Ugly Numbers - uva136 - priority_queue与set

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence1, 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.InputThere is no input to this program.OutputOutput should consist of a single line as shown below, with ‘’ replaced by the numbercomputed.Sample OutputThe 1500'th ugly number is .
分析：
x=1;2x,3x,5x都是丑数，应该考虑怎样才能顺序地构造丑数使得第1500个所求的数正好是第1500个丑数
构造一个集合和优先队列
优先队列每次弹出一个最小的数，从这个数开始构造新的丑数，入队的时候应该考虑是否已经出现过这个数，所以用一个set集合同步判断（a.count(x)!=0),在队列和集合的数不是顺序的丑数，被弹出的才是#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<queue>

using namespace std;
typedef long long LL;

set<LL> a;
priority_queue<LL,vector<LL>,greater<LL> > pq;

int f[3]={2,3,5};
int main(){
a.insert(1);
pq.push(1);

for(int i=1;;i++){
LL x=pq.top();pq.pop();
if(i==1500){
cout<<"The 1500'th ugly number is "<<x<<"."<<endl;
break;
}
for(int j=0;j<3;j++){
LL y=f[j]*x;
if(!a.count(y)){a.insert(y);pq.push(y);}
}
}

}
回顾优先队列priority_queue<int> q;//默认从大到小
priority_queue<int,vector<int>,greater<int> > pq//从小到大

//自定义排序

bool cmp{
bool operator()(const int a,const int b)const{
return a%10>b%10;//按个位数从小到大
}
}
priority_queue<int,vector<int>,cmp> p;
q.top()取队首