丑数定义:“丑数”(ugly number)是正数,且质数因子只包含2、3、5。例如6,8是丑数,但14不是丑数,因为它包含因子7,规定整数1为第一个丑数。
问题分析:丑数一定是有限个2、3、5的乘积,因为所有的正整数都能分解成1与一个或多个素数的乘积。如果一个数是丑数,那么反复除以2、3、5后,一定会是1;如果一个数不是丑数,那么反复除以2、3、5后,一定还会剩下了一个质数无法被2、3、5整除。
编程内容(python代码 & C++代码):
1、判断一个整数是否是丑数?
2、给定区间,输出丑数的个数或丑数。
3、求第N个丑数
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寻找丑数算法1:
(1)设置一个计数器用来统计出现的丑数的个数
(2)从1开始遍历每一个整数,判断是否是丑数,如果是丑数则计数器加1,否则遍历下一个整数。
(3)当计数器的值=N时,停止遍历,输出丑数。
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寻找丑数算法2:
从上一个丑数推断出下一个丑数,而不需要从1开始遍历再判断。从1开始的10个丑数分别为1,2,3,4,5,6,8,9,10,12。可以发现除了1以外,丑数都是由某个丑数*2或者*3或者*5得到的。如2是丑数1*2得到的,3是丑数1*3得到的,4是丑数1*4得到的,5是丑数1*5得到的,6是丑数2*3得到的……
具体算法步骤:
(1)从第一个丑数1开始,求出1*2=2 ,1*3=3 ,1*5 = 5;
(2)取上面乘积中大于1的最小值2,作为第二个丑数(丑数是个递增序列,所以第i+1个丑数一定比第i个丑数大)
(3)求丑数2之前的丑数与2、3、5的乘积:1*2=2 ,1*3=3 ,1*5 = 5; 2*2 = 4; 2*3 = 6; 2*5 =10;
(4)取上面乘积中大于2的最小值3,作为第三个丑数
……
(i)取出丑数i之前的丑数分别与2、3、5的乘积
(i+1)取乘积中大于i的最小值作为丑数
(i+2)重复(i)(i+1)的步骤直到计数器等于N
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问题一:判断一个数是否为丑数
python代码:
# -*- coding: utf-8 -*- import sys # 判断一个数是否为丑数 def isUglyNumber(num): if num<=0: return False elif num>=1: for i in [2,3,5]: while num%i==0: num = num / i if num == 1: return True else: return False if __name__ == '__main__': try: while True: arr = int(input()) print(isUglyNumber(arr)) except: pass
运行结果:
C++代码:
#include <iostream> #include <vector> using namespace std; //判断一个数是否为丑数 bool isUglyNumber(int num){ vector<int> index = { 2, 3, 5 }; if (num <= 0) return false; else if (num >= 1) { for (int i = 0; i < 3; i++) { while (num%index[i] == 0) { num /= index[i]; } } if (num == 1) return true; else return false; } } int main(){ int num; while (1){ cin >> num; bool index = isUglyNumber(num); if (index == true) cout << "True" << endl; else cout << "False" << endl; } system("pause"); return 0; }
运行结果:
问题二:给定区间,输出丑数的个数或丑数
python 代码:
# -*- coding: utf-8 -*- import sys # 判断一个数是否为丑数 def isUglyNumber(num): if num<=0: return False elif num>=1: for i in [2,3,5]: while num%i==0: num = num / i if num == 1: return True else: return False def getUglyNumber(lowIndex,highIndex): res = [] if highIndex<=0: #如果区间右端点小于或等于0,则返回空列表 return res elif lowIndex>=highIndex: #如果区间左端点大于或等于右端点,则返回空列表 return res elif lowIndex<=0 and highIndex>0: #如果区间左端点小于或等于0,右端点大于0 for i in range(1,highIndex+1): if isUglyNumber(i)==True: res.append(i) return res elif lowIndex>0 and highIndex>0: #如果区间左端点大于0,右端点大于0,且右端点大于左端点 for i in range(lowIndex,highIndex+1): if isUglyNumber(i)==True: res.append(i) return res if __name__ == '__main__': try: while True: arr = [int(t) for t in sys.stdin.readline().split()] # 等价于 arr = [int(t) for t in input("").split()] res = getUglyNumber(arr[0],arr[1]) # print(res) 输出格式为:如[1,2,3,4,5,6] if len(res)>=1: print(" ".join(str(i) for i in res)) # 输出列表元素,以空格为分隔符,且最后一个无空格 else: print('no') except: pass
运行结果:
C++代码:
#include <iostream> #include <vector> using namespace std; //判断一个数是否为丑数 bool isUglyNumber(int num){ vector<int> index = { 2, 3, 5 }; if (num <= 0) return false; else if (num >= 1) { for (int i = 0; i < 3; i++) { while (num%index[i] == 0) { num /= index[i]; } } if (num == 1) return true; else return false; } } // 获取第N个丑数 vector<int> getUglyNumber(int lowIndex, int highIndex){ vector<int> res; // 如果区间右端点小于或等于0,则返回空向量 if (highIndex <= 0) return res; // 如果区间左端点大于或等于右端点,则返回空列表 else if (lowIndex >= highIndex) return res; // 如果区间左端点小于或等于0,右端点大于0 else if (lowIndex <= 0 && highIndex > 0){ for (int i = 1; i < highIndex + 1; i++){ if (isUglyNumber(i) == true){ res.push_back(i); } } } // 如果区间左端点大于0,右端点大于0 else if (lowIndex > 0 && highIndex > 0){ for (int i = lowIndex; i < highIndex + 1; i++){ if (isUglyNumber(i) == true){ res.push_back(i); } } } return res; } int main(){ vector<int> index(2),myRes; while (1){ for (int i = 0; i < 2; i++){ cin >> index[i]; } myRes = getUglyNumber(index[0],index[1]); int len = myRes.size();// 求myRes长度 if (len >= 1) { for (int j = 0; j < len; j++){ cout << myRes[j] << " "; } cout << endl; } else{ cout << "no" << endl; } } system("pause"); return 0; }运行结果:
