丑数简单的优先队列和set

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本文介绍了一种使用优先队列和集合数据结构来找出序列中第N个丑数的方法，丑数定义为仅包含2、3、5作为质因数的整数。

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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, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.

Input

Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.

Output

For each line, output the n’th ugly number .:Don’t deal with the line with n=0.

Sample Input

Sample Output

1
2
10

题意大概是 1个数列其中的数的质因数只能包含2,3,5问第n个这样的数是多少。
优先队列priority_queue 和set用于判断新生成的数是否重复，比如2*3=6 3*2=6

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <cctype>
#include <ctime>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
#define N 10005
#define eps 1e-5
#define pai acos(-1)
#define MOD 1e9+7
ll num[3]={2,3,5};
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    ll _begin=clock();
    #endif
    int n;
    while(scanf("%d",&n)&&n)
    {
        priority_queue < ll , vector<ll> ,greater<ll> > s;
        set<ll> ugly;
        s.push(1);
        ugly.insert(1);
       for(int i=1;;i++)
       {
           ll x=s.top(); s.pop();
           if(i==n)
            {
                printf("%lld\n",x);
                break;
            }
           for(int j=0;j<=2;j++)
           {
               ll x2=x*num[j];
               if(!ugly.count(x2))
               {
                   s.push(x2);
                   ugly.insert(x2);
               }
           }
       }
    }
    #ifndef ONLINE_JUDGE
    ll _end=clock();
    printf("%lldms\n",_end-_begin);
    #endif
    return 0;
}