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.
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#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; } |