ZeroJudge C203: DPA Numbers I

同題：UVa 13185 – DPA Numbers I

一個整數 B 如果可以被另一個整數 A 整除 (在這裡 A > B），我們稱 B 是 A 的一個因數。

A perfect number is a positive integer that is equal to the sum of its proper divisors (excluding itself).

For example, 6 and 28 are both perfect numbers because 6 = 1 + 2 + 3, and 28 = 1 + 2 + 4 + 7 + 14.

如果一個正整數不是 Perfect，那他就是 Deficient 或者是 Abundant，根據其所有因數的和是小於或大於這個數本身。因此，9 是 Deficient 因為 1 + 3 < 9。而 12 是 Abundant 因為 1 + 2 + 3 + 4 + 6 > 12。

Please write a program to determine whether a given number is Perfect, Deficient, or Abundant.

Sample Inputs/Outputs

Sample Input(s)Sample Output(s)
The first line of input contains a positive integer t, representing the number of test cases to follow. (1 <= T <= 500)
Each test case consists of a single line containing a positive integer N. (2 <= N <= 1000)		For each test case, output a single line indicating whether N is Perfect, Deficient, or Abundant.
10
5
6
16
18
21
28
29
30
40
43		deficient
perfect
deficient
abundant
deficient
perfect
deficient
abundant
abundant
deficient

Thought Process

Use a for loop to iterate over each number as a potential divisor. If the remainder of N divided by the current number (i) is 0, then i is a divisor of N. After summing up the divisors, determine whether it is less than, equal to, or greater than N.

Sample Code－ZeroJudge C203: DPA Numbers I

#include <iostream>
using namespace std;

int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    for (int i = 0; i<T; i++)
    {
        int N;
        cin >> N;
        int count = 0;
        for (int j = 1; j<N; j++)
        {
            if (N % j == 0)
            {
                count += j;
            }
        }
        if (count == N) cout << "perfect\n";
        else if (count > N) cout << "abundant\n";
        else cout << "deficient\n";
    }
}

//ZeroJudge C203
//Dr. SeanXD
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