Beautiful Number

最新推荐文章于 2022-02-20 23:44:40 发布

weixin_30294021

最新推荐文章于 2022-02-20 23:44:40 发布

阅读量336

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7125843.html

本文介绍了一道算法题——求第N个美丽数。美丽数定义为能被3或5整除的正数。文章提供了完整的C++代码实现，并采用预处理方法，通过map映射来快速获取第N个美丽数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2829

Beautiful Number

Time Limit: 2 Seconds Memory Limit: 65536 KB

Mike is very lucky, as he has two beautiful numbers, 3 and 5. But he is so greedy that he wants infinite beautiful numbers. So he declares that any positive number which is dividable by 3 or 5 is beautiful number. Given you an integer N (1 <= N <= 100000), could you please tell mike the Nth beautiful number?

Input

The input consists of one or more test cases. For each test case, there is a single line containing an integer N.

Output

For each test case in the input, output the result on a line by itself.

Sample Input

1
2
3
4

Sample Output

3
5
6
9

题目思路，刚开始我以为只有能被3或5整除的数为beautiful number，英语翻译过来后是只要能被3或5整除的数为beautiful number，那么就简单了,

预处理一下，map映射随便做。注意，最大是第100000个数，而不是beautiful number最大是100000。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
using namespace std;
typedef long long ll;
map<ll,ll>m;
void getbeautifulnumber()
{
    int p=1;
    for(int i=1; ;i++)
    {
        if(p>100000) break;
        if(i%3==0 || i%5==0)
            m[p++]=i;
    }
}
int main()
{
    int n;
    getbeautifulnumber();
    while(scanf("%d",&n)!=EOF)
    {
        printf("%d\n",m[n]);
    }
    return 0;
}