535A - Tavas and Nafas - CodeForces Solution


brute force implementation *1000

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Python Code:

num2words1 = {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'}

num2words2 = ['twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety']


n = int(input())
if(n<=19):
    print(num2words1[n])
else:
    k=int(str(n)[0])
    l=int(str(n)[1])
    if(n%10==0):
        print(num2words2[k-2])
    else:
        print(str(num2words2[k-2])+"-"+str(num2words1[l]))

C++ Code:

#include<bits/stdc++.h>
#include<iostream>
using namespace std;

int main()
{  
    int n;
    cin>>n;
    string o[10]={"zero", "one", "two", "three", "four",
                    "five", "six", "seven", "eight", "nine"};
    string d[10]={"ten", "eleven", "twelve", "thirteen", "fourteen",
                    "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
    string t[10]={"", "", "twenty", "thirty", "forty",
                    "fifty", "sixty", "seventy", "eighty", "ninety"};
    if(n<10){
        cout<<o[n]<<endl;
    }
    else if(n>=10 && n<=19){
        cout<<d[n-10]<<endl;
    }
    else if(n%10==0){
        cout<<t[n/10]<<endl;
    }
    else{
        cout<<t[n/10]<<"-"<<o[n%10]<<endl;
    }
    return 0;
}


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