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原创于 2017-03-13 13:02:27 发布 · 249 阅读

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本文介绍了一道算法题目，要求找出第n个是3或5的倍数的美丽数。通过分析，发现每7个数会出现一次循环，并提供了一个简单的C++实现代码。

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Description

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

题意：

找出第n个是3或5的倍数

分析：

7个一循环，n/7就是至少多少15再加上第n%7数就是第n个数的值

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long int a[200000],n;
    a[1]=3;
    a[2]=5;
    a[3]=6;
    a[4]=9;
    a[5]=10;
    a[6]=12;
    a[7]=15;
    while(cin>>n)
    {
        a[n]=n/7*15+a[n%7];
        cout<<a[n]<<endl;
    }
}

感受：

就是一个水题