POJ 1338 Ugly Numbers

Ugly Numbers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 20294   Accepted: 9018 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 Source New Zealand 1990 Division I,UVA 136

刚开始这样做，为什么不对啊~

#include <iostream>
#include <algorithm> 
#include <stdlib.h>
using namespace std;
int cmp(const void *a,const void *b)
{
return *(int *)a-*(int *)b;
}
int main()
{
    int n;
	while(cin>>n&&n)
	{
		long int i,j,k,t=0,a[1505];
		long int limit=100000000;
		for(i=1;i<limit;i=i*2)
			for(j=1;i*j<limit;j=j*3)
				for(k=1;i*j*k<limit;k=k*5)
				{
					a[t]=i*j*k;
					t++;
				}
				qsort(a,t,sizeof(int),cmp);
				cout<<a[n-1]<<endl;
	}
    return 0;
}

更新：我的while大循环放在打表外面了。应该放在最后面。而且我的循环控制结束语句也错了，为什么要加入

i*j*k<limit

这几句呢？

附ac代码：

#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int n;
	int i,j,k,t=0,a[2000];
	int max=1000000000;
	for(i=1;i<max&&t<2000;i=i*2)
		for(j=1;i*j<max&&i==i*j/j&&t<2000;j=j*3)
			for(k=1;i*j*k<max&&i==i*j*k/j/k&&t<2000;k=k*5)
			{
				a[t]=i*j*k;
				t++;
			}
			sort(a,a+t);
			while(cin>>n&&n)
				cout<<a[n-1]<<endl;
			return 0;
}

附0MS.

#include<iostream>
using namespace std;
int main(void)
{
    int n;
    while(cin>>n&&n)
    {
		int a[1505]={0};
		int m2=0,m3=0,m5=0,i,t;
		a[0]=1;
		for(i=1;i<1500;i++)
		{
			if(2*a[m2]>3*a[m3])
				t=a[m3]*3;
			else
				t=a[m2]*2;
			if(t>a[m5]*5)
				t=a[m5]*5;
			
			if(t == 2*a[m2])
				m2++;
			if(t == 3*a[m3]) 
				m3++;
			if(t == 5*a[m5])
				m5++;
			
			a[i]=t;
		}
		cout<<a[n-1]<<endl;
    }
    return 0;
}