北理工计算机考研复试上机2012年真题

1、输入十个正整数数字从小到大排序

输入:1 2 5 7 9 10 45 67 24 26

输出:1,2,5,7,9,10,24,26,45,67

代码：

#include 

using namespace std;
vector a;
vector tmp(100);

void merge_sort(int l, int r) {
	if (l >= r) return;
	int mid = l + r >> 1;
	merge_sort(l, mid);
	merge_sort(mid+1, r);
	int k = 0, i = l, j = mid+1;
	while (i <= mid && j <= r) {
		if (a[i] < a[j]) tmp[k++] = a[i++];
		else 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]; 
}
void quick_sort(int l, int r) {
	if (l >= r) return;
	int mid = l + r >> 1, i = l-1, j = r+1;
	while (i < j) {
		do i++; while (a[i] < a[mid]);
		do j--; while (a[j] > a[mid]);
		if (i < j) swap(a[i], a[j]);
	}
	quick_sort(l, j);
	quick_sort(j+1,r);
}
int main()
{ 
	for (int i = 0; i < 10; i++) { 
		int num;
		scanf("%d,",&num);
		a.push_back(num);
	} 

	merge_sort(0, 9);
	for (auto it = a.begin(); it != a.end(); it++) {
		cout << *it;
		if (it != a.end() - 1) cout << ',';
	} 
	return 0;
}

2、学生有(学号，姓名，性别，年龄)，初始化三个学生的信息

(10, wes, f, 23) (20, ert, f, 45) (30, str, t, 89) ,然后对学生信息进行插入和删

除处理

例如: I12,rt, f, 67表示插入12,rt, f, 67

D10表示删除学号为10的学生的信息

每次操作完成以后输出所有学生的信息按学号从大到小排序

输入: I12,rt,f,67

输出: (30,str,t,89),(20,erf,f,45),(12,rt,f,67),(10,wes,f,23)

输入: D10

输出: (30,str,t,89),(20,erf,f,45),(12,rt,f,67)

代码：

#include 

using namespace std;

struct student {
	string sno;
	string name;
	string gender;
	int age;
};
vector stu;

bool del(string sno) {
	for (auto it = stu.begin(); it != stu.end(); it++) {
		if (it->sno == sno) {
			stu.erase(it);
			return true;
		}
	}
	return false;
}
bool cmp(student s1, student s2) {
	return s1.sno > s2.sno;
}
int main()
{
	stu.push_back({"10","wes","f",23});
	stu.push_back({"20","ert","f",45});
	stu.push_back({"30","str","t",89});
	
	string s;
	while (1) {
		cout << "输入：";
		getline(cin, s);
		if (s == "#") break;
		if (s[0] == 'I') {
			s = s.substr(1);
			// sno
			int idx = s.find(',');
			string sno = s.substr(0, idx);
			s = s.substr(idx+1);
			// name
			idx = s.find(',');
			string name = s.substr(0, idx);
			s = s.substr(idx+1);
			// gender
			idx = s.find(',');
			string gender = s.substr(0,idx);
			s = s.substr(idx+1);
			// age
			int age = stoi(s);
			stu.push_back({sno, name, gender, age});
		} else if (s[0] == 'D') {
			s = s.substr(1);
			if (!del(s)) {
				cout << "无此人信息，无法删除。" << endl; 
			}
		} else cout << "输入有误。" << endl; 
		// 输出
		sort(stu.begin(), stu.end(), cmp);
		cout << "输出:"; 
		for (auto it = stu.begin(); it != stu.end(); it++) {
			cout << '(' << it->sno << ',' << it->name <<','
			<< it->gender <<','<< it->age << ')';
			if (it+1 != stu.end()) cout << ',';
		} 
		cout << endl;
	}
	
	return 0;
}

3、利用后序和中序确定前序遍历结果

示例:

输入(按后序、中序): CHBEDA CBHADE

输出: ABCHDE

注：推荐一篇力扣题解，非常棒。力扣题解

代码：

#include

using namespace std;
map m;
string pre, in, post;

struct Treenode{
    char val;
    Treenode* left;
    Treenode* right;
    Treenode(){}
    Treenode(char val):val(val),left(nullptr), right(nullptr)
    {}
};

Treenode* preAndIn(int ps, int pe, int is, int ie, string pre) {
    if (ps > pe || is > ie) return nullptr;
    Treenode* cur = new Treenode(pre[ps]);
    int ri = m[pre[ps]];
    cur->left = preAndIn(ps+1, ps+ri-is, is, ri-1, pre);
    cur->right = preAndIn(ps+ri-is+1, pe, ri+1, ie, pre);
    return cur;
}

void postcout(Treenode* root) {
    if (root == nullptr) return ;
    postcout(root->left);
    postcout(root->right);
    cout << root->val;
}

Treenode* postAndIn(int is, int ie, int ps, int pe, string post) {
    if (is > ie || ps > pe) return nullptr;
    Treenode* cur = new Treenode(post[pe]);
    int ri = m[post[pe]];
    cur->left = postAndIn(is, ri-1, ps, ps+ri-is-1, post);
    cur->right = postAndIn(ri+1, ie, ri+ps-is, pe-1, post);
    return cur;
}

void precout(Treenode* root) {
    if (root == nullptr) return ;
    cout << root->val;
    precout(root->left);
    precout(root->right);
}
int main()
{
    // 1、输入前序中序，给出后序
    // cout << "请输入前序：";
    // cin >> pre;
    // cout << "请输入中序：";
    // cin >> in;
    // for (int i = 0; i < in.size(); i++) m[in[i]] = i;
    // Treenode* root = preAndIn(0, in.size()-1, 0, in.size()-1, pre);
    // // 遍历
    // cout << "后序遍历结果：";
    // postcout(root);

    // 2、输入中序后序，给出前序
    cout << "请输入中序：";
    cin >> in;
    cout << "请输入后序：";
    cin >> post;
    for (int i = 0; i < in.size(); i++) m[in[i]] = i;
    Treenode* root = postAndIn(0, in.size()-1, 0, in.size()-1, post);
    cout << "前序遍历结果：";
    precout(root);
}