题目展示

题目一：从输入字符串中读取数据，计算百内加减法

方法一 C++代码

// 华为机考题目1//https: ///lyxleft/article/details/88698136//Autor: TOMD//Date: 7月24日 15:02:42// 方法一：取出加减号和数值，再分开进行运算#include <iostream>#include <vector>#include <string> //包涵stoi()using namespace std;int main() {cout << "Hello World!\n";string s,s_val;cin >> s;vector<char> operatorExp;vector<int> Val;int len = s.length();int i=0,j=0;//分离数值与运算符while(i<len){if (s[i]<='9' && s[i]>='0') ++i; //读取数值else if (s[i]=='+' || s[i]=='-' || s[i] == '=') //读取运算符{char temp_char = s[i];if ( temp_char != '=') operatorExp.push_back(temp_char);s_val = s.substr(j,i-j); //string substr (起始位置, 字符个数);int temp_int = stoi(s_val); 从第0位开始，将2进制的字符串转化为10进制数字,直接将 stoi(s,0,2)对其cout会报错Val.push_back(temp_int);j = ++i;//下一个必然是数值}}//进行计算len = operatorExp.size();i=0;int result=Val[0];while(i<len){if (operatorExp[i] == '+'){result = result + Val[i+1];}else if (operatorExp[i] == '-'){result = result - Val[i+1];}++i;}cout <<"The calculation is " << s << result << endl;while (1){//让控制窗一直显示}return result;}

题目一：方法二 C++代码

// 华为机考题目1//https: ///lyxleft/article/details/88698136//Autor: TOMD//Date: 7月24日 15:02:42// 方法二：负号直接将负值赋给对应数值再累加#include <iostream>#include <vector>#include <string> //包涵stoi()using namespace std;int main() {cout << "Hello World!"<< endl <<"The expression must end with <=> ! \n";string s,s_val;cin >> s;int len = s.length();int i=0,j=0,sum_val=0,temp_int=0;int flag_sgn=1;//默认为正，检测到负号后变为-1//检测表达式是否输入正确if (s[len-1] != '=') { //检测是否以等号结尾cout << "The input expression is wrong! Please input correctly again!" <<endl;return 0;}while(i<len){if ((s[i]<='9' && s[i]>='0') || s[i]=='+' || s[i]=='-' || s[i] == '=') ++i; //检测是否为数值及加减符号else {cout << "The input expression is wrong! Please input correctly again!" <<endl;return 0;}}i=0;//使用完i后置零，以便下个循环用//提取数值并运算while(i<len){if (s[i]<='9' && s[i]>='0') ++i; //读取数值else if (s[i]=='+' || s[i]=='-' || s[i] == '=') //读取运算符{s_val = s.substr(j,i-j); //string substr (起始位置, 字符个数);temp_int = flag_sgn*stoi(s_val); //用上一次的flag_sgn乘下一个读出来的数值，从第0位开始，将2进制的字符串转化为10进制数字,直接将 stoi(s,0,2)对其cout会报错temp_int = temp_int; //sum_val += temp_int; //累加计数到sum中if ( s[i] == '-') flag_sgn = -1;else if( s[i] == '+') flag_sgn = 1;j = ++i;//移动到下一个数值起始位置，默认下一个必需是数值}}cout <<"The calculation is " << s << sum_val << endl;while (1){//让控制窗一直显示}return sum_val;}

题目二：从输入字符串中读取所有的蛇形字符串，先按长度排序，长度相同的按字典序，输出

测试成功

主要思路

1.找寻所有符合条件的子字符串，并记下长度
2.存入待输出的字符串进string
3.按长度规律输出，次之按字典序
3.1 拷贝将长度排序
3.2 依次找寻对应位置
3.3 输出相应长度字符串

主要代码如下

#include <iostream>#include <vector>#include <string>#include <map>#include <set>#include <algorithm>#include <numeric> //用到求和函数using namespace std;//测试：SxxsrR^AaSs//理论输出：RrSs Aa Ssint main(){cout << "Please input a correct string!"<<endl;string s;cin >> s;// 测试区// string s = "SxxsrR^AaSs"; //已测试通过// string s = "DAFHAGFAHFIUHRjkfhbdajfhkEHFUWEIHAJSDKHncnthasrkjhaejsavbdcnskdscnjkdfhgbvsdfdnfkhgj&*(&%^&*(bdvdhjBSDZVBDFBDVH"; //已测试通过cout << "The input string is " << s << endl;map<char, int> mA, ma;//mA存大写字母，ma存小写字母vector<int> len;int count_num=1; //默认最小子字符串长为1for (auto ss : s){if(ss >= 'A' && ss <= 'Z') mA[ss]++; //ss为关键字if(ss >= 'a' && ss <= 'z') ma[ss]++;}for (auto it = mA.begin(); it != mA.end();){//mA每个关键字对应键值取mA和ma中最小if(ma.count(it->first + 32) == 1) //ascii码中A为65，a为97，相差32{it++;}else mA.erase(it++);//如果ma中没有，则在mA中删除此关键字}/*for (auto x : mA)//遍历mA中的元素cout << x.first << " " << x.second << endl;*/if(mA.empty()){cout << "Not Found" << endl; system("pause"); return 0;}//没有找到string tmp,tmp1;map<char, int>::iterator pre, cur;while (!mA.empty()){pre = mA.begin();cur = mA.begin();if(cur->second == 0){mA.erase(cur++);continue;}tmp += cur->first, tmp += cur->first + 32;cur->second--;//对应键值减一pre = cur, cur++;while (cur != mA.end() && cur->second > 0 && cur->first == pre->first + 1) //找寻连续的串{//继续往下查找的条件是没有到达结尾，当前的迭代器指向的键值不为0，并且与上一个字母是相邻的tmp += cur->first, tmp += cur->first + 32;cur->second--;pre++, cur++,count_num++;}len.push_back(count_num);count_num = 1; //每次查询后count_num置为1//cout << tmp << endl;//tmp.clear();}vector<int> len_sorted(len);vector<int> len_tmp(len);stable_sort(len_sorted.begin(), len_sorted.end());//输出检测出的蛇形字符串cout <<"All Serpentine string are " << tmp << endl;//按长度排序输出int start_pos=0;for (int i=len.size()-1; i>=0; i--){auto len_position = find( len.begin( ), len.end( ), len_sorted[i] ); //找到对应值的位置 len_position为len的迭代器int pos = distance(len.begin( ),len_position);len[pos] = 0; //找到输出一个字符串就消掉len中对应的字符串长度，置为0if (pos == 0) start_pos = 0; //若起点是0位，则无法累加前面的，直接赋值即可else { //若起点是非0位，则应累加前面到第a位以前的所有数值，得到原来字符串的起点位置for (int j=0;j<pos;j++) start_pos=2*len_tmp[j]+start_pos;}//int start_pos = 2*accumulate(len_tmp.begin(),len_position,0); //计算的区间为左闭右开 [len.begin(),len_position)//accumulate(len.begin(),len_position,0) 头两个形参指定要累加的元素范围，第三个形参则是累加的初值cout << tmp.substr(start_pos,2*len_sorted[i]) <<", The length of substring is " << 2*len_sorted[i] <<endl; //输出对应长度的子字符串start_pos=0; //运算完后，起点默认位置归零,进入下一个循环输出字符串}tmp.clear();system("pause"); //暂停系统，即无限循环return 0;}

链接参考

[1] /lyxleft/article/details/88698136

[2] /qq_28584889/article/details/88706373