#include <iostream> #include <string> #include <vector> #include <algorithm>template<typename T> void output(const T &c); void test1(); void test2(); void test3(); int main() {std::cout << "测试1: \n";test1();std::cout << "\n测试2: \n";test2();std::cout << "\n测试3: \n";test3(); }template <typename T> void output(const T &c) {for(auto &i : c)std::cout << i << ' ';std::cout << '\n'; }void test1() {using namespace std;string s0{"0123456789"};cout << "s0 = " << s0 << endl;string s1(s0);reverse(s1.begin(), s1.end()); cout << "s1 = " << s1 << endl;string s2(s0.size(), ' ');reverse_copy(s0.begin(), s0.end(), s2.begin()); cout << "s2 = " << s2 << endl; }void test2() {using namespace std;vector<int> v0{2, 0, 4, 9};cout << "v0: "; output(v0);vector<int> v1{v0};reverse(v1.begin(), v1.end());cout << "v1: "; output(v1);vector<int> v2{v0};reverse_copy(v0.begin(), v0.end(), v2.begin());cout << "v2: "; output(v2); }void test3() {using namespace std;vector<int> v0{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};cout << "v0: "; output(v0);vector<int> v1{v0};rotate(v1.begin(), v1.begin()+1, v1.end()); cout << "v1: "; output(v1);vector<int> v2{v0}; rotate(v2.begin(), v2.begin()+2, v2.end()); cout << "v2: "; output(v2);vector<int> v3{v0};rotate(v3.begin(), v3.end()-1, v3.end()); cout << "v3: "; output(v3);vector<int> v4{v0};rotate(v4.begin(), v4.end()-2, v4.end()); cout << "v4: "; output(v4); }
运行结果：
问题1：reverse的反转操作实在当前数组内进行，会直接改变当前数组的顺序，而reverse_copy是将反转后的数组拷贝到一个新的容器内，不会改变当前数组。
问题2：rotate（参数1，参数2，参数3）函数是一个循环位移函数，参数1和3分别代表要循环的部分的起点和终点，参数二代表循环位移后位于第一个位置的元素。
任务2：代码：
#include <iostream> #include <vector> #include <algorithm> #include <numeric> #include <iomanip> #include <cstdlib> #include <ctime>template<typename T> void output(const T &c);int generate_random_number(); void test1(); void test2();int main() {std::srand(std::time(0)); std::cout << "测试1: \n";test1();std::cout << "测试2: \n";test2(); }template<typename T> void output(const T &c) {for (auto i : c)std::cout << i << ' ';std::cout << '\n'; }int generate_random_number() {return std::rand() % 101; }void test1() {using namespace std;vector<int> v0(10); generate(v0.begin(), v0.end(), generate_random_number); cout << "v0: "; output(v0);vector<int> v1(v0);sort(v1.begin(), v1.end()); cout << "v1: "; output(v1);vector<int> v2(v0);sort(v2.begin()+1, v2.end()-1); cout << "v2: "; output(v2); }void test2() {using namespace std;vector<int> v0(10);generate(v0.begin(), v0.end(), generate_random_number);cout << "v0: "; output(v0);auto max_it = max_element(v0.begin(), v0.end());auto min_it = min_element(v0.begin(), v0.end());cout << "最小值: " << *min_it << endl;cout << "最大值: " << *max_it << endl;auto ans = minmax_element(v0.begin(), v0.end());cout << "最小值: " << *ans.first << endl;cout << "最大值: " << *ans.second << endl;double avg1 = accumulate(v0.begin(), v0.end(), 0.0) / v0.size();cout << "均值: " << fixed << setprecision(2) << avg1 << endl;sort(v0.begin(), v0.end());double avg2 = accumulate(v0.begin() + 1, v0.end() - 1, 0.0) / (v0.size() - 2);cout << "去掉最大值、最小值之后，均值: " << avg2 << endl; }
问题1：通过第三参数来给第一参数和第二参数中间的位置填充数据/元素
问题2：只需要遍历一次容器内的数据元素，节省了时间和空间成本
问题3：效果相同
任务3：代码：
