HDU 1323 Perfection

http://acm.hdu.edu.cn/showproblem.php?pid=1323

 

Perfection

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
Total Submission(s): 830 Accepted Submission(s):504

Problem 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 multipleof b or of c, and b or c is called a divisor or factor of a. If cis not 1/-1, b is called a proper divisor of a. Even integers,which include 0, are multiples of 2, for example, -4, 0, 2, 10; anodd 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 sumof all its positive, proper divisors; for example, 6, which equals1 + 2 + 3, and 28, which equals 1 + 2 + 4 + 7 + 14, are perfectnumbers. A positive number that is not perfect is imperfect and isdeficient 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 properdivisors 1, 2, 3, 4, 6, is abundant."
Given a number, determine if it is perfect, abundant, ordeficient.

 

Input

A list of N positive integers (none greater than 60,000), with1 < N < 100. A 0 will mark the end ofthe list.

 

Output

The first line of output should read PERFECTION OUTPUT. Thenext N lines of output should list for each input integer whetherit is perfect, deficient, or abundant, as shown in the examplebelow. Format counts: the echoed integers should be right justifiedwithin the first 5 spaces of the output line, followed by two blankspaces, followed by the description of the integer. The final lineof output should read END OF OUTPUT.

 

Sample Input

15 28 6 5660000 22 496 0

 

Sample Output

PERFECTIONOUTPUT 15 DEFICIENT 28 PERFECT 6 PERFECT 56 ABUNDANT 60000 ABUNDANT22 DEFICIENT 496 PERFECT END OF OUTPUT

 

Source

Mid-Atlantic USA 1996

 

Recommend

Ignatius.L

 

题目大意：求一个数的真因子之和，大于这个数叫ABUNDANT，小于叫DEFICIENT，等于叫PERFECT。

代码如下：

#include<iostream>
#include<math.h> 
using namespacestd;   
intfun(int);     //求一个数的真因子之和
int main()
{ 
 int n,ans;
    cout<<"PERFECTIONOUTPUT"<<endl;
 while(cin>>n)
 {
  if(n==0)
  {
   cout<<"ENDOF OUTPUT"<<endl;
   break;
  }
  ans=fun(n);
  printf("] ",n);
  if(ans==n)cout<<"PERFECT"<<endl;
  else if(ans<n)cout<<"DEFICIENT"<<endl;
  elsecout<<"ABUNDANT"<<endl;
 }
   return 0;
}
int fun(int n)
{
   int i,sum=0;
  for(i=2;i<=n/2;i++)
   if(n%i==0)
    sum+=i;
    returnsum+1;
}