重命名文件夹

algorithm/平年与闰年.c

+#include <stdbool.h>
+#include <stdio.h>
+
+bool leapYear(int year)
+{
+	return year % 4 == 0 && year % 100 != 0 \
+		|| year % 400 == 0;
+}
+
+int main()
+{
+	int year;
+	printf("ݣ");
+	scanf("%d", &year);
+	if (leapYear(year))
+		puts("");
+	else
+		puts("ƽ");
+	return 0;
+}

algorithm/年中天数.c

+#include <stdbool.h>
+#include <stdio.h>
+
+bool leapYear(int year)
+{
+	return year % 4 == 0 && year % 100 != 0 \
+		|| year % 400 == 0;
+}
+
+const unsigned dayForMonth[12] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30 };
+
+unsigned days(int year, unsigned month, unsigned day)
+{
+	int days = 0;
+	for (int index = 1; index < month; ++index)
+		days += dayForMonth[index];
+	days += day;
+	if (month >= 3 && leapYear(year))
+		++days;
+	return days;
+}
+
+int main()
+{
+	int year;
+	unsigned month, day;
+	printf("输入年月日，以斜线、中横线或者空格间隔：");
+	if (scanf("%d%*c%u%*c%u", &year, &month, &day) == 3)
+		printf("这天是此年第%u天\n", days(year, month, day));
+	return 0;
+}

algorithm/枚举法.c

+/*
+ * 枚举法又称为穷举法，先列举所有可能出现的情况，再一一进行测试，找出符合条件的所有结果。
+ * 例如计算一个古典数学问题——“百钱买百鸡”。
+ * 一百个铜钱买一百只鸡，公鸡每只5钱，母鸡每只3钱，小鸡3只1钱，问公鸡、母鸡和小鸡各买几只？
+ * 假设公鸡x只，母鸡y只，小鸡z只。根据题意可列出以下方程组：
+ * 5x+3y+z/3=100(百钱)；x+y+z=100(百鸡)
+ * 由于2个方程式之中有3个未知数，属于无法直接求解的不定方程，故采用“枚举法”进行试根。
+ * 此处x、y、z均为正整数，且z是3的倍数；由于鸡和钱的总数都是100，可以确定x,y,z的取值范围：
+ * x的取值范围为[1,20]
+ * y的取值范围为[1,33]
+ * z的取值范围为[3,99]，步长为3
+ * 逐一测试各种可能的x、y、z组合，并输出符合条件者。
+ */
+
+#include <stdio.h>
+
+int main()
+{
+	for (int x = 1, y, z; x <= 20; ++x)
+		for (y = 1; y <= 33; ++y)
+		{
+			z = 100 - x - y;
+			if (z % 3 == 0 \
+				&& 5 * x + 3 * y + z / 3 == 100)
+				printf("公鸡%d只，母鸡%d只，小鸡%d只\n", x, y, z);
+		}
+	return 0;
+}

feature/math.h/圆面积.c

+#include <math.h>
+#include <stdio.h>
+
+double PI = 0;
+
+double pi()
+{
+	if (PI == 0)
+		PI = atan(1) * 4; // 反三角函数arctan
+	return PI;
+}
+
+double area(double radius)
+{
+	return pi()*pow(radius, 2);
+}
+
+int main()
+{
+	double radius;
+	printf("半径：");
+	scanf("%lf", &radius);
+	printf("面积：%.9lf\n", area(radius));
+	return 0;
+}
\ No newline at end of file

工具库/隐藏密码.c

+#include <string.h>
+#include <stdio.h>
+#include <conio.h>
+
+static inline int max(int left, int right)
+{
+	return left > right ? left : right;
+}
+
+const char *PASSWORD = "universe";
+
+int main()
+{
+	printf("请输入密码：\n");
+	char password[10];
+	int length = strlen(PASSWORD);
+	int counter = 0;
+	int key;
+	while (key = getch())
+	{
+		password[counter] = key;
+		if (key == '\b')
+		{
+			printf("\b \b");
+			--counter;
+			counter = max(counter, 0);
+		}
+		else
+		{
+			if (key == '\r' || key == '\n')
+				break;
+			putchar('*');
+			++counter;
+			if (counter >= length)
+				break;
+		}
+	}
+	password[counter] = '\0';
+
+	if (strcmp(password, PASSWORD) == 0)
+		printf("\n验证通过！");
+	else
+		printf("\n密码错误！");
+	return 0;
+}

