题目

Atcoder

思路

代码

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110;
int n, m, A[N], B[N], t[N];
int main() {
    cin >> n >> m;
    int cnt = 0;
    t[1] = 1, t[2] = m;
    for (int i = 1; i <= m && cin >> A[i]; i++)
        if (A[i] & 1) t[++cnt] = i; // 找奇数
    if (cnt > 2) return cout << "Impossible" << endl, 0; // 判无解
    if (m == 1 && A[1] == 1) return cout << "1\n1\n1\n" << endl, 0;
    if (m == 1) return cout << A[1] << "\n2\n" << A[1] - 1 << ' ' << 1 << endl, 0; 
    // 把奇数放到头尾
    swap(A[1], A[t[1]]), swap(A[m], A[t[2]]);
    for (int i = 1; i <= m; i++) cout << A[i] << ' ';
    cout << endl << (A[m] == 1 ? m - 1 : m) << endl << A[1] + 1 << ' '; // 判断一下个数
    for (int i = 2; i <= m - 1; i++) cout << A[i] << ' ';
    if (A[m] > 1) cout << A[m] - 1 << endl; // 输出
    return 0;
}