蓝桥基础练习 报时助手 BASIC-26

Posted 2021-03-05 wz-archer

问题描述

　　给定当前的时间，请用英文的读法将它读出来。
　　时间用时h和分m表示，在英文的读法中，读一个时间的方法是：
　　如果m为0，则将时读出来，然后加上“o‘clock”，如3:00读作“three o‘clock”。
　　如果m不为0，则将时读出来，然后将分读出来，如5:30读作“five thirty”。
　　时和分的读法使用的是英文数字的读法，其中0~20读作：
　　0:zero, 1: one, 2:two, 3:three, 4:four, 5:five, 6:six, 7:seven, 8:eight, 9:nine, 10:ten, 11:eleven, 12:twelve, 13:thirteen, 14:fourteen, 15:fifteen, 16:sixteen, 17:seventeen, 18:eighteen, 19:nineteen, 20:twenty。
　　30读作thirty，40读作forty，50读作fifty。
　　对于大于20小于60的数字，首先读整十的数，然后再加上个位数。如31首先读30再加1的读法，读作“thirty one”。
　　按上面的规则21:54读作“twenty one fifty four”，9:07读作“nine seven”，0:15读作“zero fifteen”。

输入格式

　　输入包含两个非负整数h和m，表示时间的时和分。非零的数字前没有前导0。h小于24，m小于60。

输出格式

　　输出时间时刻的英文。

样例输入

0 15

样例输出

zero fifteen

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <deque>
#include <cmath>
#include <map>

using namespace std;
typedef long long ll;

#define INF 0x7fffffff
const double inf=1e20;
const int maxn=1000+10;
const int mod=1e7;
const double pi=acos(-1);

char a[60][50]={"zero","one","two","three","four",
                "five","six","seven","eight","nine",
                "ten","eleven","twelve","thirteen","fourteen",
                "fifteen","sixteen","seventeen","eighteen","nineteen",
                "twenty","","","","",
                "","","","","",
                "thirty","","","","",
                "","","","","",
                "forty","","","","",
                "","","","","",
                "fifty","","","","",
                };

int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    if(n<=20)printf("%s",a[n]);
    else{
        int n0=n/10*10;
        int n1=n%10;
        printf("%s %s",a[n0],a[n1]);
    }
    printf(" ");
    if(m==0)printf("o‘clock");
    else if(m<=20)printf("%s",a[m]);
    else{
        int m0=m/10*10;
        int m1=m%10;
        printf("%s %s",a[m0],a[m1]);
    }
    printf("
");
    return 0;
}