1005 Spell It Right (20)

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本文介绍了一道编程题目1005SpellItRight的解题方法，主要内容包括读取非负整数N，计算其各位数字之和，并用英文单词输出该和的每一位数字。通过字符串处理和循环逻辑实现，特别关注了特殊情况处理。

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 (<= 10^100^).

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
解题思路：
1.先用字符串读取，转化为int型的sum。
2.依次将sum的每一位放入ans数组，要注意用do……while，否则特例0会没有输出。或者先特判0，再用while循环，顺便计算位数cnt。
3.从cnt-1倒叙输出每位对应的单词，注意末尾空格。

#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;
const int maxn=110;
char change[10][10]={"zero","one","two","three","four","five","six","seven","eight","nine"};
int func1(char str[]){
    int len=strlen(str) ,sum=0;
    for(int i=0;i<len;i++){
        sum+=str[i]-'0';
    }
    return sum;
}

int main(){
    char str[maxn];
    scanf("%s",str);
    int sum=func1(str);
    int cnt=0,ans[maxn];
    do{                     //要写do……while，否则有一个点过不去 
        ans[cnt++]=sum%10;
        sum=sum/10;
    }while(sum);
    for(int i=cnt-1;i>=0;i--){
        printf("%s",change[ans[i]]);
        if(i>0) printf(" ");
    }
    return 0;
}