算法/二分查找.c

+#include <stdio.h>
+
+#define N 9
+
+int bsearch(int *array, int num, int key)
+{
+	int low = 0, high = num - 1, mid;
+	while (low <= high)
+	{
+		mid = (low + high) / 2;
+		int temp = array[mid];
+		if (temp == key)
+			return mid;
+		else if (temp > key)
+			high = mid - 1;
+		else
+			low = mid + 1;
+	}
+	return -1;
+}
+
+int main()
+{
+	int array[N], key;
+	printf("输入升序排列的%d个数：", N);
+	for (int index = 0; index < N; ++index)
+		scanf("%d", &array[index]);
+	printf("输入查找数：");
+	scanf("%d", &key);
+	int index = bsearch(array, N, key);
+	if (index >= 0)
+		printf("array[%d] = %d\n", index, array[index]);
+	else
+		puts("不存在查找数");
+	return 0;
+}

算法/冒泡排序.c

+#include <stdio.h>
+
+#define N 9
+
+void sort(int *array, int num)
+{
+	for (int i = num - 1; i > 0; --i)
+		for (int j = 0; j < i; ++j)
+			if (array[j] > array[j + 1])
+			{
+				int temp = array[j + 1];
+				array[j + 1] = array[j];
+				array[j] = temp;
+			}
+}
+
+int main()
+{
+	int array[N], x;
+	printf("%d\n", N);
+	for (int i = 0; i < N; ++i)
+		scanf("%d", &array[i]);
+	sort(array, N);
+	for (int i = 0; i < N; ++i)
+		printf("%d ", array[i]);
+	return 0;
+}

算法/平年与闰年.c

+#include <stdbool.h>
+#include <stdio.h>
+
+bool leapYear(int year)
+{
+	return year % 4 == 0 && year % 100 != 0 \
+		|| year % 400 == 0;
+}
+
+int main()
+{
+	int year;
+	printf("ݣ");
+	scanf("%d", &year);
+	if (leapYear(year))
+		puts("");
+	else
+		puts("ƽ");
+	return 0;
+}

算法/年中天数.c

+#include <stdbool.h>
+#include <stdio.h>
+
+bool leapYear(int year)
+{
+	return year % 4 == 0 && year % 100 != 0 \
+		|| year % 400 == 0;
+}
+
+const unsigned dayForMonth[12] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30 };
+
+unsigned days(int year, unsigned month, unsigned day)
+{
+	int days = 0;
+	for (int index = 1; index < month; ++index)
+		days += dayForMonth[index];
+	days += day;
+	if (month >= 3 && leapYear(year))
+		++days;
+	return days;
+}
+
+int main()
+{
+	int year;
+	unsigned month, day;
+	printf("输入年月日，以斜线、中横线或者空格间隔：");
+	if (scanf("%d%*c%u%*c%u", &year, &month, &day) == 3)
+		printf("这天是此年第%u天\n", days(year, month, day));
+	return 0;
+}

算法/排列数.c

+#include <stdio.h>
+
+//// ݹ
+//unsigned permutate(unsigned n, unsigned m)
+//{
+//	if (m > n)
+//		return 0;
+//	else if (m == 0)
+//		return 1;
+//	else if (m == 1)
+//		return n;
+//	else
+//		return n * permutate(n - 1, m - 1);
+//}
+
+// 
+unsigned permutate(unsigned n, unsigned m)
+{
+	if (m > n)
+		return 0;
+	unsigned result = 1;
+	for (; m > 0; --m, --n)
+		result *= n;
+	return result;
+}
+
+int main()
+{
+	unsigned m, n;
+	printf("n = ");
+	scanf("%u", &n);
+	printf("m = ");
+	scanf("%u", &m);
+	printf("A(%u,%u) = %u\n", n, m, permutate(n, m));
+	return 0;
+}

