POJ 1338 Ugly Numbers

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

分析:

这个题的关键是模拟出来规律，然后用筛法离线求出1500个丑数。

一开始用了一种比较慢的算法模拟了一下

代码如下

//第一时间想到的低效算法
#include <iostream>
#include <cstring>
#include <string>

using namespace std;

int ans[1505];
const int tt[3]={2,3,5};

bool ugly(int x)
{
while(x!=1)
{
if(0==x%2)
x/=2;
else if(0==x%3)
x/=3;
else if(0==x%5)
x/=5;
else
return false;
}

return true;
}

int main()
{
int n,i,k;
k=0;
ans[k]=1;
for(i=2;k<1500;i++)
{
if(ugly(i))
ans[++k]=i;
}

while(cin>>n, n>0)
{
cout<<ans[n-1]<<endl;
}
return 0;
}

之后用筛法模拟出1500个丑数AC的

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

int ans[1505];

int main()
{
int n,i,j,k,t;

memset(ans,0,sizeof(ans));
ans[1]=1;
i=1,j=1,k=1;
for(t=2; t<=1500; t++)
{
ans[t]=min(min(ans[i]*2,ans[j]*3),ans[k]*5);
if(ans[i]*2==ans[t])
++i;
if(ans[j]*3==ans[t])
++j;
if(ans[k]*5==ans[t])
++k;
}

while(cin>>n, n>0)
cout<<ans
<<endl;

return 0;
}