leecode 解题总结：263. Ugly Number

#include <iostream>#include <stdio.h>#include <vector>#include <string>using namespace std;/*问题:Write a program to check whether a given number is an ugly number.Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.Note that 1 is typically treated as an ugly number.分析:丑数是只包含质数银子为2,3,5的。1也是丑数。判定一个 给定的数是否是丑数。最简单的方法：先将该数除以2，直到不能除以2后，如果最后变成1，说明该数是丑数再不断除以3，如果变成1，说明是丑数；否则再不断除以5，如果变成1说明是丑数，如果最后除不尽5，那么就不是质数。丑数是正整数,因此<=0，不是丑数输入:012 3 5614900901输出:falsetruetruetruetruetruefalsetruefalse关键:1 不断累除2,3,5中间不断累除某个数的过程中，只要变成1，说明就是丑数；否则累除完3个数还不为1，就不是丑数*/class Solution {public:    bool isUgly(int num) {if(num <= 0){return false;}int primeArr[3] = {2,3,5};int i = 0;while(num != 1){//如果数字大于质数，并且对质数取余没有余数(可以除尽)，继续做。这里等于也要处理while(num >= primeArr[i] && (0 == num % primeArr[i]) ){while(1 == num){return true;}num /= primeArr[i];}//判断是否可以处理下一个质数,如果下标超过2，但是当前数字不为1，//说明不是丑数if(i >= 2){if(1 != num){return false;}else{return true;}}else{i++;}}return true;    }};void print(vector<int>& result){if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;}void process(){ int num; Solution solution; while(cin >> num ) { bool result = solution.isUgly(num); if(result) { cout << "true" << endl; } else { cout << "false" << endl; } }}int main(int argc , char* argv[]){process();getchar();return 0;}