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Ugly Numbers

2024-07-11 22:57:10   223人阅读

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



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;
}

#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;
}







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