【UVA - 136】Ugly Numbers（set）

Ugly Numbers

Descriptions:

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, 15, ... 
shows the first 11 ugly numbers. By convention, 1 is included. 
Write a program to find and print the 1500’th ugly number.

Input

There is no input to this program

Output

Output should consist of a single line as shown below, with ‘’ replaced by the number computed. 

Sample Output

The 1500'th ugly number is <number>.

题意

丑数是指不能被2，3，5以外的其他素数整除的数。把丑数从小到大排列起来，结果如下：
1,2,3,4,5,6,8,9,10,12,15……
求第1500个丑数

输入

没有输入

输出

The 1500'th ugly number is <number>.

题目链接：

https://vjudge.net/problem/UVA-136

不难发现2,3,5之后的丑数全是有这三个数乘上{2,3,5}之后得来的，这就好办了，因为不可重复，还需要按顺序来，所以选择<set>是没有问题的，用<set>存入这些丑数，记录 it 的大小，当it为1500-1（数组从0开始计数）时就是我们要找的这个丑数

AC代码：

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define mod 1000000007
#define ll long long
#define INF 0x3f3f3f3f
#define ME0(x) memset(x,0,sizeof(x))
using namespace std;
set<ll> num;//存入丑数
set<ll>::iterator it;
int main()
{
    int a[]= {,,};
    num.insert();
    int sum=;//这里面肯定会有重复的，所以要重新设一个计数器
    it=num.begin();//从0开始计数
    while(sum<-)//当然sum的值可以更大一点，这也没事，主要是去重复
    {
        for(int i=; i<; i++)
            num.insert(*it * a[i]);
        it++;//记录当前为第几个丑数
        sum++;
    }
    printf("The 1500'th ugly number is %llu.\n", *it);
}