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转载请注明出处：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.
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;
}
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