From 441a7c1ba88dcb52b4d9cd0d88fcb969a1290eb0 Mon Sep 17 00:00:00 2001
From: flykhan <liubo666666@outlook.com>
Date: Sat, 8 Jul 2023 16:16:58 +0800
Subject: [PATCH] =?UTF-8?q?=E7=AC=AC5=E5=A4=A9=E4=BD=9C=E4=B8=9A?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 day5/homework/h1.c  | 27 +++++++++++++++++++
 day5/homework/h10.c | 15 +++++++++++
 day5/homework/h11.c | 22 +++++++++++++++
 day5/homework/h12.c | 29 ++++++++++++++++++++
 day5/homework/h13.c | 34 +++++++++++++++++++++++
 day5/homework/h14.c | 66 +++++++++++++++++++++++++++++++++++++++++++++
 day5/homework/h15.c | 24 +++++++++++++++++
 day5/homework/h16.c | 23 ++++++++++++++++
 day5/homework/h2.c  |  8 ++++++
 day5/homework/h3.c  | 13 +++++++++
 day5/homework/h4.c  |  8 ++++++
 day5/homework/h5.c  | 13 +++++++++
 day5/homework/h6.c  | 12 +++++++++
 day5/homework/h7.c  | 12 +++++++++
 day5/homework/h8.c  | 20 ++++++++++++++
 day5/homework/h9.c  | 27 +++++++++++++++++++
 16 files changed, 353 insertions(+)
 create mode 100644 day5/homework/h1.c
 create mode 100644 day5/homework/h10.c
 create mode 100644 day5/homework/h11.c
 create mode 100644 day5/homework/h12.c
 create mode 100644 day5/homework/h13.c
 create mode 100644 day5/homework/h14.c
 create mode 100644 day5/homework/h15.c
 create mode 100644 day5/homework/h16.c
 create mode 100644 day5/homework/h2.c
 create mode 100644 day5/homework/h3.c
 create mode 100644 day5/homework/h4.c
 create mode 100644 day5/homework/h5.c
 create mode 100644 day5/homework/h6.c
 create mode 100644 day5/homework/h7.c
 create mode 100644 day5/homework/h8.c
 create mode 100644 day5/homework/h9.c

diff --git a/day5/homework/h1.c b/day5/homework/h1.c
new file mode 100644
index 0000000..1194f7a
--- /dev/null
+++ b/day5/homework/h1.c
@@ -0,0 +1,27 @@
+// 请编程，定义char变量并赋值‘\101’, 输出这个变量的十进制数和字符内容。
+// 【提示】\ddd 代表8进制， % c 输出字符
+#include <stdio.h>
+
+void printBinary(char num)
+{
+    int size = sizeof(num);
+    int i = size * 8 - 1;
+    while (i--)
+    {
+        int bit = (num >> i) & 1;
+        printf("%d", bit);
+    }
+    printf("\n");
+}
+
+int main()
+{
+    char x = '\101';
+    printf("十进制数: %d\n", x);
+    printf("二进制数: ");
+    printBinary(x);
+    printf("八进制数: %o\n", x);
+    printf("十六进制数: 0x%hhx\n", x);
+    printf("字符内容: %c\n", x);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h10.c b/day5/homework/h10.c
new file mode 100644
index 0000000..e5cd3b9
--- /dev/null
+++ b/day5/homework/h10.c
@@ -0,0 +1,15 @@
+// 请编程， 输入一个圆的半径，输出圆的周长和面积
+#include <stdio.h>
+
+#define PI 3.14 // 定义 PI
+
+int main()
+{
+    int r; // 定义半径
+    printf("请输入圆的半径：");
+    scanf("%d", &r);
+    printf("圆的周长为：%.2f\n", 2 * PI * r);
+    printf("圆的面积为：%.2f\n", PI * r * r);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h11.c b/day5/homework/h11.c
new file mode 100644
index 0000000..5614881
--- /dev/null
+++ b/day5/homework/h11.c
@@ -0,0 +1,22 @@
+// 请编程， 输入两个整数， 输出两个整数所有公约数。
+// 例如， 输入 12 和 18， 输出 1 2 3 6。
+#include <stdio.h>
+
+int main()
+{
+    int a, b;
+    printf("请输入两个整数：");
+    scanf("%d%d", &a, &b);
+    int min = a < b ? a : b; // 取两个数中较小的一个
+    while (min)
+    {
+        if (a % min == 0 && b % min == 0)
+        {
+            printf("%d ", min);
+        }
+        min--;
+    }
+    printf("\n");
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h12.c b/day5/homework/h12.c
new file mode 100644
index 0000000..502b4ba
--- /dev/null
+++ b/day5/homework/h12.c
@@ -0,0 +1,29 @@
+// 请编程， 输入一个整数n， 输出斐波那契数列中第n位上的数值
+// 【提示】斐波那契数列： 1 1 2 3 5 8 13 ...
