ACM-數論之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

這題也挺簡單，題意為 輸入一個數求它的因子和，并判斷因子和與該數的大小關系，如果因子和大于該數輸出
DEFICIENT，如果等于輸出PERFECT，如果小于輸出ABUNDANT,題意很簡單，注意格式，輸出前端有：       
PERFECTION OUTPUT，末尾輸出 END OF OUTPUT。還有數字寬度為5，數字與後面之間有兩個空格，廢話不多說，代碼：
       
#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;
}
           
        

代碼中，每一次sum都會判斷一次 sum是否大于n，是以時間上可能慢一點。