C++程序练习-1338:Ugly Numbers-set的应用-优快云博客

本文介绍了一种使用C++ set容器高效寻找特定位置的丑数的方法。丑数定义为只包含质因数2、3和5的正整数。通过预先计算并存储丑数，实现了快速查找功能。

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描述
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, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.

输入
Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.
输出
For each line, output the n’th ugly number .:Don’t deal with the line with n=0.
样例输入
1
2
9
0
样例输出
1
2
10

分析：
建立数据表，然后查找数据表，输出结果。

思路：
可以用C++中set，且它的元素唯一，有序。

//冰非寒(binfeihan@126.com) //260kB 0ms 1196 B G++ #include <cstdio> #include <set> using namespace std; //global variables const int N =1601; set<unsigned long>uglynum; //functions void init(); void print(int); //main function int main(){ int n; init(); while(scanf("%d",&n) == 1&&n){ print(n); } return 0; } //initialize ugly numbers void init(){ int j(1); pair<set<unsigned long>::iterator,bool> ptr; uglynum.insert(1); for(set<unsigned long>::iterator it = uglynum.begin();j < N;++ it){ ptr = uglynum.insert(*it * 2); if(ptr.second == 1){ j ++; } ptr = uglynum.insert(*it * 3); if(ptr.second == 1){ j ++; } ptr = uglynum.insert(*it * 5); if(ptr.second == 1){ j ++; } } } //print result void print(int n){ int j(1); for(set<unsigned long>::iterator it = uglynum.begin();j <= n && it != uglynum.end();++ it){ if(j == n){ printf("%u\n",*it); } j ++; } }