N位水仙花数 Java

原创于 2023-02-24 10:40:20 发布 · 709 阅读

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#java #算法

先上代码:
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int inputNum = in.nextInt();

    //取位数
    int pos = 0;
    int temp = inputNum;
    while (temp / 10 != 0) {
        ++pos;
        temp = temp / 10;
    }
    for (int poss = pos; poss > 1; poss--) {
        for (int index = inputNum; index > 10; index--) {
            List<Double> nums = new ArrayList<>();
            //存放各个位数
            for (int i = 0; i <= poss; i++) {
                nums.add(Math.floor(index / Math.pow(10, poss - i) % 10));
            }
            //计算和
            double sum = 0.0;
            for (Double num : nums) {
                sum += Math.pow(num, poss + 1);
            }
            //4150和4151是特殊情况
            //当pos=5时，4^5+1^5+5^5+1^5=4150
            //其实4150当时pos=4才对，所以把这两个数排除
            if (sum == index) {
                if (sum == 4150 || sum == 4151)
                    continue;
                System.out.println(index);
            }
        }
    }
}

pass:该题效率较低，欢迎大佬提供建议~