题目：读入一个自然数n，计算其各位数字之和，用汉语拼音写出和的每一位数字。
输入格式：每个测试输入包含1个测试用例，即给出自然数n的值。这里保证n小于10^100^。
输出格式：在一行内输出n的各位数字之和的每一位，拼音数字间有1 空格，但一行中最后一个拼音数字后没有空格。
输入样例：
1234567890987654321123456789
输出样例：
yi san wu


我的代码：
#include <cstdio>
int main()
{
    int n,sum=0;
    int b=10;
    scanf("%d",&n);
    while (n/10.0!=0)
    {
        sum=n%10+sum;
        n=n/10;
    }
    while(sum/b!=0)
    {
        b=b*10;
    }
    b=b/10;
    while (b!=1)
    {
        switch (sum/b)
        {
            case 1: printf("yi ");break;
            case 2: printf("er ");break;
            case 3: printf("san ");break;
            case 4: printf("si ");break;
            case 5: printf("wu ");break;
            case 6: printf("liu ");break;
            case 7: printf("qi ");break;
            case 8: printf("ba ");break;
            case 9: printf("jiu ");break;
            case 0: printf("ling ");break;
        }
        sum=sum%b;
        b=b/10;
    }
    switch (sum)
        {
            case 1: printf("yi");break;
            case 2: printf("er");break;
            case 3: printf("san");break;
            case 4: printf("si");break;
            case 5: printf("wu");break;
            case 6: printf("liu");break;
            case 7: printf("qi");break;
            case 8: printf("ba");break;
            case 9: printf("jiu");break;
            case 0: printf("ling");break;
        }
    return 0;
}

数值过大
C语言中int的取值范围为：-2147483648 ~ 2147483647

int 的取值范围是-2147483648~2147483647

超长整数的话，建议用数组来实现

#include <stdio.h>
#include<string.h>

int main()
{
    char n[3000];
    int sum=0;
    int b=10,i;
    scanf("%s",n);
    for(i=0;i<strlen(n);i++)
        sum=(n[i]-48)+sum;  //n[i]-48,输入的数字对应的是ASCII码，需要-48
    /*while (n/10.0!=0)
    {
        sum=n%10+sum;
        n=n/10;
    }*/
    while(sum/b!=0)
    {
        b=b*10;
    }
    b=b/10;
    while (b!=1)
    {
        switch (sum/b)
        {
            case 1: printf("yi ");break;
            case 2: printf("er ");break;
            case 3: printf("san ");break;
            case 4: printf("si ");break;
            case 5: printf("wu ");break;
            case 6: printf("liu ");break;
            case 7: printf("qi ");break;
            case 8: printf("ba ");break;
            case 9: printf("jiu ");break;
            case 0: printf("ling ");break;
        }
        sum=sum%b;
        b=b/10;
    }
    switch (sum)
        {
            case 1: printf("yi");break;
            case 2: printf("er");break;
            case 3: printf("san");break;
            case 4: printf("si");break;
            case 5: printf("wu");break;
            case 6: printf("liu");break;
            case 7: printf("qi");break;
            case 8: printf("ba");break;
            case 9: printf("jiu");break;
            case 0: printf("ling");break;
        }
    return 0;
}

666，谢了