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c++ 笔记
辗转相除法
求最小公倍数
冒泡排序
1、辗转相除法求最大公约数，简写为gcd。应用：化简分数，即分子分母同时除以最大公约数。
#include <iostream>#include <cstdio>using namespace std;int gcd(int, int);int main(){ cout << gcd(6, 9) << endl; // output=3 return 0; }int gcd(int m, int n) { for (int r; (r=m%n) != 0; m=n, n=r); return n;}
2、最小公倍数,least common multiple，简写为lcm。
#include <iostream>#include <cstdio>using namespace std;int gcd(int, int);int lcm(int, int);int main(){ cout << gcd(6, 9) << endl; // output=3 cout << lcm(6, 9) << endl; // output=18 return 0; }int gcd(int m, int n) { for (int r; (r=m%n) != 0; m=n, n=r); return n;}int lcm(int m, int n) { return m*n / gcd(m, n);}
3、排序之对数字
#include <cstdio>#include <cstdlib>#include <ctime>#define N 1000void swap(int& a, int& b) { a = a + b; b = a - b; a = a - b;}void sort(int *arr, int len) { for (int i = 0; i != len-1; ++i) // len个元素的数组需要排(len-1)趟 for (int j = 0; j != len-i-1; ++j) if (arr[j] > arr[j+1]) swap(arr[j], arr[j+1]);}int arr[N];int main(){ srand(time(0)); for (int i = 0; i < N; i++) printf("%4d",arr[i]=rand()%100+1); // 生成N个1-100的随机整数。 printf("\n"); sort(arr, N); for (i = 0; i < N; ++i) printf("%4d", arr[i]); printf("\n"); return 0;}
#include <cstdio>#include <cstdlib>#include <ctime>#define N 100void swap(int &a, int &b) { a = a + b; b = a - b; a = a - b;}void sort(int *arr, int len) { for (int i = 0; i != len - 1; ++i) for (int j = 0; j != len - i - 1; ++j) if (arr[j] > arr[j+1]) swap(arr[j], arr[j+1]);}int main(){ /* int a = 3, b = 4; swap(a, b); printf("%d %d\n", a, b); */ srand((int)time(0)); int arr[N]; for (int i = 0; i != N; ++i) { printf("%4d", arr[i] = rand()%100 + 1); } printf("\n"); sort(arr, N); for (i = 0; i != N; ++i) { printf("%4d", arr[i]); } printf("\n"); return 0;}
排序对象， c++创建对象的方式：http://blog.csdn.net/azhexg/article/details/14225545
#include <cstdio>#include <cstdlib>#include <ctime>#include <iostream>#include <string>using namespace std;#define N 10/*void swap(int &a, int &b) { a = a + b; b = a - b; a = a - b;}void sort(int *arr, int len) { for (int i = 0; i != len - 1; ++i) for (int j = 0; j != len - i - 1; ++j) if (arr[j] > arr[j+1]) swap(arr[j], arr[j+1]);}*/class Student {friend ostream& print(ostream&, const Student&);friend void sort(Student **arr, int len);public: Student(const string& name, int score): name(name), score(score) {}private: string name; int score;};void swap(Student* &left, Student* &right) { Student *temp; temp = left; left = right; right = temp;}void sort(Student **arr, int len) { for (int i = 0; i != len - 1; ++i) for (int j = 0; j != len - i - 1; ++j) if (arr[j]->score > arr[j+1]->score) swap(arr[j], arr[j+1]);}ostream& print(ostream& os, const Student& stu) { os << stu.name << " " << stu.score; return os;}int main(){ /* int a = 3, b = 4; swap(a, b); printf("%d %d\n", a, b); */ srand((int)time(0)); /* int arr[N]; for (int i = 0; i != N; ++i) { printf("%4d", arr[i] = rand()%100 + 1); } printf("\n"); sort(arr, N); for (i = 0; i != N; ++i) { printf("%4d", arr[i]); } printf("\n"); */ Student *stu[N]; for (int i = 0; i != N; ++i) { print(cout, *(stu[i] = new Student("jack", rand()%100 + 1))) << endl; } printf("******************************************\n"); sort(stu, N); for (i = 0; i != N; ++i) { print(cout, *stu[i]) << endl; } return 0;}
c++ 笔记
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