实验三

task-1

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<windows.h>
#define N 80

void printText(int line, int col, char text[]);//函数声明
void printSpaces(int n);//函数声明
void printBlankLines(int n);//函数声明

int main()
{
    int line, col, i;
    char text[N] = "hi,May"; 

    srand(time(0));//以当前时间作为随机种子

    for (i = 1; i <= 10; i++)
    {
        line = rand() % 25;
        col = rand() % 80;
        printText(line, col, text);
        Sleep(1000);//暂停1000毫秒
    }
    return 0;
}

//打印n个空格
void printSpaces(int n)
{
    int i;
    for (i = 1; i <= n; ++i)
        printf(" ");
}

//打印n行空白行
void printBlankLines(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 printText(int line, int col, char text[])
{
    printBlankLines(line - 1);//打印n-1行空行
    printSpaces(col - 1);//打印n-1列空格
    printf("%s", text);
}

功能：在0-24行和0-79列随机输出10个hi,May

task-2.1

#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;
}

#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;
    printf("%lld\n", p);
    p = p * n;
    return p;
}

task-2.2

#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;
}

 task-3

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

task-4

#include<stdio.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()
{
    unsigned int n,m;
    while (scanf_s("%u", &n) != EOF)
    {
        hanoi(n, 'A', 'B', 'C');
        m = pow(2, n) - 1;
        printf("一共移动了%d次\n\n", m);
    }
    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("第%u个盘子:%c-->%c\n", n, from, to);
}

task-5

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int is_prime(int n);
int main()
{
    int m,j;
    for (m = 4; m <= 20; ++m)
    {
        if (m % 2 != 0)
            continue;
        else
        {
            for (j = 2; j < m - 1; ++j)
            {
                if (is_prime(j) && is_prime(m - j))
                {
                    printf("%d=%d+%d\n", m, j, m - j);
                    break;
                }
            }
        }
    }
    return 0;
}
int is_prime(int n)
{
    int i;
    if (n == 0 || n == 1)
        return 0;
    else
    {
        for (i = 2; i < n - 1; ++i)

        {
            if (n % i != 0)
                return 1; break;
        }
        while (i == n - 1)
            return 0;
    }
}

 task-6

#include<stdio.h>
long fun(long s);

int main()
{
    long s, t;
    printf("Enter a number:");
    while (scanf_s("%ld", &s) != EOF)
    {
        t = fun(s);
        printf("new number is:%ld\n\n", t);
        printf("Enter a number:");
    }
    return 0;
}
long fun(long s)
{
    long i, p = 0, n = 1;
    while (s != 0)
    {
        i = s % 10;
        if (i % 2 != 0)
        {
            p = n * i + p;
            n *= 10;
        }
        s = s / 10;
    }
    return p;
}