POJ1338 Ugly Numbers 堆优化+模拟

1.题目描述：

Ugly Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 23957 Accepted: 10597

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

1290

Sample Output

1210

Source

New Zealand 1990 Division I,UVA 136

2.题意概述：

定义丑数是质因子只有2、3、5的数，简单说就是可以写成2^a*3^b^5^c（a,b,c>=0）要你求第n个丑数

3.解题思路：

可以考虑用一个小顶堆去维护丑数，一个数组来记录前N个丑数，每次拿出堆顶，判断是否重复，然后在分别乘以二、三、五放回堆中，为了避免重复计算，预处理一下再查询，这样查询就是线性的196K0MS

4.AC代码：

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <functional>#include <cmath>#include <vector>#include <queue>#include <deque>#include <stack>#include <map>#include <set>#include <ctime>#define INF 0x7fffffff#define maxn 100100#define N 1555#define eps 1e-6#define pi acos(-1.0)#define e 2.718281828459#define mod (int)1e9 + 7using namespace std;typedef long long ll;ll a[N];void solve(){  int cnt = 1;  a[0] = INF;  priority_queue<ll, vector<ll>, greater<ll>> heap;  heap.push(1);  while (cnt <= 1500)  {    ll temp = heap.top();    heap.pop();    if (temp == a[cnt - 1])      continue;    a[cnt++] = temp;    heap.push(temp * 2);    heap.push(temp * 3);    heap.push(temp * 5);  }}int main(){#ifndef ONLINE_JUDGE  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout);  long _begin_time = clock();#endif  int n;  solve();  while (~scanf("%d", &n), n)    printf("%lld\n", a[n]);#ifndef ONLINE_JUDGE  long _end_time = clock();  printf("time = %ld ms.", _end_time - _begin_time);#endif  return 0;}