实验3

task1

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <windows.h>
#define N 80
void print_text(int line, int col, char text[]); // 函数声明
void print_spaces(int n); // 函数声明
void print_blank_lines(int n); // 函数声明
int main() {
	int line, col, i;
	char text[N] = "hi, November~";
	srand(time(0)); // 以当前系统时间作为随机种子
	for (i = 1; i <= 10; ++i) {
		line = rand() % 25;
		col = rand() % 80;
		print_text(line, col, text);
		Sleep(1000); // 暂停1000ms
	}
	return 0;
}
// 打印n个空格
void print_spaces(int n) {
	int i;
	for (i = 1; i <= n; ++i)
		printf(" ");
}
// 打印n行空白行
void print_blank_lines(int n) {
	int i;
	for (i = 1; i <= n; ++i)
		printf("\n");
}
// 在第line行第col列打印一段文本
void print_text(int line, int col, char text[]) {
	print_blank_lines(line - 1); // 打印(line-1)行空行
	print_spaces(col - 1); // 打印(col-1)列空格
	printf("%s", text); // 在第line行、col列输出text中字符串
}

　　

task2.1

 

// 利用局部static变量的特性，计算阶乘
#include <stdio.h>
long long fac(int n); // 函数声明
int main() {
	int i, n;
	printf("Enter n: ");
	scanf_s("%d", &n);
	for (i = 1; i <= n; ++i)
		printf("%d! = %lld\n", i, fac(i));
	return 0;
}
// 函数定义
long long fac(int n) {
	static long long p = 1;
	p = p * n;
	return p;
}

　　

task2.2

// 练习：局部static变量特性
#include <stdio.h>
int func(int, int); // 函数声明
int main() {
	int k = 4, m = 1, p1, p2;
	p1 = func(k, m); // 函数调用
	p2 = func(k, m); // 函数调用
	printf("%d, %d\n", p1, p2);
	return 0;
}
// 函数定义
int func(int a, int b) {
	static int m = 0, i = 2;
	i += m + 1;
	m = i + a + b;
	return m;
}

　　

task3

#include <stdio.h>
long long func(int n); // 函数声明
int main() {
	int n;
	long long f;
	while (scanf_s("%d", &n) != EOF) {
		f = func(n); // 函数调用
		printf("n = %d, f = %lld\n", n, f);
	}
	return 0;
}
// 函数定义
long long func(int n) 
{
	int i =1, ans;
	if (n == 1)
		ans = 1;
	else
		ans = 2 * func(n - 1) + 1;
   return(ans);
}

　　

task4

#include <stdio.h>
int func(int n, int m);
int main() {
	int n, m;
	while (scanf_s("%d%d", &n, &m) != EOF)
		printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m));
	return 0;
}
// 函数定义
int func(int n, int m)
{
	int ans, time = 0;
	if (n < m)
		ans = 0;
	else if (m == 1)
		ans = n + time;
	else if (m == 0)
		ans = 1;
	else
		ans = n * func(n - 1, m - 1) / m;
	time = time + 1;

	return(ans);

}

　　

task5

#include <stdio.h>
int mul(int n, int m);
int main() {
	int n, m;
	while (scanf_s("%d%d", &n, &m) != EOF)
		printf("%d * %d = %d\n", n, m, mul(n, m));
	return 0;
}
// 函数定义
int mul(int n, int m) {
	int ans = 0;

	if (n > m) {
		if (m == 1)
			ans = n;
		else if (m == 0)
			ans = 0;
		else
			ans = n + mul(n, m - 1);
	}
	else if (n <= m) {
		if (n == 1)
			ans = m;
		else if (n == 0)
			ans = 0;
		else
			ans = n + mul(n - 1, m);
	}

	return(ans);
}

　　

task6

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
void hanoi(unsigned int n, char from, char temp, char to);
void moveplate(unsigned int n, char from, char to);
int main()
{
	int time;
	unsigned int n;
	scanf_s("%u", &n);
	hanoi(n, 'A', 'B', 'C');
	time = pow(2, n) - 1;
	printf("一共移动了%d次", time);
	system("pause");
	return 0;
}
void hanoi(unsigned int n, char from, char temp, char to)
{
	if (n == 1)
		moveplate(n, from, to);
	else
	{
		hanoi(n - 1, from, to, temp);
		moveplate(n, from, to);
		hanoi(n - 1, temp, from, to);
		
	}
}
void moveplate(unsigned int n, char from, char to)
{
	printf("%u:%c-->%c\n", n, from, to);
	
}

　　

task7

#include<stdio.h>
#include<math.h>
int is_prime(long int a);
int main()
{
	int i, j;
	long int n;
	while (1) {
		scanf_s("%ld", &n);
		for (i = 2; i <= n / 2; i++)
		{
			if (is_prime(i) && is_prime(n - i))
			{
				printf("%ld=%d+%d\n", n, i, n - i); break;
			}
		}
	}
	return 0;
}
int is_prime(long int a)
{
	int j;
	double k;
	k = sqrt(1 * a);
	for (j = 2; j <= k; j++)
		if (a % j == 0)
			return 0;
	if (j > k)
		return 1;
}

　　

 

task8

 

#include <stdio.h>
#include<math.h>
long fun(long s); // 函数声明
int main() {
	long s, t;
	printf("Enter a number: ");
	while (scanf_s("%ld", &s) != EOF) {
		t = fun(s); // 函数调用
		printf("new number is: %ld\n\n", t);
		printf("Enter a number: ");
	}
	return 0;
}
// 函数定义
long fun(long s) {
	int n = 1, m, t = 0;
	while (s > 0)
	{
		m = s % 10;
		if (m % 2 != 0)
		{
			t = t + n * m;
			n = n * 10;
		}
		s = s / 10;
     }

	return t;
}