Ugly Numbers, UVa 136

Ugly Numbers, UVa 136

题目

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 1500th ugly number.

Input

There is no input to this program.

Output

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 >.

题意

丑数是指不能被2,3,5以外的其他素数整除的数，本题要求从小到排列的丑数中第1500个。

分析

书中方法定义了一个“越小的整数优先级越大的优先队列”，定义方法为：

priority_queue<int,vector<int>,greater<int> > pq;

最后两个“>”不要写在一起，以防编译器认作“>>”。

因为2x,3x,5x仍为丑数，每次将pq中最小丑数出队并乘2,3,5形成新的丑数加入队列，以求出第1500个丑数。

代码实现

#include<cstdio>
#include<iostream>
#include<sstream>
#include<algorithm>

//STL
#include<string>
#include<map>
#include<set>
#include<vector>
#include<iterator>
#include<stack>
#include<queue>
#include <functional>
using namespace std;

typedef long long LL;
const int coeff[3] = { 2,3,5 };

int main()
{
priority_queue<LL, vector<LL>, greater<LL> > pq;
set<LL> s;
pq.push(1);
s.insert(1);
for (int i = 1;; i++)
{
LL x = pq.top(); pq.pop();
if (i == 1500)
{
cout << "The 1500'th ugly number is " << x <<".\n";
break;
}
for (int j = 0; j < 3; j++)
{
LL x2 = x*coeff[j];
if (!s.count(x2)
4000
) { s.insert(x2); pq.push(x2); }
}
}
return 0;
}

遇到的问题

使用

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

时报错“greater”不是模板，加入

#include <functional>

后解决。