sicily 1691. Abundance

Description
An abundant number is a positive integer n for which Sigma(n) – 2n > 0, Where Sigma(n) is defined as the sum of all the divisors of n. And the quantity Sigma(n) – 2n is called abundance.

Given the range of n, you should find out the maximum abundance value that can be reached. For example, if the range is [10,12], then the only abundant number is 12, and the maximum abundance value is Sigma(12) – 2 * 12 = 4.

Input
Input may contain several test cases. The first line is a positive integer, T (T<=20), the number of test cases below. Each test case contains two positive integers x, y, (1<= x <= y <= 1024), indicating the range of n

Output
For each test case, output the maximum abundance value that can be reached in the range of n. If there is no abundant number n in the given range, simply output -1.

水题，枚举并更新记录最大值即可

有一个小技巧，每个case中都将max初始化为-1，如果没有abundant number，max就不会被更新，这样最后就会直接输出-1，省去了判断语句

#include<stdio.h>
int sigma( int n )
{
    int i, sum = 0;
    
    for ( i = 1; i <= n; i++ )
        if ( n % i == 0 )
            sum += i;
    
    return sum;
}
int main()
{
    int t, x, y, i, max, abun;
    
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &x, &y);
        max = -1;
        
        for ( i = x; i <= y; i++ )
        {
            abun = sigma(i) - 2 * i;
            if ( abun > 0 )
                if ( abun > max )
                    max = abun;
        }
        
        printf("%d\n", max);
    }
    
    return 0;
}