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  • 北京理工大学复试上机--2012

    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
    #include <iostream>
    #include <vector>
    #include <algorithm>
    using namespace std;
    int main()
    {
        int a, i;
        vector<int> v;
        for (i = 0; i < 10; i++)
        {
            scanf("%d", &a);
            v.push_back(a);
        }
        sort(v.begin(), v.end());
        for (i = 0; i < 10; i++)
        {
            cout << v[i];
            if (i < 9)
                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)
    #include <iostream>
    #include <vector>
    #include <algorithm>
    using namespace std;
    
    struct student
    {
        int age, id;
        string name, gender;
    };
    
    bool cmp(student t1, student t2)
    {
        if (t1.id != t2.id)
            return t1.id > t2.id;
    }
    
    int main()
    {
        int a, i, j;
        string s;
        student stu;
        vector<student> v;
        student st[3];
        st[0].id = 10; st[0].name = "wes"; st[0].gender = "f"; st[0].age = 23;
        st[1].id = 20; st[1].name = "erf"; st[1].gender = "f"; st[1].age = 45;
        st[2].id = 30; st[2].name = "str"; st[2].gender = "t"; st[2].age = 89;
        v.push_back(st[0]);
        v.push_back(st[1]);
        v.push_back(st[2]);
        while (getline(cin, s)) {
            string str[4];
            if (s[0] == 'I') {
                int n = (s[1] - '0') * 10 + (s[2] - '0');
                stu.id = n;
                n = s[s.length() - 1] - '0' + (s[s.length() - 2] - '0') * 10;
                stu.age = n;
                string ss = "";
                for (i = 4; i < s.length(); i++) {
                    if (s[i] == ',') {
                        stu.name = ss;
                        ss = "";
                        break;
                    }
                    ss += s[i];
                }
                for (j = i + 1; j < s.length(); j++) {
                    if (s[j] == ',') {
                        stu.gender = ss;
                        break;
                    }
                    ss += s[j];
                }
                v.push_back(stu);
                sort(v.begin(), v.end(), cmp);
                for (i = 0; i < v.size(); i++) {
                    cout << "(" << v[i].id << "," << v[i].name << "," << v[i].gender << "," << v[i].age << ")";
                    if (i < v.size() - 1)
                        cout << ",";
                }
            }
            if (s[0] == 'D') {
                int n = (s[1] - '0') * 10 + (s[2] - '0');
                for (i = 0; i < v.size(); i++) {
                    if (v[i].id == n)
                    {
                        v.erase(v.begin() + i);
                        break;
                    }
                }
                sort(v.begin(), v.end(), cmp);
                for (i = 0; i < v.size(); i++) {
                    cout << "(" << v[i].id << "," << v[i].name << "," << v[i].gender << "," << v[i].age << ")";
                    if (i < v.size() - 1)
                        cout << ",";
                }
            }
        }
    
        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
     
    这个题目前还没完全搞清楚,先空着
     
     
    PS:突入其来的又不会的题,甚是心慌!!!  要踏实一点了,加油!
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  • 原文地址:https://www.cnblogs.com/ache/p/12571658.html
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