CCC加拿大高中生信息学奥赛
其余来源
CODEVS[3312]——CCC 1996 01 Deficient, Perfect, and Abundant ——http://codevs.cn/problem/3312/
POJ[1928]——Perfection——http://poj.org/problem?id=1528
CODEVS描述——中文题目
题目描述 Description
读入一个正整数n，判断整数是完数，亏数还是盈数。
•如果它的约数的和等于它本身，那它便是一个完数(Perfect)（约数包含1，但不包含它本身）。
•如果它的约数的和小于它本身，那它便是一个亏数(Deficient)（约数包含1，但不包含它本身）。
•如果它的约数的和大于它本身，那它便是一个盈数(Abundant)（约数包含1，但不包含它本身）。
输入描述 Input Description
输入文件共两行，
第一行为一个正整数n，
第二行为n个正整数，中间用空格隔开。
输出描述 Output Description
输出为n行，分别为第1..n个数的类别。
完数：Perfect
亏数：Deficient
盈数：Abundant
样例输入 Sample Input
3
4 6 12
样例输出 Sample Output
4 is a deficient number.
6 is a perfect number.
12 is an abundant number.
数据范围及提示 Data Size & Hint
1<=n<=2^31-1
POJ描述（英文）
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
思路
利用数论知识快速计算一个数的约数和，详见程序函数。
样例
CODEVS：
var t,x:longint;
function ans(n:longint):longint;
var i:longint;
begin
ans:=0;
for i:=1 to n do
begin
if (i*i=n) then
begin
ans:=ans+i;
break;
end;
if (i*i>n) then break;
if (n mod i=0) then ans:=ans+i+n div i;
end;
end;
procedure main;
var anss,n:longint;
begin
read(n);
anss:=ans(n)-n;
if anss<n then writeln(n,‘ is a deficient number.‘);
if anss=n then writeln(n,‘ is a perfect number.‘);
if anss>n then writeln(n,‘ is an abundant number.‘);
end;
begin
readln(t);
for x:=1 to t do main;
end.
POJ：
var t,x,n:longint;
function ans(n:longint):longint;
var i:longint;
begin
ans:=0;
for i:=1 to n do
begin
if (i*i=n) then
begin
ans:=ans+i;
break;
end;
if (i*i>n) then break;
if (n mod i=0) then ans:=ans+i+n div i;
end;
end;
procedure main;
var anss,x,i:longint;s:ansistring;
begin
anss:=ans(n)-n;
str(n,s);
x:=length(s);
for i:=1 to 5-x do write(‘ ‘);
if anss<n then writeln(n,‘ DEFICIENT‘);
if anss=n then writeln(n,‘ PERFECT‘);
if anss>n then writeln(n,‘ ABUNDANT‘);
end;
begin
writeln(‘PERFECTION OUTPUT‘);
while true do
begin
read(n);
if n=0 then
begin
writeln(‘END OF OUTPUT‘);
halt;
end;
main;
end;
end.