当然,你要是把int范围内的5个Perfect Number(第5个是33550336)都找出来了,直接做判断也行。 publicbooleancheckPerfectNumber3(intnum){int[] primes = {2,3,5,7,13};for(intp: primes) {if((1<< (p -1)) * ((1<< p) -1) == num) {returntrue; } }returnfalse; } 05 小结 算法专题目前...
for (int i=2; i<=Math.sqrt(num); i++) { if (num%i == 0) { sum += i + num/i; } } return sum == num; } 1 2 3 4 5 6 7 8 9 10 11 12 04 第三种解法 最小的Perfect Number是6,接着是28,然后是496 6: 2x3 true ...
Time Complexity: O(Math.sqrt(num)). Space: O(1). AC Java: 1publicclassSolution {2publicbooleancheckPerfectNumber(intnum) {3if(num == 1){4returnfalse;5}67intsum = 1;8for(inti = 2; i<Math.sqrt(num); i++){9if(num%i == 0){10sum += i+num/i;11}12}13returnsum ==num;14...
Note: The input number n will not exceed 100,000,000. (1e8) 题意:判断一个数是不是完全数。 思路: 1、遍历会超时,稍微优化一***意num=0,1的特殊情况 class Solution { public: bool checkPerfectNumber(int num) { if (num == 0) return false; int sum = 0; for (int i = 1; i...
bool checkPerfectNumber(int num) { if (num == 1) return false; int sum = 1; for (int i= 2; i <= (int)sqrt(num); i++) { if (num%i == 0) { sum += i; if (i != num / i) sum += num / i; } } return sum == num; ...
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完美数 https://leetcode-cn.com/problems/perfect-number/ 难度:简单 解题思路: 设计一个判断单个数是否是自除数的函数,比如取名为IsSelfDIv() IsSelfDIv函数的要点是: 排除数位中含有0的数 排除原数不是末位数字倍数的数 遍历原数列,对每一个数调用一次进行IsSelfDIv函数,将满足要求的加入List中即可 ...
process to make it seamless, convenient, and effective. The software can compute the bill amount on the basis of units of electricity consumed in a month. Electricity Billing System is considered one of the best java project ideas for beginners. The app should have the below-mentioned features...
7. Perfect square number @Test public void test_PerfectSquare() { for (long l = 1L; l < 100000; l++) { if (Math.sqrt((l + 100)) % 1 == 0) { if (Math.sqrt((l + 268)) % 1 == 0) { System.out.println(l + "加100是一个完全平方数,再加168又是一个完全平方数"); }...
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