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 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>.
这道题用到了优先队列。STL的queue头文件中提供了优先队列,用"priority_queue<int> s"方式定义。
用push()和pop()进行元素的入队和出队操作,
top()取队首元素。
priority_queue<int,vector<int>,greater<int> >pq表示越小的整数优先级越大。
看书上的解题的时候,一开始不明白为什么要循环1500次,想了一下,原来是因为每次都取的是最小的元素啊,取上1500次,就是第1500个丑数了。
还有觉得比较神奇的地方是,丑数和丑数相乘,结果依然是丑数。
然后用到了typedef,感觉挺新鲜~
渣渣一枚,坚持!
#include<iostream> #include<vector> #include<queue> #include<set> using namespace std; typedef long long LL; const int coeff[3]={2,3,5}; int main(){ priority_queue<LL,vector<LL>,greater<LL> >pq;//优先队列 set<LL> s;//集合 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<<". "; break; } for(int j=0;j<3;j++){ LL x2=x*coeff[j]; if(!s.count(x2)){ s.insert(x2); pq.push(x2); } } } return 0; }