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[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; }
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Answer: |
21124 |
（Problem 17）Number letter counts
原文：http://www.cnblogs.com/acutus/p/3545399.html