Ugly Numbers(UVa 136)优先队列

来自《算法竞赛入门经典第二版》

1.题目原文

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>.

2.解题思路

按照从小到大的顺序生成丑数。最小的丑数是1，而对于任意丑数x，2x,3x,5x都是丑数，就可以利用一个优先队列保存已经生成的丑数，每次去除最小的丑数，生成三个新的丑数。注意一个丑数可能有多个生成方式，因此需要判断是否已生成过。

3.AC代码

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include<cmath>
#include<bitset>
using namespace std;

typedef long long ll;
const int a[]={2,3,5};

int main()
{
     priority_queue<ll,vector<ll>,greater<ll> > que;
     set<ll> s;
     que.push(1);
     s.insert(1);
     for(int i=1;;i++){
        ll x=que.top();
        que.pop();
        if(i==1500){
            cout<<"The 1500'th ugly number is "<<x<<".\n";
            break;
        }
        else{
            for(int j=0;j<3;j++){
                ll x2=x*a[j];
                if(!s.count(x2)){
                    s.insert(x2);
                    que.push(x2);
                }
            }
        }
     }
    return 0;
}