XVII Open Cup named after E.V. Pankratiev. Eastern Grand Prix. Problem F. Buddy Numbers 贪心数论构造

Posted 2022-12-15 ProLightsfxjh

XVII Open Cup named after E.V. Pankratiev. Eastern Grand Prix.

Problem F. Buddy Numbers
Input le: standard input
Output le: standard output
Time limit: 1 second
Memory limit: 256 mebibytes

Polycarp says that two positive integers are buddies, if one is divisible by another. For example, 2 and 4
are buddies, and 10 and 3 are not. Note that 1 is buddy with every positive integer.
Given integer n nd out if it is possible to arrange all consecutive integers from 1 to n in a row such that
any pair of neighbor numbers are buddies. If it is possible, print any of those arrangements.

Input
The only line of the input contains one integer n (1 n 1000).

Output
If it is impossible to arrange numbers from 1 to n in such a way, printany of them.

Examples
standard input standard output
2                                             2 1
3                                             3 1 2

Source

XVII Open Cup named after E.V. Pankratiev. Eastern Grand Prix.

My Solution

题意：给出一个n，问能不能把1~n这n个数排出一个序列，使得任意相邻的两个数，一个数是另一个数的因数。

贪心、数论、构造

1   1

2   2  1

3   3 1 2

4   3 1 2 4

然后5的时候，也就是当1~n里素数的个数大于2的时候就不好办了，

此外这里n == 6的时候还是可以构造出来了， 3 6 2 4 1 5.

所以对于n == 1、2、3、4、6的时候是有答案的，直接输出即可，其它的时候都构造不出来，为-1。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
LL n;
LL a[100][100];
int main () 
    int n;
    cin>>n;
    if(n==1)printf("1");
    else
    if(n==2)printf("1 2");
    else
    if(n==3)printf("3 1 2");
    else
    if(n==4)printf("3 1 2 4");
    else
    if(n==6)printf("5 1 3 6 2 4");
    else printf("-1");

  Thank you！

                                                                                                                                             ------from ProLights