1751: Ugly Numbers

最新推荐文章于 2019-08-20 15:29:52 发布

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本文介绍了一个寻找第1500个丑数的算法实现，丑数是指只包含2、3和5这三个质因数的整数。通过使用C++中的堆排序，该算法高效地生成了所需的丑数。

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	8192K	1038	322	Standard

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 and Output

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

Sample output

The 1500'th ugly number is <number>.

This problem is used for contest: 112

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
   int i;
   double x,ugly[1501];
   vector<double> heap(1);
   heap[0]=-1;
   for(i=1;i<=1500;i++)
   {
      x=heap[0],ugly[i]=-x;
      pop_heap(heap.begin(),heap.end());
      heap.pop_back();
      if(find(heap.begin(),heap.end(),x*2)==heap.end())
      {
    heap.push_back(x*2);
    push_heap(heap.begin(),heap.end());
   }
   if(find(heap.begin(),heap.end(),x*3)==heap.end())
   {
    heap.push_back(x*3);
    push_heap(heap.begin(),heap.end());
   }
   if(find(heap.begin(),heap.end(),x*5)==heap.end())
   {
    heap.push_back(x*5);
    push_heap(heap.begin(),heap.end()) ;
   }
}
    printf("The 1500'th ugly number is %.0lf./n",ugly[1500]) ;
    return 0 ;
}