第 45 届国际大学生程序设计竞赛（ICPC）亚洲区域赛（南京）K Co-prime Permutation

题目

Kotori is very good at math (really?) and she loves playing with permutations and primes.

One day, she thinks of a special kind of permutation named kk co-prime permutation. A permutation (p_1,p_2,cdots,p_n)

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of nn is called a kk co-prime permutation of nn if there exists exactly kk integers ii such that (1 le i le n1≤i≤n ,and ext{gcd}(p_i,i)=1,gcd(p i,i)=1, where, ext{gcd}(x,y)gcd(x,y)) indicates the greatest common divisor of xx and yy.

Given nn and kk, please help Kotori construct a kk co-prime permutation of nn or just report that there is no such permutation.

Recall that a permutation of nn is a sequence of length nn containing all integers from 11 to nn.
输入描述:
There is only one test case in each test file.

The first and only line contains two integers nn and kk ((1 le n le 10^6,1≤n≤10^6 , 0 le k le n ,0≤k≤n).)
输出描述:
Output one line containing nn integers (p_1, p_2, cdots, p_n)

separated by one space, indicating the permutation satisfying the given constraints. If no such permutation exists output "-1" (without quotes) instead. If there are multiple valid answers you can print any of them.

Please, DO NOT output extra spaces at the end of each line, otherwise your answer may be considered incorrect!
示例1
输入
5 3
输出
1 4 5 2 3
示例2
输入
1 0
输出
-1

题意

给定一个长度为n的数列和k，让求出一个(gcd(a[i],i)=1) 个数等于k的排列

思路

尝试多写几组样例可得
5 0
-1
5 1
1 2 3 4 5
1 2 3 4 5

5 2
2 1 3 4 5
1 2 3 4 5

5 3
1 2 3 5 4
1 2 3 4 5

5 4
2 1 4 3 5
1 2 3 4 5

5 5
5 1 2 3 4
1 2 3 4 5

k=0的时候无解，因为1和1互质
其他情况都有解
k是奇数的时候从非1的数里面两两调换，这样就可以多出来偶数对(gcd(a[i],i)=1)
k是奇数的时候从1开始的数里面两两调换，这样就可以多出来奇数对(gcd(a[i],i)=1)

代码

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1000000+100;
int a[maxn];
int main()
{
	int n,k;
	cin>>n>>k;
	for(int i=1;i<=n;i++) a[i] = i;
	if(k==0) return cout<<-1,0;
	else
	{
		if(k==1)
		{
			for(int i=1;i<=n;i++)
			cout<<i<<" ";
		}
		else if(k==n)
		{
			cout<<n<<" ";
			for(int i=1;i<=n-1;i++)
			cout<<i<<" ";
		}
		else
		{
			if(k%2==0)
			{
				int now = k/2;
				int f = 1;
				int cnt = 0;
				for(int i=1;i<=n;i++)
				{
					if(now)
					{
						cnt++;
						if(f)
						cout<<i+1<<" " , f = !f; 
						else
						cout<<i-1<<" " , f = !f;
						if(cnt==2) now-- , cnt = 0;
					}
					else
					cout<<i<<" ";
				}	
			}
			else
			{
				a[1] = 1;
				for(int i=2;i<=k;i+=2)
				{
					a[i] = i+1;
					a[i+1] = i;
				}
				for(int i=1;i<=n;i++) cout<<a[i]<<" ";
			}
		}
	} 
}

/*
5 0 
-1
5 1
1 2 3 4 5
1 2 3 4 5

5 2
2 1 3 4 5
1 2 3 4 5

5 3
1 2 3 5 4
1 2 3 4 5 

5 4
2 1 4 3 5
1 2 3 4 5

5 5
5 1 2 3 4
1 2 3 4 5
*/