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2013
12-09

HDU 1323 Perfection[解题报告] C++

Perfection

问题描述 :

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.

输入:

A list of N positive integers (none greater than 60,000), with 1 < N < 100. A 0 will mark the end of the list.

输出:

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.

样例输入:

15 28 6 56 60000 22 496 0

样例输出:

PERFECTION OUTPUT
   15  DEFICIENT
   28  PERFECT
    6  PERFECT
   56  ABUNDANT
60000  ABUNDANT
   22  DEFICIENT
  496  PERFECT
END OF OUTPUT

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1323

题意:某数的真因子加起来比它大叫做ABUNDANT,比它小叫做DEFICIENT,相等叫做PERFECT。输入一个数,输出它是哪种情况。直接暴。

代码:

# include <stdio.h>


int main ()
{
    int n, ans, i ;
    puts ("PERFECTION OUTPUT") ;
    while (~scanf ("%d", &n),n)
    {
        for(i = 1,ans=0 ; i < n && ans <= n; i++)
            if (n%i==0)ans+=i;
        printf("%5d  ",n) ;
        if(ans>n)puts("ABUNDANT");
        else if(ans==n)puts("PERFECT");
        else puts("DEFICIENT") ;
    }
    puts("END OF OUTPUT") ;
    return 0 ;
}

解题报告转自:http://www.cnblogs.com/lzsz1212/archive/2012/02/16/2353689.html


  1. “再把所有不和该节点相邻的节点着相同的颜色”,程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的