Codeforces Round #728 (Div. 2)A. Pretty Permutations

给n个数字, 从1开始到n, 求距离最小的每个数都不在自己位置上的数组.

因为要距离最小, 所有前后swap一下即可. 注意如果是奇数,要多swap一下.

#include "bits/stdc++.h"
using namespace std;   
const auto fr = [](){
    std::ios_base::sync_with_stdio(0); std::cin.tie(0);
    std::cout << std::fixed << std::setprecision(12);
    return 1;
}();

template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v);
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.first << ", " << p.second << ")"; };
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
	cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) {
	cin >> p.first;
	return cin >> p.second;
}

// vars:
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vi = std::vector<int>;
using vl = std::vector<ll>;
using vvi = std::vector<vi>;
using vvl = std::vector<vl>;
using pii = std::pair<int,int>;
using pil = std::pair<int,ll>;
using pli = std::pair<ll,int>;
using pll = std::pair<ll,ll>;
using vpii = std::vector<pii>;
using vvpii = std::vector<vpii>;
int main() { 
    fr; 
    int T;
    cin >> T;
    while (T --)
    {
        int N;
        cin >> N;
        int A[N];
        for (int i = 1; i <= N; i++)
        {
            A[i - 1] = i;
        }
        for (int i = 1; i < N; i+=2)
        {
            swap(A[i],A[i - 1]);
        }   
        if(N % 2 != 0)
        {
            swap(A[N - 1], A[N - 2]);
        }
        for (int i = 0; i < N; i++)
        {
            cout << A[i] << " ";
        }
        cout << endl;
        
    }
   
    return 0; 
}