更新 README.md

README.md

@@ -1,3 +1 @@
 **_本仓库为私人仓库_**
-
- **b667c019058732967262412c7aee4475** 

class/e_de_jin_se_zhi.cpp

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+#include<iostream>
+#include<iomanip>
+
+using namespace std;
+
+int main(){
+    double e = 1.0;
+    double factorial = 1.0;
+
+    for (int i = 1; i <= 20; i++)
+    {
+        factorial *= i;
+        e += 1.0 / factorial;
+    }
+
+    cout << fixed << setprecision(3) << e << endl;
+    
+    return 0;
+}

class/fen_zi_xue_lie_qiu_he.cpp

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+//分子序列求和
+#include<iostream>
+#include<iomanip>
+
+using namespace std;
+
+int main(){
+    int n;
+    double a=2,b=1,c,d;
+    cin >>n;
+    
+    if (n > 100)
+    {
+        cout <<"输入的数字太大了!"<<endl;
+        return 1;
+    }
+
+    double sum=0;
+    for (int i = 1; i <= n; i++)
+    {
+        sum += a/b;
+        c = a;
+        d = b;
+        a = a+d;
+        b = c;
+    }
+    cout <<fixed<<setprecision(2)<<sum<<endl;
+    return 0;
+}

class/zui_da_gong_yue_shu.cpp

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+#include <iostream>
+using namespace std;
+
+int main(){
+    int n,m,x;
+
+    cout << "请输入两个数:";
+    cin >> n >> m;
+
+    if (m>n)
+    {
+        x = n;
+    }else
+    {
+        x = m;
+    }
+
+    while ( n%x!=0 || m%x!=0)
+    {
+        x--;
+    }
+
+    cout << "最大公约数为:" << x << endl;
+    return 0;
+}

old_homework/3721.cpp

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+#include <iostream>
+using namespace std;
+
+int main() {
+    // 遍历1到200的所有数
+    for (int i = 1; i <= 200; i++) {
+        // 判断该数是否满足"3721"数的条件
+        if (i % 3 == 2 && i % 7 == 1) {
+            cout << i << " ";  // 输出满足条件的数
+        }
+    }
+    return 0;
+}

test.cpp

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+#include <iostream>
+
+using namespace std;
+
+// 函数：计算最大公约数
+int gcd(int a, int b) {
+    while (b != 0) {
+        int temp = b;
+        b = a % b;
+        a = temp;
+    }
+    return a;
+}
+
+// 函数：计算最小公倍数
+int lcm(int a, int b) {
+    return (a / gcd(a, b)) * b;
+}
+
+int main() {
+    int M, N;
+    cin >> M >> N;
+    cout << lcm(M, N) << endl;
+    return 0;
+}