ACM-数论之Perfection——hdu1323

Perfection

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

这题也挺简单，题意为 输入一个数求它的因子和，并判断因子和与该数的大小关系，如果因子和大于该数输出
DEFICIENT，如果等于输出PERFECT，如果小于输出ABUNDANT,题意很简单，注意格式，输出前端有：

PERFECTION OUTPUT，末尾输出 END OF OUTPUT。还有数字宽度为5，数字与后面之间有两个空格，废话不多说，代码：

#include <iostream>
#include <iomanip>

#include <string.h>
using namespace std;
int main()
{
int n,sum;
int i;
char dey[8]="PERFECT",day[9]="ABUNDANT",xy[10]="DEFICIENT";
printf("PERFECTION OUTPUT\n");
while(scanf("%d",&n) && n)
{

sum=0;
for(i=1;i<n;++i)
{
if(sum>n)
{
cout<<setw(5)<<n<<"  "<<day<<endl;
break;
}
if(n%i==0)
sum+=i;
}
if(sum==n)
cout<<setw(5)<<n<<"  "<<dey<<endl;
else if(sum<n)
cout<<setw(5)<<n<<"  "<<xy<<endl;
}
printf("END OF OUTPUT\n");
return 0;
}

代码中，每一次sum都会判断一次 sum是否大于n，所以时间上可能慢一点。