蓝桥杯基础练习VIP-报时助手-题解（C语言代码）普通版解法//每一步都有注释哦

解题思路:

先建三个数组，分别是贮存 一.1-9数字;二.0-24数字；三.20-60整数；

再获得输入，可以判断输入是否满足要求，不满足可以结束它；

后判断

结束

注意事项:

注意点我在代码中注释；

参考代码:

#define _CRT_SECURE_NO_WARNINGS

#include<stdio.h>

int main()

{

char no[9][10] = { "one","two","three","four","five","six","seven","eight","nine" }; //1--9

char str[25][15] = { "zero","one","two","three","four","five","six","seven","eight","nine" ,"ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty","twenty one","twenty two","twenty three","twenty four" };  //0--24

char tnt[5][10] = { "twenty","thirty","forty","fifty","sixty" };//20--60

int n1, n2;//n1是：小时；n2是：分钟；

int i, j;//i是用于确定分钟的 十位；j是用于确定分钟的 个位；

scanf("%d%d", &n1, &n2);//获取输入

if (n1 > 24 || n1 < 0 || n2>60 || n2 < 0)//不满足条件的直接结束；

{

return -1;

}

else {

printf("%s", str[n1]);//小时的直接输出；

if (n2 == 0)

{

printf(" o'clock");//分钟为0的情况；

}

else {

if (n2 <= 24)//分钟小于24直接用第二个数组输出；

{

printf(" %s", str[n2]);

}

else {

i = n2 / 10;//i是用于确定分钟的 十位；j是用于确定分钟的 个位；

j = n2 % 10;

printf(" %s", tnt[i - 2]);

printf(" %s", no[j - 1]);

}

}

}

printf("\n");

return 0;

}