PAT1002

//10的100次方一共101位数
#include <stdio.h>


int main()
{
    int indexNum = 0;int sum = 0;char num[101] = {0};char result[10] = {0};int indexResult = 0;int aa = 0;
    scanf("%s",&num);
    char pinyin[10][5] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};//构造一个10*10的矩阵另外多加一行
    while(num[indexNum] != '\0')
    {
        sum += num[indexNum] - '0';//字符转数字
        indexNum++;
    }
    while(sum != 0)
    {
        result[indexResult] = sum % 10;//注意先得到的是个位，依次..
        sum /= 10;
        indexResult++;
    }
    for (int i = indexResult - 1;i >= 0 ;i--)
    {
        printf("%s",pinyin[result[i]]);
        if(i!=0)
            printf(" ");
    }
}