实验1 现代c++编程
任务1
代码:
#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); }
截图:

问题1:reverse是原地翻转;reverse_copy是翻转并拷贝到别处,源不变。
问题2:rotate改变元素顺序:元素变化规则(左旋)把区间切成两段:[first, middle) 和 [middle, last)旋转之后顺序 = [middle, last) + [first, middle)。rotate的三个参数:1.first:整个待旋转区间的起点 2.middle:旋转完成后,middle 指向的元素成为整个区间新的第一个元素 3.last:整个待旋转区间的末尾(不包含该位置)
任务2:
代码:
#include <iostream> #include <vector> #include <algorithm> #include <numeric> #include <iomanip> #include <cstdlib> #include <ctime> // 模板函数声明 template<typename T> void output(const T &c); int generate_random_number(); void test1(); void test2(); int main() { std::srand(std::time(0)); // 添加随机种子 std::cout << "测试1: \n"; test1(); std::cout << "\n测试2: \n"; test2(); } // 输出容器对象c中的元素 template <typename T> void output(const T &c) { for(auto &i: c) std::cout << i << ' '; std::cout << '\n'; } // 返回[0, 100]区间内的一个随机整数 int generate_random_number() { return std::rand() % 101; } // 测试1:对容器类对象指定迭代器区间赋值、排序 void test1() { using namespace std; vector<int> v0(10); // 创建一个动态数组对象v0, 对象大小为10 generate(v0.begin(), v0.end(), generate_random_number); // 生成随机数填充v0 cout << "v0: "; output(v0); vector<int> v1{v0}; sort(v1.begin(), v1.end()); // 对整个vector排序 cout << "v1: "; output(v1); vector<int> v2{v0}; sort(v2.begin()+1, v2.end()-1); // 只对中间部分排序,不包含首尾元素 cout << "v2: "; output(v2); } // 测试2:对容器类对象指定迭代器区间赋值、计算最大值/最小值/均值 void test2() { using namespace std; vector<int> v0(10); generate(v0.begin(), v0.end(), generate_random_number); cout << "v0: "; output(v0); // 求最大值和最小值 auto min_iter = min_element(v0.begin(), v0.end()); auto max_iter = max_element(v0.begin(), v0.end()); cout << "最小值: " << *min_iter << endl; cout << "最大值: " << *max_iter << endl; // 同时求最大值和最小值 auto ans = minmax_element(v0.begin(), v0.end()); cout << "最小值: " << *(ans.first) << endl; cout << "最大值: " << *(ans.second) << endl; // 求平均值 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; }
截图:

问题1:generate 对 [first, last) 区间内每个元素,调用第 3 个参数指定的生成器,把返回值依次赋进去。区别于 fill填固定值,它每次可以产生不同的值。
问题2:只需要扫一遍数组,比较次数更少;一次调用返回 pair,.first/.second 直接拿到最小、最大值迭代器,代码更短。
问题3:改用 lambda 后效果完全等同,lambda 就是现写的匿名函数。
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 << "\n测试2: 字符变换\n"; test2(); } unsigned char func(unsigned char c) { if(c == 'z') return 'a'; if(c == 'Z') return 'A'; if(std::isalpha(c)) return static_cast<unsigned char>(c+1); return c; } void test1() { std::string s1{"Hello World 2049!"}; std::cout << "s1 = " << s1 << '\n'; std::string s2; for(auto c: s1) s2 += std::tolower(c); std::cout << "s2 = " << s2 << '\n'; std::string s3; for(auto c: s1) s3 += std::toupper(c); std::cout << "s3 = " << s3 << '\n'; } void test2() { std::string s1{"I love cosmos!"}; std::cout << "s1 = " << s1 << '\n'; std::string s2(s1.size(), ' '); std::transform(s1.begin(), s1.end(), s2.begin(), func); std::cout << "s2 = " << s2 << '\n'; }
截图:

问题:
1.把字母换成下一个字母(循环移位)
2.tolower(c):若 c 是大写字母,返回对应小写;否则原样返回。
toupper(c):若 c 是小写字母,返回对应大写;否则原样返回。
都是 里的单字符处理函数,一次只处理一个字符,不是整串。
3.s1.begin():源区间起点
s1.end():源区间终点(不含)
s2.begin(): 结果写入的起点
func:变换函数,对每个源元素调用一次
区别:结果就地写回 s1 自身,s,而 s2 保持初始的 9 个空格不变。
任务4:
#include <iostream> #include <string> #include <algorithm> #include <cctype> 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"; } } // is_palindrome定义 bool is_palindrome(const std::string &s) { std::size_t i = 0; std::size_t j = s.size(); while(i + 1 < j) { if(s[i] != s[j - 1]) return false; ++i; --j; } return true; } // is_palindrome_ignore_case定义 bool is_palindrome_ignore_case(const std::string &s) { std::size_t i = 0; std::size_t j = s.size(); while(i + 1 < j) { unsigned char left = static_cast<unsigned char>(s[i]); unsigned char right = static_cast<unsigned char>(s[j - 1]); if(std::tolower(left) != std::tolower(right)) return false; ++i; --j; } return true; }

问题:可以把cin >> s 替换为 getline(cin,s),getline(cin,s)读取正行输入,可接收空格
任务5:
#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"; } } // 函数dec2n定义 std::string dec2n(int x, int n) { if(x == 0) return "0"; std::string ans; while(x > 0) { int rem = x % n; if(rem < 10) ans += static_cast<char>('0' + rem); else ans += static_cast<char>('A' + rem - 10); x /= n; } std::reverse(ans.begin(), ans.end()); return ans; }

任务6:
#include <iostream> #include <iomanip> int main() { std::cout << " "; for (char c = 'a'; c <= 'z'; ++c) { std::cout << c << ' '; } std::cout << '\n'; for (int i = 1; i <= 26; ++i) { std::cout << std::right << std::setw(2) << i << ' '; for (int j = 0; j < 26; ++j) { char ch = 'A' + (j + i) % 26; std::cout << ch << ' '; } std::cout << '\n'; } return 0; }

任务7:
#include <iostream> #include <cstdlib> #include <ctime> #include <iomanip> #include <algorithm> using namespace std; int main() { srand(time(0)); int correct_count = 0; const int total_questions = 10; for (int i = 0; i < total_questions; ++i) { int op = rand() % 4; int a, b, ans; char oper; if (op == 0) { oper = '+'; a = rand() % 10 + 1; b = rand() % 10 + 1; ans = a + b; } else if (op == 1) { oper = '-'; a = rand() % 10 + 1; b = rand() % 10 + 1; if (a < b) swap(a, b); ans = a - b; } else if (op == 2) { oper = '*'; a = rand() % 10 + 1; b = rand() % 10 + 1; ans = a * b; } else { oper = '/'; b = rand() % 10 + 1; int max_q = 10 / b; int q = rand() % max_q + 1; a = b * q; ans = a / b; } cout << a << oper << b << "="; int user_ans; cin >> user_ans; if (user_ans == ans) { correct_count++; } } cout << fixed << setprecision(2); cout << "正确率:" << (correct_count * 100.0) / total_questions << "%" << endl; return 0; }


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