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Турниры > CEN112 Homeworks 2013-2015 > задача:


50388 - Number of Armstrong Numbers - 1

CEN112 Homeworks 2013-2015

Старт: 15.дек.2013 в 12:00:00
Финиш: 15.дек.2013 в 17:00:00
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Задачи турнира

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• 50375 - Area of Circles
• 50388 - Number of Armstrong N...
• 50393 - Palindromic Numbers
• 50339 - The Largest Rectangle
• 50345 - Orders
• 50758 - National Elections - Revisited
• 50754 - Team Standings
• 12-Spr1-40. 50327 - Parallel Lines
• 13-Fall2-20. 50395 - Page "Like" Av...
• 14-Spr1-30. 50492 - Contest Score...

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Лимит времени 2000/4000/4000/4000 мс. Лимит памяти 65000/65000/65000/65000 Кб.
Prepared by Ibrahim Mesecan.

Number of Armstrong Numbers

Number 153 has 3 digits and if you take the 3rd power of every digit and calculate the sum of them, you will have the same number again.
153 = 13 + 53 + 33
     = 1 + 125 + 27 = 153
Thus, for 1634 you should take the fourth powers of every digit (because the number has 4 digits) and calculate the sum. Because the sum is equal to the number itself, then, we say that this number is an Armstrong Number.

Question: Assume that you are given the following functions:

  • int numDigits (int num); // a function which calculates the number of digits in the given number (num)
  • int power(int n, int p); // calculates np
  • int getDigit(int num, int i);// gets ith digit of the given number num. And digits start from 1.

Using the functions above, in your program, calculate the number of Armstrong numbers between the given numbers n and m.

Input specification
There will be two numbers between 1 ≤ n < m ≤ 100000.
Note: For C and C++, you can include the functions using the following statement.
  #include "utils/armstrong.h" Download armstrong.h.

Output specification
show one integer: the number of Armstrong Numbers. If no Armstrong numbers exists, then show 0 (zero).

Input I   Input II  
 100 200  400 2000
Output 1   Output 2  
  1
  2

Для отправки решений необходимо выполнить вход.

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