codeforces 535A Tavas and Nafas

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#codeforces #水题

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本文介绍了一个简单的程序，用于将整数分数转换为英文文字形式，特别适用于0到99之间的数字。程序通过逐位分析输入的整数，并将其转换为对应的英文单词。

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Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.

His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.

$\text{[math]}$

He ate coffee mix without water again, so right now he's really messed up and can't think.

Your task is to help him by telling him what to type.

Input

The first and only line of input contains an integer s (0 ≤ s ≤ 99), Tavas's score.

Output

In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.

   Examples
 
     input 
    
      Copy
    
6

     output 
    
      Copy
    
six

     input 
    
      Copy
    
99

     output 
    
      Copy
    
ninety-nine

     input 
    
      Copy
    
20

     output 
    
      Copy
    
twenty

Note

You can find all you need to know about English numerals in http://en.wikipedia.org/wiki/English_numerals .

水题，特别水，就是比较麻烦！其实完全可以写出所有情况

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int main()
{
    // freopen("shuju.txt", "r", stdin);
    int n;
    cin >> n;
    if(n == 0)
        printf("zero\n");
    else if(n == 1)
        printf("one\n");
    else if(n == 2)
        printf("two\n");
    else if(n == 3)
        printf("three\n");
    else if(n == 4)
        printf("four\n");
    else if(n == 5)
        printf("five\n");
    else if(n == 6)
        printf("six\n");
    else if(n == 7)
        printf("seven\n");
    else if(n == 8)
        printf("eight\n");
    else if(n == 9)
        printf("nine\n");
    else if(n == 10)
        printf("ten\n");
    else if(n == 11)
        printf("eleven\n");
    else if(n == 12)
        printf("twelve\n");
    else if(n == 13)
        printf("thirteen\n");
    else if(n == 14)
        printf("fourteen\n");
    else if(n == 15)
        printf("fifteen\n");
    else if(n == 16)
        printf("sixteen\n");
    else if(n == 17)
        printf("seventeen\n");
    else if(n == 18)
        printf("eighteen\n");
    else if(n == 19)
        printf("nineteen\n");
    else if(n == 20)
        printf("twenty\n");
    else if(n == 30)
        printf("thirty\n");
    else if(n == 40)
        printf("forty\n");
    else if(n == 50)
        printf("fifty\n");
    else if(n == 60)
        printf("sixty\n");
    else if(n == 70)
        printf("seventy\n");
    else if(n == 80)
        printf("eighty\n");
    else if(n == 90)
        printf("ninety\n");
    else
    {
        int a = n / 10, b = n % 10;
        if(a == 2)
            printf("twenty-");
        else if(a == 3)
            printf("thirty-");
        else if(a == 4)
            printf("forty-");
        else if(a == 5)
            printf("fifty-");
        else if(a == 6)
            printf("sixty-");
        else if(a == 7)
            printf("seventy-");
        else if(a == 8)
            printf("eighty-");
        else if(a == 9)
            printf("ninety-");
        if(b == 1)
            printf("one\n");
        else if(b == 2)
            printf("two\n");
        else if(b == 3)
            printf("three\n");
        else if(b == 4)
            printf("four\n");
        else if(b == 5)
            printf("five\n");
        else if(b == 6)
            printf("six\n");
        else if(b == 7)
            printf("seven\n");
        else if(b == 8)
            printf("eight\n");
        else if(b == 9)
            printf("nine\n");
    }
    return 0;
}