问题三:求第N个丑数
python代码1:
# -*- coding: utf-8 -*- import sys # 判断一个数是否为丑数 def isUglyNumber(num): if num<=0: return False elif num>=1: for i in [2,3,5]: while num%i==0: num = num / i if num == 1: return True else: return False # 获取第N个丑数 def getUglyNumber(N): count = 0 #用于计数 if N<=0: return 0 else: num = 1 while (count<N): if isUglyNumber(num) == True: count = count + 1 num = num + 1 return num-1 if __name__ == '__main__': try: while True: arr = int(input()) print(getUglyNumber(arr)) except: pass
运行结果:
C++代码1:
#include <iostream> #include <vector> using namespace std; //判断一个数是否为丑数 bool isUglyNumber(int num){ vector<int> index = { 2, 3, 5 }; if (num <= 0) return false; else if (num >= 1) { for (int i = 0; i < 3; i++) { while (num%index[i] == 0) { num /= index[i]; } } if (num == 1) return true; else return false; } } //获取第N个丑数 int getUglyNumber(int N){ int count = 0; if (N <= 0) return 0; else{ int num = 1; while (count < N){ if (isUglyNumber(num) == true){ count += 1; } num += 1; } return num - 1; } } int main(){ int num; while (1){ cin >> num; cout << getUglyNumber(num) << endl; } system("pause"); return 0; }
运行结果:
python代码2:
import sys def getUglyNumber(N): if N < 1: return 0 res = [1] t2 = t3 = t5 = 0 next = 1 while next < N: min_num = min(res[t2] * 2, res[t3] * 3, res[t5] * 5) res.append(min_num) if res[t2] * 2 <= min_num: t2 += 1 if res[t3] * 3 <= min_num: t3 += 1 if res[t5] * 5 <= min_num: t5 += 1 next += 1 return res[N - 1] if __name__ == '__main__': try: while True: arr = int(input()) print(getUglyNumber(arr)) except: pass
运行结果如上,该方法在优于方法1。因此,在剑指offer上的剑指offer——丑数的提交python代码应为方法2,方法1不通过。
即:
# -*- coding:utf-8 -*- class Solution: def GetUglyNumber_Solution(self, index): # write code here if index < 1: return 0 res = [1] t2 = t3 = t5 = 0 next = 1 while next < index: min_num = min(res[t2]*2, res[t3]*3, res[t5]*5) res.append(min_num) if res[t2]*2 <= min_num: t2 += 1 if res[t3]*3 <= min_num: t3 += 1 if res[t5]*5 <= min_num: t5 += 1 next += 1 return res[index-1]
提交结果:
C++代码:
#include <iostream> #include <vector> using namespace std; //返回三个数字里面的最小值 int min(int num1, int num2, int num3) { int min = num1 < num2 ? num1 : num2; min = min < num3 ? min : num3; return min; } //获取第N个丑数 int getUglyNumber(int N){ if (N < 1) return 0; vector<int> res = { 1 }; int t2 = 0; int t3 = 0; int t5 = 0; int next = 1; while (next < N){ int min_num = min(res[t2] * 2, res[t3] * 3, res[t5] * 5); res.push_back(min_num); if (res[t2] * 2 <= min_num) t2 += 1; if (res[t3] * 3 <= min_num) t3 += 1; if (res[t5] * 5 <= min_num) t5 += 1; next += 1; } return res[N - 1]; } int main(){ int num; while (1){ cin >> num; int res = getUglyNumber(num); cout << res << endl; } system("pause"); return 0; }
同理,在剑指offer上的剑指offer——丑数的提交C++代码:
class Solution { public: int min(int num1, int num2, int num3){ int min = num1 < num2 ? num1 : num2; min = min < num3 ? min : num3; return min; } int GetUglyNumber_Solution(int index) { if (index < 1) return 0; vector<int> res = { 1 }; int t2 = 0; int t3 = 0; int t5 = 0; int next = 1; while (next < index){ int min_num = min(res[t2] * 2, res[t3] * 3, res[t5] * 5); res.push_back(min_num); if (res[t2] * 2 <= min_num) t2 += 1; if (res[t3] * 3 <= min_num) t3 += 1; if (res[t5] * 5 <= min_num) t5 += 1; next += 1; } return res[index - 1]; } };
提交结果:
参考网址:
1、https://blog.csdn.net/u013632190/article/details/52036119
2、https://blog.csdn.net/lzuacm/article/details/51336420
3、https://blog.csdn.net/my_mao/article/details/24366291