• (Problem 21)Amicable numbers


    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.

    d(n)定义为n 的所有真因子(小于 n 且能整除 n 的整数)之和。 如果 d(a) = b 并且 d(b) = a, 且 a ≠ b, 那么 a 和 b 就是一对相亲数(amicable pair),并且 a 和 b 都叫做亲和数(amicable number)。

    例如220的真因子是 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 和 110; 因此 d(220) = 284. 284的真因子是1, 2, 4, 71 和142; 所以d(284) = 220.

    计算10000以下所有亲和数之和。

    // (Problem 21)Amicable numbers 
    // Completed on Wed, 24 Jul 2013, 06:07
    // Language: C
    //
    // 版权所有(C)acutus   (mail: acutus@126.com) 
    // 博客地址:http://www.cnblogs.com/acutus/
    #include<stdio.h>
    
    int FactorSum(int n)  //计算n的所有小于n的因素和
    {
        int i;
        int sum=1;
        for(i=2; i<=n/2; i++)
        {
            if(n%i==0)
                sum+=i;
        }
        return sum;
    }
    
    int main()
    {
        int t,i=2;
        int sum=0;
        while(i<10000)
        {
            t=FactorSum(i);
            if(t!=i && FactorSum(t)==i) 
                sum+=i;
            i++;
        }
        printf("%d
    ",sum);
        return 0;
    }
    Answer:
    31626
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  • 原文地址:https://www.cnblogs.com/acutus/p/3545413.html
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