[Project Euler] 来做欧拉项目练习题吧: 题目017

                [Project Euler] 来做欧拉项目练习题吧: 题目017

                              周银辉

 

题目描述:

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.

 

问题分析:

这个题目比较简单哈，用铅笔和纸也比如容易算出啦。比如计算单个数字的出现次数，"hundred"单词出现的次数等等。

而用程序计算嘛，关键在于创建一个数字和对应单词长度的映射，也就是上面numbers那个数组。

然后分离出百位数、十位数、个位数，以及处理各种特殊情况就可以了。

 

#include <stdio.h>

int numbers[91];

void ini_numbers()

{

//为了节省空间，用hash_map也可以

numbers[0] = 0;//no pronunciation for zero 

numbers[1] = 3;//"one"

numbers[2] = 3;//"two"

numbers[3] = 5;//"three"

numbers[4] = 4;//"four"

numbers[5] = 4;//"five"

numbers[6] = 3;//"six"

numbers[7] = 5;//"seven"

numbers[8] = 5;//"eight"

numbers[9] = 4;//"nine"

numbers[10] = 3;//"ten"

numbers[11] = 6;//"eleven"

numbers[12] = 6;//"twelve"

numbers[13] = 8;//"thirteen"

numbers[14] = 8;//"fourteen"

numbers[15] = 7;//"fifteen"

numbers[16] = 7;//"sixteen"

numbers[17] = 9;//"seventeen"

numbers[18] = 8;//"eighteen"

numbers[19] = 8;//"nineteen"

numbers[20] = 6;//"twenty"

numbers[30] = 6;//"thirty"

numbers[40] = 5;//"forty"

numbers[50] = 5;//"fifty"

numbers[60] = 5;//"sixty"

numbers[70] = 7;//"seventy"

numbers[80] = 6;//"eighty"

numbers[90] = 6;//"ninety"

}

int get_length(int n)

{

int a=0; //hundreds'digit

int b=0; //ten's digit

int c=0; //units' digit

int length = 0;

a = n/100;

n = n%100;

b = n/10;

c = n%10;

if(a!=0)

{

length += numbers[a] + 7; // 7 for "hundred"

if(b!=0 || c!=0)

{

length += 3; //3 for "and"

}

}

if(b!=0)

{

if(b==1)

{

length += numbers[b*10+c];

return length;

}

else

{

length += numbers[b*10];

}

}

if(c!=0)

{

length += numbers[c];

}

return length;

}

int main()

{

int i, length=0;

ini_numbers();

for(i=1; i<=999; i++)

{

length += get_length(i);

}

length += 11; //11 for "one thousand"

printf("total length: %d\n", length);

return 0;

}