+// 【示例】n = 1, 输出1; n = 7, 输出13.
+#include <stdio.h>
+
+int main()
+{
+    int a = 1, b = 1, c, n;
+    printf("请输入一个整数 n : ");
+    scanf("%d", &n);
+    if (n >= 1)
+        c = a;
+    if (n >= 2)
+        c = b;
+
+    int temp = n - 2;
+    while (n > 2 && temp > 0)
+    {
+        c = a + b;
+        // printf("%d ", c);
+        a = b;
+        b = c;
+        temp--;
+    }
+
+    printf("%d\n", c);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h13.c b/day5/homework/h13.c
new file mode 100644
index 0000000..78f006e
--- /dev/null
+++ b/day5/homework/h13.c
@@ -0,0 +1,34 @@
+// 请编程，实现以下操作： 定义unsigned char 型变量a，赋值0x7f，
+// 将其第1、5、7位置1，2、3、6位清0，其它位保持不变。
+#include <stdio.h>
+
+void printBinary(char num)
+{
+    int size = sizeof(num);
+    for (int i = size * 8 - 1; i >= 0; i--)
+    {
+        int bit = (num >> i) & 1;
+        printf("%d", bit);
+    }
+    printf("\n");
+}
+
+int main()
+{
+    unsigned char a = 0x7f; // 十六进制
+    printf("原样输出: ");
+    printBinary(a);
+
+    a |= 0b1010001; // 0b 表示二进制数的标识符
+    printf("1、5、7位置1输出: ");
+    printBinary(a);
+
+    a &= ~0b100110;
+    printf("2、3、6位清0输出: ");
+    printBinary(a);
+
+    printf("结果的十六进制输出: ");
+    printf("0x%hhx\n", a); // 0x59
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h14.c b/day5/homework/h14.c
new file mode 100644
index 0000000..d1546e9
--- /dev/null
+++ b/day5/homework/h14.c
@@ -0,0 +1,66 @@
+// 请编程，根据个人收入计算个人所得税，免税额为3500
+
+//     > 7级超额累进税率： >> 全月应纳税所得额 税率 速算扣除数(元) >> 全月应纳税额不超过1500元 3 % 0 >> 全月应纳税额超过1500元至4500元 10 % 105 >> 全月应纳税额超过4500元至9000元 20 % 555 >> 全月应纳税额超过9000元至35000元 25 % 1005 >> 全月应纳税额超过35000元至55000元 30 % 2755 >> 全月应纳税额超过55000元至80000元 35 % 5505 >> 全月应纳税额超过80000元 45 % 13505 >> 例如：你月收入 : 3000,
+//     低于免税额3500，不必缴税；
+//         >> 月收入 : 4000,
+//     4000 - 3500 = 500 < 1500, 属于第一档，缴税 = (4000 - 3500) * 0.03 - 0 == 15 月收入 : 15000, 15000 - 3500 = 11500, 属于第四档, 缴税 = (15000 - 3500) * 0.25 - 1005 == 1870 >> 要求：输入你的税前收入，输出：你最后的所得以及缴纳的税额
+#include <stdio.h>
+
+int main()
+{
+    int pretexIncome;   // 税前收入
+    double rate;        // 税率
+    int deduct;         // 速算扣除数
+    int tex;            // 缴纳税额
+    int aftertexIncome; // 税后收入
+    printf("请输入你的税前收入: ");
+    scanf("%d", &pretexIncome);
+
+    int diff = pretexIncome - 3500; // 征税差值
+    if (diff <= 0)
+    {
+        rate = 0;
+        deduct = 0;
+    }
+    else if (diff > 0 && diff <= 1500)
+    {
+        rate = 0.03;
+        deduct = 0;
+    }
+    else if (diff > 1500 && diff <= 4500)
+    {
+        rate = 0.1;
+        deduct = 105;
+    }
+    else if (diff > 4500 && diff <= 9000)
+    {
+        rate = 0.2;
+        deduct = 555;
+    }
+    else if (diff > 9000 && diff <= 35000)
