欧拉计划问题二十三matlab实现

Problem 23 ：Non-abundant sums

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

思路 ：

找出28123以下所有的盈数，对其进行加和；最后用1到28123所有的总数和再减去所有盈数之和即为所求结果；

我们可先设立两个矩阵，一个是用于存放盈数的矩阵A，另一个是存放1到28123的矩阵B；若矩阵B中的数满足盈数条件，则将其存入矩阵A中，将该数在原矩阵B中的位置替换成0；最后就只需要输出矩阵B的总和即为结果。

代码 ：

clear,clc;
tic
A = [];         %null matrix
B = (1:28123)'; %save number
for i = 12:28123
    k = 1:sqrt(i);   %similar with prime
    k = k(mod(i,k)==0);
    s = sum(k)+sum(i./k);
    if k(end) == sqrt(i)
        s = s - k(end);
    end
    if s > 2*i
        A = [A,i];
        t = A + i;
        B(t(t<=28123))=0;  %replace to zero
    end    
end
result = sum(B);
result
toc

结果 ：4179871

     也可以像求素数那样做，方法很多！