HDU 1323 Perfection（公因子）

Perfection

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2336    Accepted Submission(s): 1420

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 multiple of b or of c, and b or c is called a divisor or factor of a. If c is 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; 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."

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 ABUNDANT

60000 ABUNDANT

22 DEFICIENT

496 PERFECT

END OF OUTPUT

Source

​​Mid-Atlantic USA 1996 ​​

題解：

給你一個數，如果這個數的所有因子加起來等于這個數，就輸出PERFECT...

如果小于這個數，輸出DEFICIENT....

如果大于這個數，輸出ABUNDANT...

AC代碼：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
typedef long long LL;
using namespace std;
int main()
{
  int a[110],num=1,i,j;
  while(~scanf("%d",&a[num]),a[num])
  {
    num++;
  }
  printf("PERFECTION OUTPUT\n");
  for(i=1;i<num;i++)  //i不用等于num，因為最後輸入的數為0 
  {
    int sum=0;
    for(j=1;j<=a[i]/2;j++)
    {
      if(a[i]%j==0)     //求出比該數小的所有公約數，比如8有1,2,4,此時為DEFICIENT
      {
        sum+=j;
      }
    } 
    if(sum==a[i])
    printf("%5d  PERFECT\n",a[i]);
    else if(sum>a[i])
    printf("%5d  ABUNDANT\n",a[i]);
    else 
    printf("%5d  DEFICIENT\n",a[i]);
  }
  printf("END OF OUTPUT\n");
  return 0;

}