Solution of ZOJ 2095 Divisor Summation （Online Version）

Give a natural number n (1 <= n <= 500000), please tell the summation of all its proper divisors.

Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.

e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

 

Input

An integer stating the number of test cases, and that many lines follow each containing one integer between 1 and 500000.

Output

One integer each line: the divisor summation of the integer given respectively.

Sample Input

3
2
10
20

Sample Output

1
8
22

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Source: ZOJ Monthly, March 2004

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分析：此题关键点在I/O部分，如果是用C++标准库中的cin和cout对象，则将超时。网上提供的大部分算法都是基于查表的方法，这里给出一个我实现的可Accept的online算法，所占内存只需188KB。

 

#include <iostream>#include <cstdio>#include <cmath>using namespace std;int proper_div_sum( int n ){int sum = 1;for( int i = (int)sqrt(n * 1.0); i > 1; --i ){if( n % i == 0 ) { sum += i;if( n / i != i )sum += n / i;}}if( n == 1 || n == 0) {printf("0/n");//sum = 0;}elseprintf( "%d/n", sum );return sum;}int main(){int n;scanf("%d", &n );while( scanf("%d", &n ) != EOF ){proper_div_sum(n);}return 0;}