POJ 1338 Ugly Numbers

对构造目标函数还是不熟悉。
自己每次都是想用排除法去构造，于是  时间和空间复杂度提高。
不如采用直接构造  数列 的方法         或者直接找出目标数    或 用i j  k 循环 等构造


Ugly Numbers
Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17602		Accepted: 7805
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




#include<iostream>
using namespace std;

int a[1510];

int main()
{
    int p2,p3,p5;
    p2=p3=p5=a[1]=1;
    int n;
    for(int i=2;i<=1500;i++)
    {
          int t2=a[p2]*2;
          int t3=a[p3]*3;
          int t5=a[p5]*5;
          if(t2<t3&&t2<t5)  {a[i]=t2;p2++;}
          else if(t3<t2&&t3<t5) {a[i]=t3;p3++;}
          else if(t5<t2&&t5<t3)   {a[i]=t5;p5++;}
          else if(t2==t3&&t2<t5)  {a[i]=t2;p2++;p3++;}
          else if(t2==t5&&t2<t3)  {a[i]=t2;p2++;p5++;}
          else if(t3==t5&&t3<t2)  {a[i]=t3;p3++;p5++;}
          else {a[i]=t2;p2++;p3++;p5++;}
    }
    while(cin>>n&&n!=0)
    {
         cout<<a[n]<<endl;
    }

    return 0;
}