Uva 136 Ugly Numbers(丑数)

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
There is no input to this program.
Output
Output should consist of a single line as shown below, with ‘’ replaced by the number
computed.
Sample Output
The 1500’th ugly number is .

[分析1]
学习优先队列，基本可以说代码就是照着书打的了。不过就是想记录一下优先队列让我蛋疼的对比重载方法。

这个greater对比模板包含于头文件（不知道为什么code block不需要头文件，vs需要），反正照着打就对了。。。。。。。。还是要深入学习一下。

[代码1]

#include<cstdio>#include<queue>#include<vector>#include<set>#include<functional>typedef long long ll;using namespace std;int pri[3] = { 2,3,5 };int main(){    priority_queue<ll, vector<ll>, greater<ll> >q;    set<ll>s;    q.push(1);    s.insert(1);    for (int i = 1;; i++)    {        ll x = q.top(); q.pop();        if (i == 1500)        {            printf("The 1500'th ugly number is %lld.\n", x);            break;        }        for (int j = 0; j < 3; j++)        {            ll x2 = pri[j] * x;            if (s.count(x2) == 0)            {                s.insert(x2);                q.push(x2);            }        }    }}

[分析2]
自定义的比较要像这样写，建议多用这种，习惯了以后做起来快。

[代码2]

#include<cstdio>#include<queue>#include<vector>#include<set>#include<functional>typedef long long ll;using namespace std;int pri[3] = { 2,3,5 };struct cmp{    bool operator()(const ll a, const ll b)const    {        return a > b;    }};int main(){    priority_queue<ll,vector<ll>,cmp>q;    set<ll>s;    q.push(1);    s.insert(1);    for (int i = 1;; i++)    {        ll x = q.top(); q.pop();        if (i == 1500)        {            printf("The 1500'th ugly number is %lld.\n", x);            break;        }        for (int j = 0; j < 3; j++)        {            ll x2 = pri[j] * x;            if (s.count(x2) == 0)            {                s.insert(x2);                q.push(x2);            }        }    }    getchar();}