Aliquot Sum
시간 제한8초메모리 제한1024 MB
최대 100만 개의 수(각 100만 이하)에 대해 진약수의 합과 자기 자신을 비교해 abundant, deficient, perfect로 분류한다.
문제
A divisor of a positive integer n is an integer d where m = n d is an integer. In this problem, we define the aliquot sum s(n) of a positive integer n as the sum of all divisors of n other than n itself. For examples, s(12) = 1 + 2 + 3 + 4 + 6 = 16, s(21) = 1 + 3 + 7 = 11, and s(28) = 1 + 2 + 4 + 7 + 14 = 28.
With the aliquot sum, we can classify positive integers into three types: abundant numbers, deficient numbers, and perfect numbers. The rules are as follows.
- A positive integer x is an abundant number if s(x) > x.
- A positive intewer y is a deficient number if s(y) < y.
- A positive integer z is a perfect number if s(z) = z.
You are given a list of positive integers. Please write a program to classify them.
입력
The first line of the input contains one positive integer T indicating the number of test cases. The second line of the input contains T space-separated positive integers n1, ..., nT.
출력
Output T lines. If ni is an abundant number, then print abundant on the i-th line. If ni is a deficient number, then print deficient on the i-th line. If ni is a perfect number, then print perfect on the i-th line.
제한
- 1 ≤ T ≤ 106
- 1 ≤ ni ≤ 106 for i ∈ {1, 2, ..., T}.