RunningLetters

level1/p01_runningLetter/runningletter.c

@@ -0,0 +1,39 @@
+# include <stdio.h>
+# include <stdlib.h>
+# include <Windows.h>
+# include <conio.h>
+
+# define WAITTIME 40
+
+void print_space(int cnt);
+int main(void)
+{
+	//获取屏幕宽度 
+	HANDLE handle_out;                              //定义一个句柄  
+	CONSOLE_SCREEN_BUFFER_INFO screen_info;         //定义窗口缓冲区信息结构体  
+	handle_out = GetStdHandle(STD_OUTPUT_HANDLE);   //获得标准输出设备句柄  
+	GetConsoleScreenBufferInfo(handle_out, &screen_info);   //获取窗口信息 
+	const int LENGTH = screen_info.dwSize.X - 1;        //screen_info.dwSize.X(缓冲区宽度，也就是横坐标最大值加1)
+	//功能实现 
+	int remainder, flag, steps;
+	for(int cnt = 0;; cnt++)
+	{
+		//主体逻辑部分, 实现循环  remainder和flag共同控制方向 
+		remainder = cnt % LENGTH;
+		flag = (cnt / LENGTH) % 2;
+		steps = abs(LENGTH * flag - remainder);
+		print_space(steps);
+
+		printf("R");
+		Sleep(WAITTIME);
+		system("cls");
+	}
+	CloseHandle(handle_out);    //关闭标准输出设备句柄
+}
+void print_space(int steps)
+{
+	for(int i = 0; i < steps; i++)
+	{
+		printf(" ");
+	}
+}

level1/p02_isPrime/isprime.c

@@ -0,0 +1,36 @@
+# include <stdio.h>
+# include <math.h>
+
+void is_prime(int n);
+int main(void)
+{
+	long long int n;
+	printf("请输入一个整数:");
+	if(scanf("%lld", &n))
+	{
+		is_prime(n);
+	}
+	else
+	{
+		printf("Oops!你应该输入一个整数!");
+	}
+	return 0;
+}
+void is_prime(int n)
+{
+	int i;
+	int sqrt_of_n = (int)sqrt(n);
+	for(i = 2; i <= sqrt_of_n; i++)
+	{
+		if(n % i == 0) break;
+	}
+	if(i > sqrt_of_n)
+	{
+		printf("您所输入的%d是素数\n", n);
+	}
+	else
+	{
+		printf("您输入的不是素数\n");
+	}
+
+}

level1/p03_Diophantus/diophantus.c

@@ -0,0 +1,14 @@
+# include <stdio.h>
+# define UPPERLIMIT 250
+
+int main(void)
+{
+	for(double i = 0; i < UPPERLIMIT; i++)
+	{
+		if((i / 12 + i / 6 + i / 7 + 5 + i / 2 + 4) == i)
+		{
+			printf("当儿子死时, 丢番图的年龄为%.0f\n", i - 4);
+			break;
+		}
+	}
+}

level1/p04_ narcissus/narcissus.c

@@ -0,0 +1,31 @@
+# include <stdio.h>
+# include <math.h>
+
+# define LOWERLIMIT 100
+# define UPPERLIMIT 1000
+
+int main(void)
+{
+	printf("3位水仙花数为:\n");
+	int number[3] = {0};
+	for(int i = LOWERLIMIT; i < UPPERLIMIT; i++)
+	{
+		int j = 0, sum = 0;
+		int temp = i;
+		do
+		{
+			number[j++] = temp % 10;
+			temp /= 10;
+		} while(temp);
+		for(j = 0; j < 3; j++)
+		{
+			sum += pow(number[j], 3);
+		}
+		if(sum == i)
+		{
+			printf("%d\t", i);
+		}
+	}
+	printf("\n");
+	return 0;
+}

level1/p05_allPrimes/allprimes.c

@@ -0,0 +1,48 @@
+# include <stdio.h>
+# include <math.h>
+# include <time.h>
+
+# define LOWERLIMIT 2
+# define UPPERLIMIT 1000
+
+int is_prime(int i);
+int main(void)
+{
+	//测量时间 
+	clock_t start, finish;
+	double  duration;
+	start = clock(); 
+	//打印素数 
+	printf("2-1000内所有的素数为:\n");
+	for(int i = LOWERLIMIT; i <= UPPERLIMIT; i++)
+	{
+		if(is_prime(i))
+		{
+			printf("%d\t", i);
+		}
+	}
+	//测量时间 
+	finish = clock();
+	duration = (double)(finish - start) / CLOCKS_PER_SEC;
+	printf("\n该程序运行总时间为: %f seconds\n", duration);
+	return 0;
+}
+
+int is_prime(int i)
+{
+	int sqrt_of_i = (int)sqrt(i);
+	int j, if_is_prime;
+	for(j = 2; j <= sqrt_of_i; j++)
+	{
+		if(i % j == 0) break;
+	}
+	if(j > sqrt_of_i)
+	{
+		if_is_prime = 1;
+	}
+	else
+	{
+		if_is_prime = 0;
+	}
+	return if_is_prime;
+}

level1/p06_Goldbach/goldbach.c

@@ -0,0 +1,54 @@
+//Goldbach
+# include <stdio.h>
+# include <math.h>
+
+# define LENGTH 25
+# define LOWERLIMIT 2
+# define UPPERLIMIT 100
+# define LOWERLIMIT1 4
+# define UPPERLIMIT1 100
+
+int is_prime(int n);
+int main(void)
+{
+	int primes[LENGTH] = {0};
+	for(int i = LOWERLIMIT, j = 0; i <= UPPERLIMIT;  i++)
+	{
+		if(is_prime(i))
+		{
+			primes[j++] = i;
+		}
+	}
+	for(int i = LOWERLIMIT1;i <= UPPERLIMIT; i += 2)
+	{
+		for(int j = 0; j < LENGTH; j++)
+		{
+			for(int k = 0; k < LENGTH; k++)
+			{
+				if(primes[j] + primes[k] == i)
+				{
+					printf("%d == %d + %d\n", i, primes[j], primes[k]);
+				}
+			}
+		}
+	}
+	return 0;
+}
+int is_prime(int n)
+{
+	int sqrt_of_n = (int)sqrt(n);
+	int i, if_is_prime;
+	for(i = 2; i <= sqrt_of_n; i++)
+	{
+		if(n % i == 0) break;
+	}
+	if(i > sqrt_of_n)
+	{
+		if_is_prime = 1;
+	}
+	else
+	{
+		if_is_prime = 0;
+	}
+	return if_is_prime;
+}

level1/p08_hanoi/hanoi.c

@@ -0,0 +1,29 @@
+/*
+解决思想
+1.将A上面的n-1个通过B挪到C上
+2.将A上的另一个盘移到B上
+3.重复步骤即可实现
+4.采用递归实现
+*/
+# include <stdio.h>
+
+void move(int steps, char A, char B, char C);
+int main(void)
+{
+	int steps;
+	printf("请输入要移动的块数：");
+	scanf("%d", &steps);
+	move(steps, 'A', 'B', 'C');
+}
+void move(int steps, char A, char B, char C)
+{
+	if(steps == 1)
+	{
+		printf("%c --> %c\n", A, C);//当steps只有一个的时候直接从a移动到b上 
+	} else
+	{
+		move(steps - 1, A, C, B);//第n-1个要从a通过b移动到c
+		printf("%c --> %c\n", A, C);
+		move(steps - 1, B, A, C);//n-1个移动自后c变成了初始，c通过a移动到c 
+	}
+}