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 OUTPUT15  DEFICIENT28  PERFECT6  PERFECT56  ABUNDANT
60000  ABUNDANT22  DEFICIENT496  PERFECT
END OF OUTPUT

Source

Mid-Atlantic 1996

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;
beginans:=0;for i:=1 to n dobeginif (i*i=n) thenbeginans:=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;
beginread(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;beginreadln(t);for x:=1 to t do main;
end.

View Code

　　POJ：

var t,x,n:longint;function ans(n:longint):longint;
var i:longint;
beginans:=0;for i:=1 to n dobeginif (i*i=n) thenbeginans:=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;
beginanss:=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;beginwriteln('PERFECTION OUTPUT');while true dobeginread(n);if n=0 thenbeginwriteln('END OF OUTPUT');halt;end;main;end;
end.