Acwing perfect number

Title description:

A whole number is called a perfect number if the sum of all its divisors except itself is equal to that number.

For example, 66 is a perfect number because the sum of its divisors other than itself is 1+2+3=61+2+3=6.

Now, given your NN integers, please determine whether these numbers are perfect in turn.

Input format:

The first line contains the integer NN, indicating a total of NN test cases.

The next NN lines, each containing an integer XX that requires your judgment.

Output format:

Output one result per test case, one line per result.

If the test data is perfect, output X is perfect, where XX is the test data.

If the test data is not perfect, output X is not perfect, where XX is the test data.

Data range:

1≤N≤1001≤N≤100,
1≤X≤1081≤X≤108

Sample input:

36528

Sample output:

6 is perfect5 is not perfect28 is perfect

Analysis:

This question itself is not difficult, and it can be done in double loops, but when a number is very large, the time will exceed the time limit. At this time, we must think about the optimization method: this is the test of mathematics, if a isA divisor of b, then b/a is also a divisor of b.For example: 2 is a divisor of 12, and 6 is also a divisor of 12.We only need to enumerate from 1 to b/a, that is, a*a

#include using namespace std;int panduan(int n){int sum=0;for(int i=1;i>n;while(n--){int a;cin>>a;if(panduan(a)==1){cout<

This is the result: (obviously)

Let's take a look at the optimized:

#include using namespace std;int panduan(int n){int sum=0;for(int i=1;i*i<=n;i++){if(n%i==0){if(i>n;while(n--){int a;cin>>a;if(panduan(a)==1){cout<

Pass!