1005. Spell It Right (20) -PAT

1005. Spell It Right (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

我的代码

#include<stdio.h>

int main() {
    int sum = 0;
    char input[1000];
    int i = 0;
    while ((input[i] = getchar()) != '
') {
        i++;
    }
    int k = i;
    for (i = 0; i < k; i++) {
        sum = sum+input[i]-'0';
    }
    printf("%d ", sum);
    int a = sum / 100;
    int b = (sum - 100*a) / 10;
    int c = sum - a*100 - b*10;
    printf("%d %d %d 
", a, b, c);
    switch (a) {
    case 9:printf("nine "); break;
    case 8:printf("eight "); break;
    case 7:printf("seven "); break;
    case 6:printf("six "); break;
    case 5:printf("five "); break;
    case 4:printf("four "); break;
    case 3:printf("three "); break;
    case 2:printf("two "); break;
    case 1:printf("one "); break;
    case 0:break;
    }
    switch (b) {
    case 9:printf("nine "); break;
    case 8:printf("eight "); break;
    case 7:printf("seven "); break;
    case 6:printf("six "); break;
    case 5:printf("five "); break;
    case 4:printf("four "); break;
    case 3:printf("three "); break;
    case 2:printf("two "); break;
    case 1:printf("one "); break;
    case 0:break;
    }
    switch (c) {
    case 9:printf("nine"); break;
    case 8:printf("eight"); break;
    case 7:printf("seven"); break;
    case 6:printf("six"); break;
    case 5:printf("five"); break;
    case 4:printf("four"); break;
    case 3:printf("three"); break;
    case 2:printf("two"); break;
    case 1:printf("one"); break;
    case 0:printf("zero"); break;
    }
    return 0;
}

 

   感悟：1、读取特大数字的时候 可考虑用字符数组来储存该数 ，第5行。

   2、((input[i] = getchar()) != ' ');     红色括号一定要加！ 类似的还有 scanf("%d", &n) != EOF

   3、如何求一个较小的数的各个位上的数字 假设sum 是不超过999的一个数.

  百位  int a = sum / 100;  
  十位  int b = (sum - 100*a) / 10;
  个位  int c = sum - a*100 - b*10;
 4、switch case 语句的运用