POJ 1528: 完全數

——完全數
原題傳送門

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.

Data

Input
A list of N positive integers (none greater than 60,000). 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.
1<=N<100

	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

思路

定性:一道純粹的數論題
其實還是一道比較簡單的題目.?
完全數的定義見題目解釋,這裏不再贅述.
主要是看比較基礎的算法:求約數

求約數

其實就一句話很簡單
設原數爲X
因爲約數都小於根號X
所以枚舉 i:1~根號X,判斷x%i 是否爲0即可.

Code

// 離線
void work(long long x)
{
	for (long long i=1;i<=x/2;i++)
		if (x%i==0)
			sum+=1LL*i;
} 
int main()
{
	int cnt=0;
	long long x;
	while (scanf("%lld",&x)&&x) n[++cnt]=x;
	printf("PERFECTION OUTPUT\n");
	for (int i=1;i<=cnt;i++)
	{
		sum=0;
		work(n[i]);
		printf("%5lld  ",n[i]);
		if (sum<n[i]) printf("DEFICIENT\n");
		else if (sum>n[i]) printf("ABUNDANT\n");
		else printf("PERFECT\n");
	} 
	printf("END OF OUTPUT\n"); 
}

感謝奆老關注 qwq ?