算法/枚举法.c

+/*
+ * 枚举法又称为穷举法，先列举所有可能出现的情况，再一一进行测试，找出符合条件的所有结果。
+ * 例如计算一个古典数学问题——“百钱买百鸡”。
+ * 一百个铜钱买一百只鸡，公鸡每只5钱，母鸡每只3钱，小鸡3只1钱，问公鸡、母鸡和小鸡各买几只？
+ * 假设公鸡x只，母鸡y只，小鸡z只。根据题意可列出以下方程组：
+ * 5x+3y+z/3=100(百钱)；x+y+z=100(百鸡)
+ * 由于2个方程式之中有3个未知数，属于无法直接求解的不定方程，故采用“枚举法”进行试根。
+ * 此处x、y、z均为正整数，且z是3的倍数；由于鸡和钱的总数都是100，可以确定x,y,z的取值范围：
+ * x的取值范围为[1,20]
+ * y的取值范围为[1,33]
+ * z的取值范围为[3,99]，步长为3
+ * 逐一测试各种可能的x、y、z组合，并输出符合条件者。
+ */
+
+#include <stdio.h>
+
+int main()
+{
+	for (int x = 1, y, z; x <= 20; ++x)
+		for (y = 1; y <= 33; ++y)
+		{
+			z = 100 - x - y;
+			if (z % 3 == 0 \
+				&& 5 * x + 3 * y + z / 3 == 100)
+				printf("公鸡%d只，母鸡%d只，小鸡%d只\n", x, y, z);
+		}
+	return 0;
+}

算法/组合数.c

+#include <stdio.h>
+
+unsigned combine(unsigned, unsigned);
+
+int main()
+{
+	unsigned m, n;
+	printf("n = ");
+	scanf("%u", &n);
+	printf("m = ");
+	scanf("%u", &m);
+	printf("C(%u,%u) = %u\n", n, m, combine(n, m));
+	return 0;
+}
+
+unsigned permutate(unsigned n, unsigned m)
+{
+	if (m > n)
+		return 0;
+	unsigned result = 1;
+	for (; m > 0; --m, --n)
+		result *= n;
+	return result;
+}
+
+unsigned factorial(unsigned n)
+{
+	unsigned result = 1;
+	for (; n > 1; --n)
+		result *= n;
+	return result;
+}
+
+unsigned combine(unsigned n, unsigned m)
+{
+	return permutate(n, m) / factorial(m);
+}

算法/阶乘.c

+#include <stdio.h>
+
+//// 递归
+//unsigned factorial(unsigned n)
+//{
+//	if (n <= 1)
+//		return 1;
+//	else
+//		return factorial(n - 1)*n;
+//}
+
+// 递推
+unsigned factorial(unsigned n)
+{
+	unsigned result = 1;
+	for (; n > 1; --n)
+		result *= n;
+	return result;
+}
+
+int main()
+{
+	printf("输入一个正整数：");
+	unsigned x;
+	scanf("%u", &x);
+	printf("%u! = %u\n", x, factorial(x));
+	return 0;
+}

语言特性/ctype.h/大小写字母转换.c

+#include <ctype.h>
+#include <stdio.h>
+
+int main()
+{
+	printf("输入大写字母：");
+	printf("转为小写字母：%c\n", tolower(getchar()));
+	scanf("%*[^\r\n]%*c");
+	printf("输入小写字母：");
+	printf("转为大写字母：%c\n", toupper(getchar()));
+	return 0;
+}

语言特性/math.h/圆面积.c

+#include <math.h>
+#include <stdio.h>
+
+double PI = 0;
+
+double pi()
+{
+	if (PI == 0)
+		PI = atan(1) * 4; // 反三角函数arctan
+	return PI;
+}
+
+double area(double radius)
+{
+	return pi()*pow(radius, 2);
+}
+
+int main()
+{
+	double radius;
+	printf("半径：");
+	scanf("%lf", &radius);
+	printf("面积：%.9lf\n", area(radius));
+	return 0;
+}
\ No newline at end of file