1005. Spell It Right (20)

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']);}