（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。

#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 Sun, 17 Nov 2013, 16:30