Ugly Numbers(UVa 136)优先队列
2016-09-29 11:03
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来自《算法竞赛入门经典第二版》
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.
1.题目原文
Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence1, 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>.2.解题思路
按照从小到大的顺序生成丑数。最小的丑数是1,而对于任意丑数x,2x,3x,5x都是丑数,就可以利用一个优先队列保存已经生成的丑数,每次去除最小的丑数,生成三个新的丑数。注意一个丑数可能有多个生成方式,因此需要判断是否已生成过。3.AC代码
#include <algorithm> #include <cctype> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <string> #include <set> #include <vector> #include<cmath> #include<bitset> using namespace std; typedef long long ll; const int a[]={2,3,5}; int main() { priority_queue<ll,vector<ll>,greater<ll> > que; set<ll> s; que.push(1); s.insert(1); for(int i=1;;i++){ ll x=que.top(); que.pop(); if(i==1500){ cout<<"The 1500'th ugly number is "<<x<<".\n"; break; } else{ for(int j=0;j<3;j++){ ll x2=x*a[j]; if(!s.count(x2)){ s.insert(x2); que.push(x2); } } } } return 0; }
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