UVA 136 Ugly Numbers

寻找第1500个丑数

最新推荐文章于 2021-08-22 21:48:34 发布

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本文介绍了一个寻找第1500个丑数的算法实现，丑数定义为仅包含质因数2、3和5的整数。通过使用优先队列和集合来避免重复计算，实现了高效查找特定位置的丑数。

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 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>.

#include<iostream>
#include<vector>
#include<queue>
#include<set>
using namespace std;

const int c[3] = {2, 3, 5};
priority_queue<long long,vector<long long>,greater<long long> >pq;
set<long long> s;

int main() {
    pq.push(1);
    s.insert(1);
    long long x;
    for(int i = 1; i<=1500; i++) {
            x = pq.top(); pq.pop();
    for(int j = 0; j < 3; j++)
        {
            long long x2 = x * c[j];
            if(s.count(x2)==0) {
                    s.insert(x2); pq.push(x2);
            }
        }
    }
    cout << "The 1500'th ugly number is " << x<< "."<<endl;
    return 0;
}