#include <iostream> #include <string> #include <algorithm> #include <cctype>unsigned char func(unsigned char c); void test1(); void test2();int main() {std::cout << "测试1：字符串大小写转换\n";test1();std::cout << "测试2：字符替换\n";test2(); }unsigned char func(unsigned char c) {if (c == '1')return 'a';if (c == '2')return 'A';if (std::isalpha(c))return static_cast<unsigned char>(c + 1);return c; }void test1() {std::string s = "Hello World 2048!";std::cout << "s = " << s << "\n";std::string s1;for (char c : s)s1 += std::tolower(c);std::cout << "s1 = " << s1 << "\n";std::string s2;for (char c : s)s2 += std::toupper(c);std::cout << "s2 = " << s2 << "\n"; }void test2() {std::string s1 = "I love C++!";std::cout << "s1 = " << s1 << "\n";std::string s2(s1.size(), '*');std::transform(s1.begin(), s1.end(), s2.begin(), func);std::cout << "s2 = " << s2 << "\n"; }
问题1：把字符 1 转换为a，把字符 2 转换为A，把字母字符转换为其在字母表中后一个位置的字母。
问题2：将指定内容的大写字母全部转换为小写；将指定内容的小写字母全部转换为大写。
问题3：1：指定区域的起始位置，2：指定区域的结束位置，3：生成的数据开始存储的位置，4：数据转化/处理规则
任务4：代码：
#include <iostream> #include <string> #include <algorithm>bool is_palindrome(const std::string &s); bool is_palindrome_ignore_case(const std::string &s);int main() {using namespace std;string s;// 多组输入，直到按下Ctrl+Z结束测试while(cin >> s) {cout << boolalpha<< "区分大小写: " << is_palindrome(s) << "\n"<< "不区分大小写: " << is_palindrome_ignore_case(s) << "\n\n";} }bool is_palindrome(const std::string &s){int left=0 ,right=s.size()-1 ;while(left<right){if(s[left]!=s[right]){return false ;}left++ ;right-- ;}return true ; }bool is_palindrome_ignore_case(const std::string &s){std::string s0 ;int left=0 ,right=s.size()-1 ;for(char c : s){s0 += tolower(c) ;}while(left<right){if(s0[left]!=s0[right]){return false ;}left++ ;right-- ;}return true ; }
思考：
任务5：代码：
#include <iostream> #include <string> #include <algorithm>std::string dec2n(int x, int n = 2);int main() {int x;while(std::cin >> x) {std::cout << "十进制: " << x << '\n'<< "二进制: " << dec2n(x) << '\n'<< "八进制: " << dec2n(x, 8) << '\n'<< "十二进制: " << dec2n(x, 12) << '\n'<< "十六进制: " << dec2n(x, 16) << '\n'<< "三十二进制: " << dec2n(x, 32) << "\n\n";} }std::string dec2n(int x, int n){std::string result ;char str ;if(x == 0){return "0" ;}while(x>0){int remainder = x % n ;if( remainder < 10 ){str = '0' + remainder ;}else{str = 'A' + remainder - 10 ;}result.insert(0 , 1, str) ;x = x/n ;}return result ; }
任务6：代码：
#include<iostream> #include<string> #include<cctype> #include<iomanip> #include<algorithm> using namespace std ;int main(){string str = " a b c d e f g h i j k l m n o p q r s t u v w x y z" ;cout << " " << str << endl ;for (char &c : str) {if (c != ' ') {c = toupper(c);}}for(int i = 1 ; i <= 26 ; i++){rotate(str.begin(), str.begin() + 2, str.end()) ; cout << setw(2) << i << str << endl ;}return 0 ; }
任务7：代码：
#include<iostream> #include<cstdlib> #include<ctime> #include<iomanip> using namespace std ;int main() {srand(static_cast<unsigned int>(time(0))) ;int correctCount = 0 ;for (int i = 0; i < 10; i++) {int num1 = rand() % 10 + 1 ;int num2 = rand() % 10 + 1 ;int op = rand() % 4 ;int userAnswer, correctAnswer ;char operation ;if (op == 1) {if (num1 < num2) {swap(num1, num2) ;}operation = '-' ;correctAnswer = num1 - num2 ;} else if (op == 3) {do {num1 = rand() % 10 + 1 ;} while (num1 % num2 != 0) ;operation = '/' ;correctAnswer = num1 / num2 ;} else if (op == 0) {operation = '+' ;correctAnswer = num1 + num2 ;} else {operation = '*' ;correctAnswer = num1 * num2 ;}cout << num1 << " " << operation << " " << num2 << " = " ;cin >> userAnswer ;if (userAnswer == correctAnswer) {correctCount++ ;}}double accuracy = static_cast<double>(correctCount) * 100 / 10 ;cout << "正确率: " << fixed << setprecision(1) << accuracy << "%" << endl ;return 0 ; }
