Codeforce - 361 - B. Levko and Permutation

B. Levko and Permutation
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, … , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input
The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output
In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples
input
4 2
output
2 4 3 1
input
1 1
output
-1
Note
In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.

題意：有 n 個數，從 1 - n 排序，若其數值與位置的最大公因數大於 1 ，則爲 good ，現要求這串數中只能有 k 個 good 的數。

AC代碼思路：易知，相鄰兩個數的最大公因數只能是 1 。所以，調換兩相鄰數的位置，可以減少 2 個 good 。三點注意：1) 當 k 爲單數時， n-1 也爲單數時，和 n-1 爲雙數時。
2) 當 k 爲雙數時， n-1 爲單數，和 n-1 爲雙數。3) n-1

#include <iostream>
using namespace std;

int main()
{
    int n,k;
    cin>>n>>k;
    if (n-1<k) return cout<<-1,0;
    int tm=(n-1-k)/2;
    if (!k)
    {
        if ((n-1)%2)
        {
            cout<<n<<" ";
            for (int i=2;i<=n-1;i++)
            {
                cout<<i+1<<" "<<i<<" ";i++;
            }
            cout<<1;
        }
        else
        {
            cout<<1<<" ";
            for (int i=2;i<=n;i++)
            {
                cout<<i+1<<" "<<i<<" ";i++;
            }
        }
    }
    else {
    if ((n-1)%2)
    {
        if (k%2)
        {
            cout<<1<<" ";
            for (int i=2;i<=n-1;i++)
            {
                if (tm>0) {cout<<i+1<<" "<<i<<" ";i++;tm--;}
                else cout<<i<<" ";
            }
            cout<<n;
        }
        else
        {
            cout<<n<<" ";
            for (int i=2;i<=n-1;i++)
            {
                if (tm>0) {cout<<i+1<<" "<<i<<" ";i++;tm--;}
                else cout<<i<<" ";
            }
            cout<<1;
        }
    }
    else
    {
        if (k%2)
        {
            cout<<n<<" ";
            for (int i=2;i<=n-1;i++)
            {
                if (tm>0) {cout<<i+1<<" "<<i<<" ";i++;tm--;}
                else cout<<i<<" ";
            }
            cout<<1;
        }
        else
        {
            cout<<1<<" ";
            for (int i=2;i<=n-1;i++)
            {
                if (tm>0) {cout<<i+1<<" "<<i<<" ";i++;tm--;}
                else cout<<i<<" ";
            }
            cout<<n;
        }
    }
    }
}