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;}