高中生程式解題系統：c032: 00382 - Perfection

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題目連結 http://zerojudge.tw/ShowProblem?problemid=c032。

此題用迴圈從判斷1到 N/2 是否為 N 的因數，若是則加總(s)。接著比較加總結果與 N的大小關係來輸出對應的訊息。(那有沒有更好的演算法呢？)

C++程式碼：

#include <iostream>
#include <iomanip>

using namespace std;

/*

15 28 6 56 60000 22 496 0

   15  DEFICIENT
   28  PERFECT
    6  PERFECT
   56  ABUNDANT
60000  ABUNDANT
   22  DEFICIENT
  496  PERFECT

*/

short checkPerfection(int n)
{
    int sum = 0;
    for( int i = 1; i <= n/2; i++ )
        if( n % i == 0 )
            sum += i;

    if( n == sum )
        return 0;       // perfect
    else if( n > sum )
        return -1;      // deficient
    else if( n < sum )
        return 1;       // abundant

}

void prnMsg(short choice)
{
    switch( choice )
    {
        case 0:
            cout << "PERFECT";
            break;
        case 1:
            cout << "ABUNDANT";
            break;
        case -1:
            cout << "DEFICIENT";
            break;
        default:
            break;
    }

    cout << endl;
}

int main()
{
    int input[100];
    int size = 0;

    while( cin >> input[size] )
    {
        if( input[size] == 0 )
            break;

        size++;
    }

    cout << "PERFECTION OUTPUT" << endl;

    for( int i = 0; i < size; i++ )
    {
        cout.width(5);
        cout << input[i] << "  ";
        prnMsg(checkPerfection(input[i]));
    }

    cout << "END OF OUTPUT";

    return 0;
}

Python程式碼：

import sys
    
msg = {0:"PERFECT", 1:"ABUNDANT", -1:"DEFICIENT"}
  
def checkPerfection(n):
    sum = 0
    for i in range(1, (n // 2) + 1, 1):
        if( n % i == 0 ):
            sum += i

    if( n == sum ):
        return 0
    elif( n > sum ):
        return -1
    elif( n < sum ):
        return 1

num = []
n = 1
for s in sys.stdin:
    tn = list(map(int,s.split()))
    for n in tn:
        if n == 0:
            break
        num.append(n)

    if n == 0:
        break
    
print('PERFECTION OUTPUT')

for n in num:
    ret = checkPerfection(n)
    print("%5d  %s" % (n, msg[ret]))

print('END OF OUTPUT')