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需求:
一个数如果恰好等于它的因子之和, 这个数就称为”完数”. 例如 6 = 1+2+3。编程找出 1000 以内的所有完数。
分析:
一千个数肯定跑不了循环的使用,常用的循环有两个,一个是for ,一个是while,因为这里知道次数,所以优先选择for循环;
其次考虑到因子,因子需要用被除数除以除数,显然这个时候被除数是不变的,而除数是一直在变化的,结合这个特点,就应该想到循环的嵌套,外面的走一次,里面的走一遍。
再通过if判断语句就可以找到,被除数的所有真因子,再根据“完数”的定义: 进行比较,得出结果,那好,我们用代码试一下:
public static void main(String[] args) { // 遍历所有的被除数 for (int i = 1; i < 1000;
i++) { // 用于存储各个符合因子的和 int sum = 0; // 遍历小于被除数的所有因子 for (int j = 1; j < i; j++)
{ // 判断是否为被除数的因子 if (i % j == 0) { // 是的话就加起来,好与被除数作比较 sum += j; } } //
当内层循环结束,可以得到所有的真因子的和再和被除数比较,输出 if (sum == i ) { System.out.println(i); } } }
得到结果:
6 28 496
路虽远,行则将至;事虽难,作则必成;