報時助手
問題描述
給定當前的時間,請用英文的讀法將它讀出來。
時間用時h和分m表示,在英文的讀法中,讀一個時間的方法是:
如果m爲0,則將時讀出來,然後加上“o’clock”,如3:00讀作“three o’clock”。
如果m不爲0,則將時讀出來,然後將分讀出來,如5:30讀作“five thirty”。
時和分的讀法使用的是英文數字的讀法,其中0~20讀作:
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讀作thirty,40讀作forty,50讀作fifty。
對於大於20小於60的數字,首先讀整十的數,然後再加上個位數。如31首先讀30再加1的讀法,讀作“thirty one”。
按上面的規則21:54讀作“twenty one fifty four”,9:07讀作“nine seven”,0:15讀作“zero fifteen”。
輸入格式
輸入包含兩個非負整數h和m,表示時間的時和分。非零的數字前沒有前導0。h小於24,m小於60。
輸出格式
輸出時間時刻的英文。
樣例輸入
0 15
樣例輸出
zero fifteen
本題的一大難點是0-60的英文單詞選取哪種數據結構來存儲,以及如何將數量不小的單詞存入數據結構之中。本題我採用的方法比較暴力,將0-19,20,30,40,50存入數組。然而數組的初始化不能過多,我被這個問題困擾了很久,而後我想到將數組升爲二維數組,這樣初始化項過多的問題就解決了
#include<iostream>
using namespace std;
int main()
{
int h,m,a,b;//a代表兩位數十位數字,b代表兩位數個位數字
cin >> h >> m;
char hour[24][24] = {"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
char minute[24][24] = {"o'clock","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
char ten[10][10] = {"","","twenty","thirty","forty","fifty"};
if(h < 20)
cout << hour[h] << " ";
else{
a = h / 10;
b = h % 10;
cout << ten[a] << " ";
if(b != 0)
cout << hour[b] << " ";
}
if(m < 20)
cout << minute[m];
else{
a = m / 10;
b = m % 10;
cout << ten[a] << " " ;
if(b != 0)
cout<< minute[b] << " ";
}
return 0;
}
本題中,我的測試用例仍然較爲特殊。應當按照正常數據,錯誤數據,邊界數據來全面的測試代碼,這樣才能及早的發現代碼的錯誤與缺陷。