PAT甲级1005. Spell It Right

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (<= 10100).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:
12345

Sample Output:
one five

题目要求：

任给一个非负整数N，计算整数的每一位的和，并输出每一位的英文；

输入：输入的N<=10100

输出：计算的每一位的和的每一位对应一个英文单词，而且英文单词之间需要有一个空格，最后一位之后没有空格

解题思路：

首先将整数的每一位进行分离，然后计算出和，计算出和后也需要将每一位分离；将整数的每一位分离的方法采用除10取余的方式；和的每一位分离时，是从整数的最小位开始，而整数输出需要从最高位开始，因此需要将分离的数位存储在一个数组中；

#include "stdlib.h"#include "stdio.h"int main() {    char ch;    int count = 0;    int words[4];    int sum = 0;    ch = getchar();    while (ch != '\n')    {        sum += ch - '0';        ch = getchar();    }    while (sum / 10 > 0)    {        words[count++] = sum % 10;        sum = sum / 10;    }    words[count++] = sum % 10;    for (int i = count - 1; i >= 0; i--)    {        switch (words[i])        {        case 0:            printf("zero");            break;        case 1:            printf("one");            break;        case 2:            printf("two");            break;        case 3:            printf("three");            break;        case 4:            printf("four");            break;        case 5:            printf("five");            break;        case 6:            printf("six");            break;        case 7:            printf("seven");            break;        case 8:            printf("eight");            break;        case 9:            printf("nine");            break;        default:            break;        }        if (i!=0)        {            printf(" ");        }    }    return 0;}