Ugly Numbers(set)

最新推荐文章于 2024-08-10 11:53:55 发布

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该程序通过集合数据结构寻找第1500个丑数，丑数是仅包含质因数2、3和5的正整数。程序首先将1插入集合，然后迭代集合中的每个元素，将其乘以2、3和5并插入集合，直到找到第1500个丑数并输出。

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ... shows the first 11 ugly numbers. By convention, 1 is included. Write a program to find and print the 1500’th ugly number.

Input

There is no input to this program. Output Output should consist of a single line as shown below, with ‘’ replaced by the number computed. Sample

Output

The 1500'th ugly number is .

利用set会自动排序的性质来写。

注意输出格式错误❌，也可能报告为wrong answer。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<set>
using namespace std;
int count=0;
int main()
{
	set<long long>s;//定义一个longlong类型的
	set<long long>::iterator it;//迭代器 
	s.insert(1);
	for(it=s.begin();it!=s.end();it++)
	{
		count++;
		if(count==1500)
		{
			printf("The 1500'th ugly number is %lld.\n",*it); 
		}
		s.insert((*it)*2);
		s.insert((*it)*3);
		s.insert((*it)*5); 
	} 
	return 0;
}