Perfection
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1748 Accepted Submission(s): 1051
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
题目大意：假设一个数的约数和大于它，就是ABUNDANT，假设等于它，就是
PERFECT。若果小于它本身，就是DEFICIENT。
思路：按题目要求和规定推断、输出。
#include<stdio.h>
int a[110],b[110];
int main()
{
int i = 0,n;
while(~scanf("%d",&n) && n)
{
a[i] = n;
int sum = 0;
for(int j = 1; j <= n/2; j++)
if(n % j == 0)
sum += j;
if(sum==n)
b[i] = 1;
else if(sum > n)
b[i] = 2;
else if(sum < n)
b[i] = 0;
i++;
}
printf("PERFECTION OUTPUT
");
for(int j = 0; j < i; j++)
{
printf("%5d ",a[j]);
if(b[j]==2)
printf("ABUNDANT
");
else if(b[j]==1)
printf("PERFECT
");
else
printf("DEFICIENT
");
}
printf("END OF OUTPUT
");
return 0;
}