poj---1338---Ugly Numbers

http://poj.org/problem?id=1338

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22968		Accepted: 10218

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

 

丑数的定义：一个数的素因数只有2， 3，或5.

因此只有这三个数来回的做乘法，按顺序存起来就OL!

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

const int maxn=1502;
const int INF=0x3f3f3f3f;

int ugly[maxn]= {0, 1};


void Init()
{
    int num2=2, num3=3, num5=5;
    int i, j, k, t;
    i=j=k=1;

    for(int x=2; x<maxn; x++)
    {
        t=min(num2, min(num3, num5));
        ugly[x]=t;

        if(t==num2)
            num2=ugly[++i]*2;

        if(t==num3)
            num3=ugly[++j]*3;

        if(t==num5)
            num5=ugly[++k]*5;
    }
}
int main()
{
    Init();
    int n;

    while(scanf("%d", &n), n)
        printf("%d\n", ugly[n]);
    return 0;
}

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<math.h>
#include<queue>
#include<set>
#include<map>
#include<algorithm>
typedef long long LL;
using namespace std;
const int maxn=10010;
const int INF=0x3f3f3f3f;
const int ugly[3]={2, 3, 5};
LL a[maxn];
void init()
{
    priority_queue<LL, vector<LL>, greater<LL> >Q;
    set<LL> s;
    Q.push(1);
    s.insert(1);
    int k=1;
    for(int i=1; ; i++)
    {
        LL x=Q.top();
        Q.pop();
        a[k++]=x;
        if(i==1520)
            break;
        for(int j=0; j<3; j++)
        {
            LL t=x*ugly[j];
            if(!s.count(t))
            {
                s.insert(t);
                Q.push(t);
            }
        }
    }
}
int main()
{
    init();
    int n;
    while(scanf("%d", &n), n)
    {
        printf("%lld\n", a[n]);
    }
    return 0;
}