1005. Spell It Right (20)_for each test case, output in one line the digits -优快云博客

本文链接：https://blog.youkuaiyun.com/Peterclass/article/details/52049130

本文解析了一个PAT算法题，任务是计算一个非负整数各位数字之和，并将该和的每位数字以英文形式输出。文章提供了两种不同的实现方法，一种是逐位计算并转换为英文，另一种是先求和再整体转换。

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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

这道题目与PAT 基础层次1002题几乎一样，所以用类似的解法解决。具体思路见PAT Basic Level1002。

#include<stdio.h>

int main ( void )
{
    char number;
    int sum = 0;
    while ( (number = getchar()) != '\n') {
        sum += ( number - '0' );
    }

    int digit[ 10] = { 0 };
    int i = 0;
    while ( sum != 0 ){
        digit[ i++ ] = sum % 10;
        sum /= 10;
    }

    int j;
    for ( j = i - 1; j >= 0; j-- ){
        if ( j == 0 ){
            switch( digit[ j ] ){
                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;
                case 0: printf("zero"); break;
            }
        }else {
            switch( digit[ j ] ){
                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;
                case 0: printf("zero "); break;
            }
        }
    }
    return 0;
}

自己做完之后，又去看了一眼别人的解法，发现思路虽然差不多，但是他的代码量比我少了好多，贴在这里：

#include<stdio.h>
int main()
{
    char num[102],word[10][6]={"zero","one","two","three","four","five","six","seven","eight","nine"};
    int sum,i;
    scanf("%s",num);
    for(i=0;num[i];i++)
    {
        sum+=(num[i]-'0');
    }
    sprintf(num,"%d\0",sum);//将整数转化为字符串
    printf("%s",word[num[0]-'0']);
    for(i=1;num[i];i++)
        printf(" %s", word[num[i]-'0']);
}