POJ 1338 Ugly Numbers （質因數分解）

Ugly Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21307		Accepted: 9513

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

解題思路：

剛開始打表做，直接WA了2次，第1500個醜數，我以爲會是7000000-9000000中的某個數，但是確實錯的，還是用三個變量分別標記目前的位置，從1開始算，

開始分別乘上2,3,5，然後得到了新的數字，再在這些數字中找到最小的一個，如果最小的數字已經在前面的計算中得到了，那麼就讓x,y,z分別進行移動。。。。

 直到打完這個1500長度的表結束。

代碼：

# include<cstdio>
# include<iostream>
# include<algorithm>
# include<cstring>
# include<string>
# include<cmath>
# include<queue>
# include<stack>
# include<set>
# include<map>

using namespace std;

# define inf 999999999

# define MAX 1500+4
# define eps 1e-7

int a[MAX];

void init()
{
    int x = 1;
    int y = 1;
    int z = 1;
    a[1] = 1;
    for ( int i = 2;i <= 1500;i++ )
    {
        int t = min(a[x]*2,a[y]*3);
        a[i] = min(t,a[z]*5);
        if ( a[i]==2*a[x] )
            x++;
        if ( a[i]==3*a[y] )
            y++;
        if ( a[i]==5*a[z] )
            z++;

    }
}


int main(void)
{
    int n;
    init();
    while ( cin>>n )
    {
        if ( n==0 )
            break;

        cout<<a[n]<<endl;

    }


	return 0;
}