题意 :
- 构造一个n(偶数)的全排列,使得全排列的前半部分最小值等于a,后半部分最大值等于b,如果不能构造,则输出-1
思路一 :
- 直接构造,将a放最前,b放最后,中间从大到小放,然后构造后判断是否合法,不合法则直接-1
- *min_element表示最小元素,记得加头文件
- cout << p[i] << " \n"[i == n - 1] 表示只有在i == n - 1才输出\n否则输出空格
#include <iostream>
#include <unordered_map>
#include <algorithm>
#include <vector>
#define pb push_back
#define fi first;
#define se second;
using namespace std;
void solve()
{
int n, a, b;
cin >> n >> a >> b;
vector<int> p{
a};
for (int i = n; i >= 1; i -- )
{
if (i == a || i == b) continue;
p.pb(i);
}
p.pb(b);
if (*min_element(p.begin(), p.begin() + n / 2) == a && *max_element(p.begin() + n / 2, p.end()) == b)
{
for (int i = 0; i < n; i ++ )
cout << p[i] << " \n"[i == n - 1];
}
else
{
cout << -1 << endl;
}
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int _ = 1;
cin >> _;
while (_ -- )
{
solve();
}
return 0;
}
思路二 :
- 就en构造,发现这个做法太累赘了,比赛的时候边界还卡了很久
- 算边界的时候,忘记减去边界中的另一个空点了
#include <iostream>
#include <unordered_map>
#include <vector>
#define pb push_back
#define fi first;
#define se second;
using namespace std;
void solve()
{
int n, a, b;
cin >> n >> a >> b;
if (a > b)
{
if (b != n / 2 || n - a + 1 != n / 2)
{
cout << -1 << endl;
return ;
}
for (int i = a; i <= n; i ++ ) cout << i << ' ';
for (int i = 1; i <= b; i ++ ) cout << i << ' ';
cout << endl;
}
else
{
if (n - a < n / 2 || b - 1 < n / 2)
{
cout << -1 << endl;
return ;
}
for (int i = a; i <= a - n / 2 + b - 1; i ++ ) cout << i << ' ';
for (int i = b + 1; i <= n; i ++ ) cout << i << ' ';
for (int i = 1; i <= a - 1; i ++ ) cout << i << ' ';
for (int i = a - n / 2 + b; i <= b; i ++ ) cout << i << ' ';
cout << endl;
}
}
int main()
{
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int _ = 1;
cin >> _;
while (_ -- )
{
solve();
}
return 0;
}