fjnu 1319 Perfection

Description

From the article Number Theory in the 1994 Microsoft Encarta: "If a, b, c are integers such that a = bc, a is called a multiple of b or of c, and b or c is called a divisor or factor of a. If c is not ?, b is called a proper divisor of a. Even integers, which include 0, are multiples of 2, for example, -4, 0, 2, 10; an odd integer is an integer that is not even, for example, -5, 1, 3, 9. A perfect number is a positive integer that is equal to the sum of all its positive, proper divisors; for example, 6, which equals 1 + 2 + 3, and 28, which equals 1 + 2 + 4 + 7 + 14, are perfect numbers. A positive number that is not perfect is imperfect and is deficient or abundant according to whether the sum of its positive, proper divisors is smaller or larger than the number itself. Thus, 9, with proper divisors 1, 3, is deficient; 12, with proper divisors 1, 2, 3, 4, 6, is abundant."

Problem Statement: Given a number, determine if it is perfect, abundant, or deficient.

Input

A list of N positive integers (none greater than 60,000),with 1 < N < 100. A 0 will mark the end of the list.

Output

The first line of output should read PERFECTION
OUTPUT. The next N lines of output should list for each input
integer whether it is perfect, deficient, or abundant, as shown in the
example below. Format counts: the echoed integers should be
right justified within the first 5 spaces of the output line, followed
by two blank spaces, followed by the description of the integer.
The final line of output should read END OF OUTPUT .

Sample Input

15 28 6 56 60000 22 496 0

 

Sample Output

PERFECTION OUTPUT   15  DEFICIENT   28  PERFECT    6  PERFECT   56  ABUNDANT60000  ABUNDANT   22  DEFICIENT  496  PERFECTEND OF OUTPUT

KEY:题清题目 ，先分解因子，然后求和，比较；

 

Source:#include<iostream>
using namespace std;

int a[100000];
int m;
int n;

void divisor()
...{
    m=0;
    for(int i=1;i<n;i++)
    ...{
        if(n%i==0) a[++m]=i;
    }
}

void fun()
...{
    int sum=0;
    for(int i=1;i<=m;i++)
        sum+=a[i];
    printf("%5d",n);
    if(sum==n)         
        cout<<"  PERFECT"<<endl;
    if(sum<n)
        cout<<"  DEFICIENT"<<endl;
    if(sum>n) 
        cout<<"  ABUNDANT"<<endl;
}

int main()
...{
//    freopen("fjnu_1319.in","r",stdin);
    cout<<"PERFECTION OUTPUT"<<endl;
    while(cin>>n&&n!=0)
    ...{
        divisor();
        fun();
    }
    cout<<"END OF OUTPUT"<<endl;
    return 0;
}