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(Problem 17)Number letter counts

By Wu Yudong on January 15, 2017

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.

题目大意:

如果用英文写出数字1到5: one, two, three, four, five, 那么一共需要3 + 3 + 5 + 4 + 4 = 19个字母。

如果数字1到1000(包含1000)用英文写出,那么一共需要多少个字母?

注意: 空格和连字符不算在内。例如,342 (three hundred and forty-two)包含23个字母; 115 (one hundred and fifteen)包含20个字母。"and" 的使用与英国标准一致。

// (Problem 16)Number letter counts
// Completed on Sun, 17 Nov 2013, 16:30
// Language: C
//
// 版权所有(C)wuyudong
// 博客地址:http://www.wuyudong.com
#include <stdio.h> 
#include <stdbool.h>

int a[101] = {0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8};

void init(void)  //初始化数组
{
    a[20] = 6;
    a[30] = 6;
    a[40] = 5;
    a[50] = 5;
    a[60] = 5;
    a[70] = 7;
    a[80] = 6;
    a[90] = 6;
    a[100] = 7;
}

int within100(void)  //计算1~99所含字母的和
{
    int i, sum, t;
    t = sum = 0;
    for(i = 1; i <= 9; i++) t += a[i];
    for(i = 1; i <= 19; i++) sum += a[i];
    for(i = 2; i <= 9; i++) {
        sum += a[i*10] * 10;
        sum += t;
    }
    return sum;
}

void solve(void)
{
    int i;
    int sum, t;
    sum = t = within100();
    for(i = 1; i < 10; i++) {
        sum += (a[i] + 10) * 99 + (a[i] + 7) + t;
    }
    sum += 11;

    printf("%d\n",sum);
}

int main(void)
{
    init();
    solve();
    return 0;
}

Answer: 21124
Completed on Mon, 18 Nov 2013, 00:30

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