+    {
+        rate = 0.25;
+        deduct = 1005;
+    }
+    else if (diff > 35000 && diff <= 55000)
+    {
+        rate = 0.3;
+        deduct = 2755;
+    }
+    else if (diff > 55000 && diff <= 80000)
+    {
+        rate = 0.35;
+        deduct = 5505;
+    }
+    else if (diff > 80000)
+    {
+        rate = 0.45;
+        deduct = 13505;
+    }
+
+    tex = diff * rate - deduct;
+    aftertexIncome = pretexIncome - tex;
+    printf("你最后的所得为 %d ，你所缴纳的税额为 %d\n", aftertexIncome, tex);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h15.c b/day5/homework/h15.c
new file mode 100644
index 0000000..3258f7b
--- /dev/null
+++ b/day5/homework/h15.c
@@ -0,0 +1,24 @@
+// 【扩展题】请编程，键盘输入一个整数， 输出它的八进制数
+// 【提示】使用int数组完成, int m[32];
+#include <stdio.h>
+
+int main()
+{
+    int inputNum;
+    printf("请输入一个整数：");
+    scanf("%d", &inputNum);
+    int m[32];
+    int i = 0;
+    while (inputNum)
+    {
+        m[i++] = inputNum % 8;
+        inputNum /= 8;
+    }
+    printf("八进制数为：");
+    while (i)
+    {
+        printf("%d", m[--i]);
+    }
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h16.c b/day5/homework/h16.c
new file mode 100644
index 0000000..4e83699
--- /dev/null
+++ b/day5/homework/h16.c
@@ -0,0 +1,23 @@
+// 【扩展题】请编程，键盘输入一个二进制数的字符串，输出转不同进制的数值
+// 【提示】char m[32] = "";
+#include <stdio.h>
+
+int main()
+{
+    char m[32] = ""; // 用于存储二进制数的字符串
+    printf("请输入一个二进制数的字符串：");
+    scanf("%s", m);
+
+    int i = 0;
+    int size = sizeof(m) / sizeof(m[0]); // 数组的大小
+    int sum = 0;                         // 用于存储转换后的十进制数
+    while (i < size && m[i] != '\0')
+    {
+        sum = sum * 2 + m[i] - '0'; // 每次会将高位的结果再次乘 2
+        i++;
+    }
+    printf("十进制数为: %d\n", sum);
+    printf("八进制数为: 0%o\n", sum);
+    printf("十六进制数为: 0x%hhx\n", sum);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h2.c b/day5/homework/h2.c
new file mode 100644
index 0000000..2acafad
--- /dev/null
+++ b/day5/homework/h2.c
@@ -0,0 +1,8 @@
+// 请编程，输出 '\0' == 0 的结果
+#include <stdio.h>
+
+int main()
+{
+    printf("'\\0' == 0 的结果为: %d\n", '\0' == 0); // \0 需要转义为 \\0
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h3.c b/day5/homework/h3.c
new file mode 100644
index 0000000..a7d01f5
--- /dev/null
+++ b/day5/homework/h3.c
@@ -0,0 +1,13 @@
+// 请计算，【提示】考虑运算符优先级 int n = 0, m = 1;
+// m = n++, n++, n++, m++;
+// 问：n, m分别为多少？
+#include <stdio.h>
+
+int main()
+{
+    int n = 0, m = 1;
+    // m = n++;
+    m = n++, n++, n++, m++; // m = 0 , 然后 m++, m = 1
+    printf("n = %d, m = %d\n", n, m);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h4.c b/day5/homework/h4.c
new file mode 100644
index 0000000..17582b3
