r = []

    i = 0

    while n > 0:

        # add "and" into the word

        if i == 0 and n > 100 and n % 100 != 0:

            r.append(part(n % t, 'and ') + words[i])

        else:

            r.append(part(n % t, "") + words[i])

        n /= t

        i += 1

    r.reverse()

    return r

if __name__ == "__main__":

    s = 0

    for i in range(1, 1001):

        for w in num2words(i):

            s += sum([len(x) for x in w.split()])

    print "total length:", s

同樣推薦樓上用BBCode發(fā)布代碼.

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4樓2011-05-28 13:57:38

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微塵、夢(mèng)想(金幣+5): 謝謝參與！ 2011-05-29 16:31:31

這個(gè)題統(tǒng)計(jì)很簡(jiǎn)單,稍有難度就是數(shù)字變單詞的算法

matlab code

CODE:

 %% How many letters would be needed to write all the numbers in words from 1 to 1000?

function result = euler17()

tic;

numstr = '';

for i=1:1000

    numstr = [numstr,' ',num2words(i)];

end

numstr = strrep(numstr,' ','');

numstr = strtrim(numstr);

result = length(numstr);

toc;

end

%% write a number in words

% called by euler 17

function result = num2words(num)

key = [1:20,30:10:90];

value = {'one','two','three','four','five','six','seven','eight','nine','ten','eleven','twelve', ...

        'thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen', ...

        'twenty','thirty','forty','fifty','sixty','seventy','eighty','ninety'};

numlist = containers.Map(1,'one'); % must add a value first, then map can hold the right key type

for i=2:length(key)

    numlist(key(i)) = value{i};

end

result = '';

switch length(num2str(num))

    case 4

        if mod(num,1000)==0

            result = [result,numlist(fix(num/1000)),' thousand',num2words(rem(num,1000))];

        else

            result = [result,numlist(fix(num/1000)),' thousand and ',num2words(rem(num,1000))];

        end

    case 3

        if mod(num,100)==0

            result = [result,numlist(fix(num/100)),' hundred',num2words(rem(num,100))];

        else

            result = [result,numlist(fix(num/100)),' hundred and ',num2words(rem(num,100))];

        end

    case 2

        if fix(num/10)==1

            result = [result,numlist(num)];

        else

            result = [result,numlist(fix(num/10)*10),' ',num2words(rem(num,10))];

        end

    case 1

        if num==0

            result = '';

        else

            result = [result,numlist(num)];

        end

    otherwise

        result = '';

end

end

matlab效率還是不高

CODE:

 % Elapsed time is 6.267496 seconds.

% ans =

%        21124

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2樓2011-05-28 03:54:36

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匿名

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本帖僅樓主可見(jiàn)

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微塵、夢(mèng)想(金幣+5): 謝謝參與！ 2011-05-29 16:32:11

這一題很無(wú)聊阿
結(jié)果

CODE:

21124

elapsed time=0.000000 seconds.

代碼

CODE:

#include 

#include 

#include 

#define TIMERSTART clock_t start_time,stop_time;double elapsed_time;start_time = clock();

#define TIMERSTOP stop_time = clock();elapsed_time=(double)(stop_time-start_time)/CLOCKS_PER_SEC;printf("elapsed time=%f seconds.\n",elapsed_time);

static int bufa[]={4,3,3,5,4,4,3,5,5,4,//zero, one, ..., nine,

                  3,6,6,8,8,7,7,9,8,8 };// ten, eleven, ..., nineteen

static int bufc[]={0,0,6,6,5,5,5,7,6,6}; //0, 0, twenty, thirty, ..., ninety

int count(int n){ 

  if(n<20)return bufa[n];

  if(n<100) {

    if(n%10==0) return bufc[n/10];

    else 

      return bufc[n/10]+bufa[n%10];

  }

  if(n<1000) {

    if(n%100==0) 

      return bufa[n/100]+7; 

    else

      return bufa[n/100]+count(n%100)+10;

  }else if(n==1000) return 11;

  return -1;

}

int euler17(int n){

  int i,sum=0;

  for(i=1;i
    sum+=count(i);

  return sum;

}

int main(void){

 TIMERSTART;

 printf("%d\n",euler17(1000));

 TIMERSTOP;

  return 0;

}

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