北京理工大学复试上机--2015

Posted 2021-03-03 ache

1、图形输出：输入 0-9 内的奇数，输出用*组成的正方形中间掏出来一个空的菱形。

我是按照这种输出的

9

*********

**** ****

***   ***

**     **

*       *

**     **

***   ***

**** ****

*********

8

 

#include <iostream>
using namespace std;
int main() {
    int n;
    while(cin >> n) {
        if(n == 0 || n %2 == 0) cout << endl;
        else {
            for(int i = 0; i < n / 2; i++) {
                for(int j = 0; j < n; j++) {
                    if(j > n / 2 - i && j < n / 2 + i) cout << " ";
                    else cout << "*";
                }
                cout << endl;
            }
            for(int i = n / 2; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    if(j > i - n / 2 && j < n + n / 2 - i - 1) cout << " ";
                    else cout << "*";
                }
                cout << endl;
            }
        }
    }
    return 0;
}

 

2、有3个字母a,b,c:你输入一个数字，要输出所有的组合字符和组合数

输入: 1    输出: a,b,c 3

输入: 2    输出: aa,ab,ac,ba,bb,bc,ca,cb,cc 9

#include<iostream>
#include<vector>

using namespace std;

vector<string> perm(vector<string> a, vector<string> b) {
    vector<string> v;
    for(int i = 0; i < a.size(); i++) {
        for(int j = 0; j < b.size(); j++) {
            v.push_back(a[i] + b[j]);
        }
    }
    return v;
}

int main() {
    int n, i, j ,k;
    char ch[3] = {‘a‘, ‘b‘, ‘c‘};
    while (cin >> n) {
        vector<string> v, res;
        v = {"a", "b", "c"};
        res = v;
        for(i = 1; i < n; i++) {
            res = perm(res, v);
        }
        for(i = 0; i < res.size(); i++) {
            cout << res[i];
            if(i < res.size() - 1) cout << ",";
        }
        cout << " " << res.size() << endl;
    }
    return 0;
}

 

3、表达式展开，比如输入a-(b+c),输出a-b-c

输入: a-(b-(((c+d))))

输出: a-b+c+d

#include <iostream>
#include <vector>
#include <stack>
using namespace std;
int main() {
    string s;
    while (cin >> s) {
        int i, l, j = 0;
        int st[100];
        stack<int> ss;
        l = s.length();
        for (i = 0; i < l; i++) {
            if (s[i] == ‘(‘) {
                if (s[i - 1] == ‘-‘) st[j++] = 1;
                else st[j++] = 0;
                ss.push(i);
            }
            else if (s[i] == ‘)‘) {
                if (st[--j] == 1) {
                    int k = ss.top();
                    while (k != i) {
                        if (s[k] == ‘-‘) s[k++] = ‘+‘;
                        else if (s[k] == ‘+‘) s[k++] = ‘-‘;
                        else k++;
                    }
                }
                ss.pop();
            }
            else continue;
        }
        for (i = 0; i < l; i++) {
            if (s[i] == ‘(‘ || s[i] == ‘)‘) continue;
            cout << s[i];
        }
        cout << endl;
    }
    return 0;
}

 

4、求字符串1与字符串2的最大公共子串的长度及此长度最大公共子串的个数。

输入: abcdefg Eebcdfg  (最大公共子串:bcd)    

输出: 3 1

输入: abcdefg abcddefg  (最大公共子串为：abcd defg)

输出: 4 2

#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main() {
    string a, b;
    while(cin >> a >> b) {
        int la = a.length();
        int lb = b.length();
        int maxn = 0, num = 0;
        map<int, vector<string> > mmp;
        string str;
        for(int i = 0; i < la; i++) {
            for(int j = 0; j < lb; j++) {
                if(a[i] == b[j]) {
                    int pi = i, pj = j;
                    while(a[pi] == b[pj] && pj < lb && pi < la) {
                        num++;
                        str += a[pi];
                        pj++;
                        pi++;
                    }
                    if(num >= maxn) maxn = num;
                    mmp[num].push_back(str);
                    str = "";
                    num = 0;
                }
            }
            
        }
        cout << maxn << " " << mmp[maxn].size() << endl;

    }
    return 0;
}