例题

1. 804.n的阶乘

Acwing 804.n的阶乘

#include<iostream>
using namespace std;

int fact(int n)
{
    if(n == 1) return 1;
    return n * fact(n - 1);
}

int main()
{
    int n;
    cin >> n;   
    cout << fact(n) <<endl;
    return 0;
}

2. 805.x和y的最大值

Acwing 805.x和y的最大值

#include<iostream>
using namespace std;

int max(int x,int y)
{
    return x>y?x:y;
}

int main()
{
    int n, m;
    cin >> n >> m;
    cout << max(n,m) << endl;
    return 0;
}

3. 808.最大公约数

Acwing 808.最大公约数

#include<iostream>
using namespace std;

int gcd(int a, int b)
{
    for(int i = 1000; i > 0; i--)// 直接进行计算
        if(a % i == 0 && b % i == 0)
            return i;
    
    return 0;
}

int main()
{
    int n, m;
    cin >> n >> m;
    
    cout << gcd(n, m) << endl;
    return 0;
}

4. 811.交换数值

Acwing 811.交换数值

#include<iostream>
using namespace std;

void swap(int &x, int &y)
{
    int t = x;
    x = y;
    y = t;
}

int main()
{
    int n, m;
    cin >>n >> m;
    
    swap(n, m);
    cout << n << ' '<< m << endl;
    return 0;
}

5. 812.打印数字

Acwing 812.打印数字

#include<iostream>
using namespace std;

const int N = 1010;
int a[N];

void print(int a[], int size)
{
    for(int i = 0; i < size;i ++)
        cout << a[i] << ' ';
    cout << endl;
}

int main()
{
    int n, size;
    cin >> n >> size;
    
    for(int i = 0; i < n; i++) cin >> a[i];
    
    print(a,size);// 这里直接写数组名就可以
    return 0;
}

6. 813.打印矩阵

Acwing 813.打印矩阵

#include<iostream>
using namespace std;

void print2D(int a[][100], int row, int col)
{
    for(int i = 0; i < row; i++)
    {
        for(int j = 0; j < col; j++)
        {
            cout << a[i][j] << ' ';
        }
        cout << endl;
    }
}

int main()
{
    int row, col;
    cin >> row >> col;
    
    int a[100][100];
    for(int i = 0; i < row; i++)
        for(int j = 0; j < col; j++)
            cin >> a[i][j];
            
    print2D(a, row, col);
    return 0;
}

7. 819.递归求阶乘

Acwing 819.递归求阶乘

#include<iostream>
using namespace std;

int fact(int n)
{
    if(n == 1) return 1;
    else return n * fact(n - 1);
}

int main()
{
    int n;
    cin >> n;
    
    cout << fact(n) << endl;
    return 0;
}

8. 820.递归求斐波那契数列

Acwing 820 .递归求斐波那契数列

• 树的深度优先遍历
  

#include<iostream>
using namespace std;

int f(int n)
{
    if(n <= 2) return 1;
    return f(n - 1) + f(n - 2);
}

int main()
{
    int n;
    cin >> n;
    
    cout << f(n) << endl;
    return 0;
}