第4天作业

day4/homework/h10.c

@@ -0,0 +1,10 @@
+// 请编程， 定义一个char类型的变量m,并初始化赋值0xf1;要求按原格式输出和十进制输出
+#include <stdio.h>
+
+int main()
+{
+    char m = 0xf1;
+    printf("原格式输出: 0x%hhx\n", m);
+    printf("十进制输出: %d\n", m);
+    return 0;
+}

day4/homework/h12.c

@@ -0,0 +1,11 @@
+// 请编程， 定义char类型的变量n，初始化值为-128,  将n值减1之后，再输出它的十六进制数值和十进制数值
+#include <stdio.h>
+
+int main()
+{
+    char n = -128;
+    n -= 1;
+    printf("十六进制数值为: 0x%hhx\n", n);
+    printf("十进制数值为: %d\n", n);
+    return 0;
+}

day4/homework/h13.c

@@ -0,0 +1,38 @@
+// 【扩展题1】请编程，输出0-1000以内的水仙花数
+// 【提示】水仙花数算法：一个数 = 它各位的立方和，例如 : 153 = 1 * 1 * 1 + 5 * 5 * 5 + 3 * 3 * 3
+#include <stdio.h>
+
+void fun1()
+{
+    int i, j, k;
+    for (i = 0; i < 10; i++)
+    {
+        for (j = 0; j < 10; j++)
+        {
+            for (k = 0; k < 10; k++)
+            {
+                if (i * 100 + j * 10 + k == i * i * i + j * j * j + k * k * k)
+                    printf("%d\n", i * 100 + j * 10 + k);
+            }
+        }
+    }
+}
+
+void fun2()
+{
+    int x = 0;
+    while (x++ < 1000 - 1) // -1 是因为 1000 不是水仙花数
+    {
+        int a = x / 100;
+        int b = x / 10 % 10;
+        int c = x % 10;
+        if (x == a * a * a + b * b * b + c * c * c)
+            printf("%d\n", x);
+    }
+}
+
+int main()
+{
+    fun2();
+    return 0;
+}

day4/homework/h14.c

@@ -0,0 +1,45 @@
+// 【扩展题2】请编程,从键盘接收年、月、日三个整数,判断是这一年的第几天。
+// 【提示】必然考虑年是否为闰年,闰年条件:能被4整除且不能被100整除,或者能被400整除
+
+#include <stdio.h>
+
+int isRun(int year)
+{
+    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
+        return 1; // 是闰年
+    return 0;
+}
+
+int main()
+{
+    int year, month, day;
+    year = scanf("%d", &year);
+    month = scanf("%d", &month);
+    day = scanf("%d", &day);
+    int days = 0;
+    int mon_2 = isRun(year) ? 29 : 28; // 二月天数: 闰年29天, 平年28天
+    if (month < 8)
+    {
+        if (month % 2 == 0)
+        {
+            days += month / 2 * 31 + (month / 2 - 1) * 30 + mon_2;
+            printf("%d\n", days);
+        }
+        else
+        {
+            days += (month / 2 + 1) * 31 + (month / 2 - 1) * 30 + mon_2;
+            printf("%d\n", days);
+        }
+    }
+    else
+    {
+        if (month % 2 == 0)
+        {
+            days += (month / 2 - 1) * 31 + month / 2 * 30 + mon_2;
+        }
+        else
+        {
+            days += (month / 2) * 31 + (month / 2) * 30 + mon_2;
+        }
+    }
+}

day4/homework/h2.c

@@ -0,0 +1,21 @@
+#include <stdio.h>
+
+int main()
+{
+    char a = 'a';
+    short b = 12;
+    int c = 34;
+    unsigned int d = 43;
+    long int e = 232;
+    float f = 1.23f;
+    double g = 12.32;
+
+    printf("sizeof a = %ld\n", sizeof(a));
+    printf("sizeof b = %ld\n", sizeof(b));
+    printf("sizeof c = %ld\n", sizeof(c));
+    printf("sizeof d = %ld\n", sizeof(d));
+    printf("sizeof e = %ld\n", sizeof(e));
+    printf("sizeof f = %ld\n", sizeof(f));
+    printf("sizeof g = %ld\n", sizeof(g));
+    return 0;
+}

day4/homework/h3.c

@@ -0,0 +1,26 @@
+// 请编程，将一个整数变量输出不同进制格式的结果
+#include <stdio.h>
+
+void printBinary(int num)
+{
+    // int size = sizeof(num) * 8; // 计算整数的位数
+    int size = 8; // 计算整数的位数
+    for (int i = size - 1; i >= 0; i--)
+    {
+        int bit = (num >> i) & 1; // 获取当前位的值
+        printf("%d", bit);        // 输出当前位的值
+    }
+    printf("\n");
+}
+
+int main()
+{
+    int a = 19;
+
+    printf("二进制输出: ");
+    printBinary(a);
+    printf("10 进制: %d\n", a);
+    printf("八进制输出: %o\n", a);
+    printf("十六进制输出: 0x%hhx\n", a);
+    return 0;
+}

day4/homework/h4.c

@@ -0,0 +1,7 @@
+#include <stdio.h>
+
+int main()
+{
+    printf("%d\t%d",3,4);
+    return 0;
+}

day4/homework/h7.c

@@ -0,0 +1,9 @@
+// 请计算 char类型的0x89输出整数结果是多少？
+#include <stdio.h>
+
+int main()
+{
+    char c = 0x89;
+    printf("%d\n", c);
+    return 0;
+}

day4/homework/h8.c

@@ -0,0 +1,9 @@
+// 请计算 0b11000001有符号整数结果是多少？
+#include <stdio.h>
+
+int main()
+{
+    signed char c = 0b11000001;
+    printf("%d\n", c);
+    return 0;
+}

day4/homework/h9.c

@@ -0,0 +1,14 @@
+// 请编程，定义一个double类型的变量n和一个char类型的变量m,请计算它们的和并输出
+#include <stdio.h>
+
+int main()
+{
+    double n = 2.3;
+    char m = 'a';
+
+    printf("a = %d\n", m);
+
+    printf("n + m = %.2lf\n", n + m);
+
+    return 0;
+}