（Problem 17）Number letter counts

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）acutus   (mail: acutus@126.com)
// 博客地址：http://www.cnblogs.com/acutus/#include <stdio.h>
#include <stdbool.h>

int a[] = {,,,,,,,,,,,,,,,,,,,};

void init(void)  //初始化数组
{
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
    a[] = ;
}

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

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

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

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

Answer:

21124