poj 1338 Ugly Numbers（丑数模拟）

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题目链接：

Description

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. 

Input

Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.

Output

For each line, output the n’th ugly number .:Don’t deal with the line with n=0.

Sample Input

1

2

9

0

Sample Output

1

2

10

PS：用一个长度为1500的数组存储这些数，另有三个游标x,y,z；

a[1]=1。x=y=z=1，代表第一个数为1，此后的数都是通过已有的数乘以2,3,5得到的，

那么x，y，z分别代表a[x],a[y],a[z]能够通过乘以2,3,5来得到新的数，i递增。每次取2*a[x], 3*a[y], 5*a[z]

中的最小值。得到a[i]后。能够将相应的x(或y,z)右移，当然假设原本通过3*2得到6，那么2*3也能得到6，

因此可能x和y都须要递增。

详见代码：

#include <iostream>

using namespace std;

int min(int a, int b, int c)

{

	return min(a,min(b,c));

}

int main()

{

	int a[1517];

	int x, y, z, i;

	x = y = z = 1, a[1] = 1;

	for(i = 2; i <= 1500; i++)

	{

		a[i] = min(2*a[x],3*a[y],5*a[z]);

		if(a[i] == 2*a[x])

		x++;

		if(a[i] == 3*a[y])

		y++;

		if(a[i] == 5*a[z])

		z++;

	}

	int n;

	while(cin >> n && n)

	{

		cout<<a[n]<<endl;

	}

	return 0;

}