Print numbers in words

QUESTION

Given a number as input, if the number is between 1 and 10 both inclusive then print the number in words otherwise print \”\”not in range\”\”.\n\nInput: \nFirst line of input contains an integer T which denotes the number of test cases. Then T test cases follows. First line of each test case contains an integer N.\n \n\nOutput: \nFor each test case, if the number is between 1 and 10 both inclusive then print the number in words in small cases otherwise print \”\”not in range\”\” \n\n\nConstraints:\n\n1<=T<=1000\n1<=N<=100.

“TESTCASE_1”: “5\n5\n6\n12\n2\n10\n###—###SEPERATOR—###—\nfive\nsix\nnot in range\ntwo\nten”, “TESTCASE_2”: “4\n9\n7 \n11\n8\n###—###SEPERATOR—###—\nnine\nseven\nnot in range\neight”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

ANSWER

#include <iostream>
using namespace std;
int main()
{
  int n,i,a[100];
 scanf("%d",&n);
  for(i=1;i<=n;i++)
    cin>>a[i];
  for(i=1;i<=n;i++)
  {
  if(a[i]>=1 && a[i]<=10)
  {
    if(a[i]==1)
      cout<<"one\n";
    else if(a[i]==2)
      cout<<"two\n";
    else if(a[i]==3)
      cout<<"three\n";
    else if(a[i]==4)
      cout<<"four\n";
    else if(a[i]==5)
      cout<<"five\n";
    else if(a[i]==6)
      cout<<"six\n";
    else if(a[i]==7)
      cout<<"seven\n";
    else if(a[i]==8)
      cout<<"eight\n";
    else if(a[i]==9)
      cout<<"nine\n";
    else if(a[i]==10)
      cout<<"ten\n";
  }
  else
    cout<<"not in range\n";
     
  
  }
	return 0;
}
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