郑州大学2022-2023第一学期算法设计与分析-实验3

1 插座问题

#include
using namespace std;

signed main() 
{
	int n, k; cin >> n >> k;
	int t = (n - 1) / (k - 1);
	cout << (t * (k - 1) == n - 1 ? t : t + 1) << '\n';
}
1
2
3
4
5
6
7
8
9

2 办事大厅排队

#include
#include
#include
using namespace std;

signed main() 
{
	int T; cin >> T;
	queue<string> q;
	string op, name;
	while(T --)
	{
		cin >> op;
		if(op[0] == 'i') cin >> name, q.push(name);
		else if(op[0] == 'o') q.pop();
		else if(op[0] == 'q')
		{
			if(q.empty()) cout << "NULL" << '\n';
			else cout << q.front() << '\n';
		}
	}
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

3 字符串的全排列

#include
#include
#include
using namespace std;

signed main() 
{
	string s; cin >> s;
	sort(s.begin(), s.end());
	
	cout << s << '\n';
	while(next_permutation(s.begin(), s.end())) 
		cout << s << '\n';
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14

4 集合的“交”与“并”

#include 
#include 
#include 
#include 
#include 
using namespace std;

signed main() 
{
	int n, m, x; cin >> n >> m;
	unordered_set<int> s1, s2;
	vector<int> vec1, vec2;
	
	for(int i = 1; i <= n; i ++)
		cin >> x, s1.insert(x);
	for(int i = 1; i <= m; i ++)
		cin >> x, s2.insert(x);
	
	for(auto x : s1)
		if(s2.count(x)) vec1.push_back(x);
		else vec2.push_back(x);
	
	for(auto x : s2) vec2.push_back(x);
	
	sort(vec1.begin(),vec1.end());
	sort(vec2.begin(),vec2.end());
	
	cout << vec1.size();
	for(auto x : vec1) cout << " " << x;
	cout << '\n';
	cout << vec2.size();
	for(auto x : vec2) cout << " " << x;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33

5 统计英文单词个数

#include 
#include 
#include 
#include 
using namespace std;

void solve()
{
	map<string,int> mp;
	int n; cin >> n;
	string s;
	while(n --) cin >> s, mp[s] ++;
	for(auto [x, c] : mp) cout << x << " " << c << '\n';
}

signed main() 
{
	int T; cin >> T;
	while(T --) solve();
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

6 括号匹配

#include 
#include 
#include 
#include 
using namespace std;
void solve()
{
    string s; cin >> s;
	stack<char> stk;
	for(int i = 0; i < s.size(); i ++)
	{
		if(s[i] == '(' || s[i] == '[' || s[i] == '{') stk.push(s[i]);
		else
		{
			if(stk.empty())  stk.push('#');
			if(stk.top() == '(' && s[i] == ')' || stk.top() == '[' && s[i] == ']' || stk.top() == '{' && s[i] == '}') 
			    stk.pop();
			else stk.push('#');
		}
	}
	if(stk.empty()) cout << "Yes" << '\n';
	else cout << "No" << '\n';
}

signed main() 
{
	int T; cin >> T;
	while(T --) solve();
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29

7 后续+中序序列构造二叉树

#include 
#include 
#include 
#include 
using namespace std;

string post, in, pre = "";
unordered_map<char,int>pos;
void dfs(int il, int ir, int pl, int pr)
{
	if(pl > pr) return ;
	int k = pos[post[pr]];
	pre += post[pr];
	dfs(il, k - 1, pl, pl + k - 1 - il);
	dfs(k + 1, ir, pl + k - 1 - il + 1, pr - 1);
}
signed main() 
{
	int n; cin >> n;
	cin >> post >> in; 
	for(int i = 0; i < n; i ++) pos[in[i]] = i;
	dfs(0, n - 1, 0, n - 1);
	cout << pre << '\n';
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24

8 找第 k 小的数

#include 
using namespace std;
const int N = 100010;
int a[N];
int find(int l, int r, int k)
{
    if(l >= r) return a[l];
    
    int i = l - 1, j = r + 1, v = a[l + r >> 1];
    while(i < j)
    {
        do i ++; while(a[i] < v);
        do j --; while(a[j] > v);
        if(i < j) swap(a[i], a[j]);
    }
    
    if(j - l + 1 >= k) return find(l, j, k);
    else return find(j + 1, r, k - (j - l + 1));
}

signed main()
{
    int n, k; cin >> n >> k;
    for(int i = 0; i < n; i ++) cin >> a[i];
    cout << find(0, n - 1, k) << '\n';
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

9 求逆序对数目

#include 
using namespace std;
#define int long long
const int N = 100010;
int a[N], tmp[N];
int res = 0;
void merge(int l, int r)
{
	if(l >= r) return ;
	int mid = l + r >> 1;
	merge(l, mid);
	merge(mid + 1, r);
	int i = l, j = mid + 1, k = 0;
	while(i <= mid && j <= r)
	{
		if(a[i] <= a[j]) tmp[k ++] = a[i ++];
		else
		{
			res += mid - i + 1;
			tmp[k ++] = a[j ++];
		}
	}
	while(i <= mid) tmp[k ++] = a[i ++];
	while(j <= r) tmp[k ++] = a[j ++];
	for(int i = l, k = 0; i <= r; i ++, k ++) a[i] = tmp[k];
}
signed main()
{
    int n; cin >> n;
    for(int i = 0; i < n; i ++) cin >> a[i];
    merge(0, n - 1);
    cout << res << '\n';
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34