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 .
思路:从一开始,对应的乘2,3,5,得到的数进行排列,每次取得到的最小的值,以此类推一直到1500个,但是要注意会有重复的数,因为你一个数乘2乘5次也就能被5整除了,同理对应乘3也是如此,所以要判断一下。我用的是优先队列解决的。
#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stack>
using namespace std;
const int maxn = 1505;
long long a, b, c;
priority_queue<long long, vector<long long>, greater<long long>> p;
int main()
{
p.push(1);
long long num = 0;
for (int i = 1; i < maxn; i++)
{
long long ans = p.top();
p.pop();
num++;
if (num == 1500)
cout << "The 1500'th ugly number is " << ans << "." << endl;
a = ans * 2;
b = ans * 3;
c = ans * 5;
if ((a % 5) && (a % 3))
p.push(a);
if ((b % 5))
p.push(b);
p.push(c);
}
return 0;
}