Project Euler #17: Number to Words | Hackerrank | Project Euler

 Number to Words

The numbers 1 to 5 written out in words are One, Two, Three, Four, Five 

First character of each word will be capital letter for example:

 104382426112 is 

One Hundred Four Billion Three Hundred Eighty Two Million Four Hundred Twenty Six Thousand One Hundred and Twelve

Given a number, you have to write it in words.

Input Format

The first line contains an integer T, i.e., number of test cases.

Next T lines will contain an integer N.

Constraints

1<=T<=10

0<=N<=10^2

Output Format

Print the values corresponding to each test case.

Sample Input

2

10

17

Sample Output

Ten 

Seventeen

Solution : Source (Hackerrank)

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

char digits[20][10] = {"","One","Two","Three","Four",

"Five","Six","Seven","Eight","Nine","Ten","Eleven",

"Twelve","Thirteen","Fourteen",

"Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};

char tens[11][10] =  {"","","Twenty","Thirty","Forty",

"Fifty","Sixty","Seventy","Eighty","Ninety"};

void word(int num){

    if(num/100)

        printf("%s Hundred ", digits[num/100]);

    if((num%100) < 20 && (num%100) > 0)

        printf("%s ",digits[num%100]);

    else if((num/10)%10){

        printf("%s ", tens[(num/10)%10]);

        if(num%10)

            printf("%s ", digits[num%10]);

    }

}

int main() {

    int t;

    scanf("%d", &t);

    while(t--){

        long long int num;

        scanf("%lld", &num);

        int tn = num / 1000000000000;

        int bn = (num / 1000000000) % 1000;

        int mn = (num / 1000000) % 1000;

        int th = (num / 1000) % 1000;

        int hd = num % 1000;

        //printf("%d %d %d %d %d\n", tn, bn, mn, th, hd);

        if((tn + bn + mn + th + hd) == 0)

            printf("Zero");

        if(tn){

            word(tn);

            printf("Trillion ");

        }

        if(bn){

            word(bn);

            printf("Billion ");

        }

        if(mn){

            word(mn);

            printf("Million ");

        }

        if(th){

            word(th);

            printf("Thousand ");

        }

        if(hd){

            word(hd);

        }

       printf("\n"); 

    }

    return 0;

}

You can also run it on an online IDE :

https://ide.geeksforgeeks.org/c658ddc7-5ffc-4db5-932a-ad8c3610f6df

Your feedback are always welcome! If you have any doubt you 

can contact me or leave a comment!  

Happy Coding!! Cheers!!!

Project Euler #2: Even Fibonacci numbers | Hackerrank | Project Euler

 Even Fibonacci Numbers

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1,2,3,5,8,13,21,3,55,89,...

By considering the terms in the Fibonacci sequence whose values do not exceed N, find the sum of the even-valued terms.

Input Format

First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer,N.

Constraints

1<=T<=10^5

10<=N<=4*10^16

Output Format

Print the required answer for each test case.

Sample Input 0

2

10

100

Sample Output 0

10

44

Explanation 0

For N=10, we have {2,8}, sum is 10.

For N=100, we have {2,8,34}, sum is 44.

Solution:

#include <math.h>

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <assert.h>

#include <limits.h>

#include <stdbool.h>

int main(){

    int t; 

    scanf("%d",&t);

    for(int a0 = 0; a0 < t; a0++){

        long n,sum=2,p1 = 1, p2 = 2, fib = 0; 

        scanf("%ld",&n);

         fib= p1 + p2;

        while (fib <= n) {

         if(fib%2==0)

            sum=sum+fib;

         p1 = p2;

         p2 = fib;

         fib = p1 + p2;

    }

        printf("%ld\n",sum);

    }

    return 0;

}

You can also run it on an online IDE: 

https://ide.geeksforgeeks.org/1fe1e528-4698-43bf-8231-014d797877a5 

Your feedback are always welcome! If you have any doubt you can contact me or leave a comment!  Happy Coding !! Cheers!!!

Related Links:

https://codepiggy.blogspot.com/2022/07/project-euler-1-multiples-of-3-and-5.html