--- /dev/null
+++ b/day5/homework/h4.c
@@ -0,0 +1,8 @@
+// 请计算, 12 ^ 3 && 12 & 3 的结果     【提示】考虑运算符优先级
+#include <stdio.h>
+
+int main()
+{
+    printf("12 ^ 3 && 12 & 3 的结果为: %d\n", 12 ^ 3 && 12 & 3);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h5.c b/day5/homework/h5.c
new file mode 100644
index 0000000..cde672f
--- /dev/null
+++ b/day5/homework/h5.c
@@ -0,0 +1,13 @@
+// 请计算, int i = 1;
+// i += --i && -5;
+// i的结果是多少？
+#include <stdio.h>
+
+int main()
+{
+    int i = 1;
+    i += --i && -5; // i = 1 + 0 && -5 = 1 + 0 = 1
+    // i = ((i + (--i)) && (-5));
+    printf("i = %d\n", i);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h6.c b/day5/homework/h6.c
new file mode 100644
index 0000000..f9799ca
--- /dev/null
+++ b/day5/homework/h6.c
@@ -0,0 +1,12 @@
+// 请计算， char n = 1;
+// n = ~n;
+// n的结果是多少？
+#include <stdio.h>
+
+int main()
+{
+    char n = 1;
+    n = ~n; // n = ~n = ~00000001 = 11111110 = -2
+    printf("n = %d\n", n);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h7.c b/day5/homework/h7.c
new file mode 100644
index 0000000..f5e30fd
--- /dev/null
+++ b/day5/homework/h7.c
@@ -0,0 +1,12 @@
+// 请计算, short n = 112;
+// n &= ~(3 << 4);
+// n的结果是多少？
+#include <stdio.h>
+
+int main()
+{
+    short n = 112;
+    n &= ~(3 << 4); // n = n & (~(3 << 4)) = n & (~48) = n & (-49) = n & 65487 = 112 & 65487 = 112
+    printf("n = %d\n", n);
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h8.c b/day5/homework/h8.c
new file mode 100644
index 0000000..b2ae48c
--- /dev/null
+++ b/day5/homework/h8.c
@@ -0,0 +1,20 @@
+// 请编程，输入一个整数，输出它的所有因数。
+#include <stdio.h>
+
+int main()
+{
+    int num;
+    printf("请输入一个整数: ");
+    scanf("%d", &num);
+    int i = 1;
+    printf("%d 的因数有: ", num);
+    while (i <= num)
+    {
+        if (num % i == 0)
+            printf("%d ", i);
+        i++;
+    }
+    printf("\n");
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day5/homework/h9.c b/day5/homework/h9.c
new file mode 100644
index 0000000..db2fb7d
--- /dev/null
+++ b/day5/homework/h9.c
@@ -0,0 +1,27 @@
+// 请编程， 输入一个整数，输出它的二进制数中1和0的个数。
+#include <stdio.h>
+
+int main()
+{
+    int inputNum;
+    int temp = 0, zeroCnt = 0, oneCnt = 0;
+
+    printf("请输入一个整数: ");
+    scanf("%d", &inputNum);
+
+    for (int i = 0; temp < inputNum; i++)
+    {
+        temp = 1 << i;
+        // if ((inputNum & temp) == temp)
+        //     oneCnt++;
+        // else if ((inputNum | ~temp) == ~temp)
+        //     zeroCnt++;
+        if (inputNum & temp)
+            oneCnt++;
+        else
+            zeroCnt++;
+    }
+
+    printf("%d 的二进制中 1 有 %d 个, 0 有 %d 个", inputNum, oneCnt, zeroCnt);
+    return 0;
+}
\ No newline at end of file