PAT. basic level. 1002

来源：互联网发布：shell编程如何保存编辑：程序博客网时间：2024/05/28 05:14

#include<stdio.h>#include<string.h>void printf_num(int num);main(){    char str[101];    int z,i,sum;    i = 0;    sum = 0;    scanf("%s",str);    i=strlen(str);     for(z=0;z<=i-1;z++){        sum=sum+str[z]-'0';    }     printf_num(sum);                return 0; }void printf_num(int num){    int z,a[101],sum,i;    sum = 0;    i = 0;    z = 1;    while (num>9){        i=num%10;        num/=10;        a[z]=i;        z++;    }       a[z]=num;    for (i=z;i>=2;i--){          switch(a[i]){            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;         }    }    if(i==1){        switch(a[i]){            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;         }    }     }

重点在于用字符串做，长整和长长整是做不了的。

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