Given a number “n”, find if it is Disarium or not. A number is called Disarium if sum of its digits powered with their respective positions is equal to the number itself.
Input : n = 135 Output : Yes 1^1 + 3^2 + 5^3 = 135 Therefore, 135 is a Disarium number Input : n = 89 Output : Yes 8^1+9^2 = 89 Therefore, 89 is a Disarium number Input : n = 80 Output : No 8^1 + 0^2 = 8
The idea is to fist count digits in given numbers. Once we have count, we traverse all digits from right most (using % operator), raise its power to digit count and decrement the digit count.
Below is the implementation of above idea.
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Improved By : rutvik_56