[Project Euler] Problem 21

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

我们直接编写函数d(n)，如果d(d(n))=n,并且n!=d(n)那么n就是符合要求的数

我们直接暴力搜索

#include <iostream>
#include <cmath>
usingnamespace std;

int d(int tmp){
    int sum =0;
    int i;
    for(i=1; i<sqrt(tmp); i++){
        if(tmp%i ==0){
            sum = sum+i+tmp/i;
        }
    }
    if(i*i == tmp){
        sum += i;
    }
    return sum-tmp;
}

int main(){
    int sum =0; 
    for(int i=1; i<10000; i++){
        if(i==d(d(i)) && i!=d(i)){
            sum += i;
        }
    }
    cout << sum << endl;
    return0;
}