实验3

task1

程序源码

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

void print_text(int line, int col, char text[]);
void print_spaces(int n);
void print_blank_lines(int n);

int main()
{
    int line, col, i;
    char text[N] = "hi, April~";
    
    srand(time(0));
    
    for(i = 1; i <= 10; ++i)
    {
        line = rand() % 25;
        col = rand() % 80;
        print_text(line, col, text);
        Sleep(1000);
    }
    
    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);
     print_spaces(col-1);
     printf("%s", text);
  } 
 
 

问题回答：这个程序实现的功能是在0~24行，0~79列随机取两个数，在第line行col列输出hi,April，重复运行十次，每出现一次hi，April则打印line-1行和col-1列空格，并且每次运行中间间隔1000ms，以此做出字符小人跑的动画效果。

 

task2

task2-1

程序源码

//利用局部static变量的特性，计算阶乘

#include <stdio.h>
long long fac(int n);

int main()
{
    int i, n;
    printf("Enter n: ");
    scanf("%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("p = %lld\n", p);
     p = p*n;
     return p;
  } 
  

程序运行截图

task2-2

程序源码

// 练习：局部static变量特性 

#include <stdio.h>
int func(int, int);

int main()
{
    int k = 4, m = 1, p1, p2;
    
    p1 = func(k, m);
    p2 = func(k, m);
    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;
 } 

程序运行截图

问题回答：static变量第特征是保持变量始终存在，再次进入该函数时，使用上次的结果。

 

task3

程序源码

#include <stdio.h>
#include <math.h>

long long func(int n);

int main()
{
    int n;
    long long f;
    
    while (scanf("%d", &n) != EOF)
    {
        f = func(n);
        printf("n = %d, f = %lld\n", n, f);
    }
    
    return 0;
}

//函数定义
long long func(int n)
{
    long long ans;
    if(n==0)
        ans=0;
    else
        ans=(func(n-1)+1)*2-1;
    
    return ans;
 }  

程序运行截图

 

 

task4

程序源码

#include <stdio.h>
int func(int n, int m);

int main()
{
    int n, m;
    
    while(scanf("%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;
    
    if(m==n||m==0)
        ans = 1;
    else if(m>n)
        ans = 0;
    else
        ans = func(n-1, m) + func(n-1, m-1);
        
    return ans;
 } 

程序运行截图

 

 

task5

task5-1

程序源码

#include <stdio.h>
double mypow(int x, int y);  //函数声明

int main()
{
    int x, y;
    double ans;
    
    while(scanf("%d%d", &x, &y) != EOF)
    {
        ans = mypow(x, y);  //函数调用
        printf("%d的%d次方：%g\n\n", x, y, ans); 
    }
    
    return 0;
 } 
 
 //函数定义
 double mypow(int x, int y)
 {
     int i;
    double ans=1;
     
    if(x==1||y==0)
         ans = 1;
     if(y > 0){
         for(i = 1; i <= y; ++i)
         ans = ans*x;}
     if(y < 0){
         for(i = -1; i >= y; --i)
         ans = ans/x;}
         
     return ans;
  } 

程序运行截图

task5-2

程序源码

#include <stdio.h>
double mypow(int x, int y);  //函数声明

int main()
{
    int x, y;
    double ans;
    
    while(scanf("%d%d", &x, &y) != EOF)
    {
        ans = mypow(x, y);  //函数调用
        printf("%d的%d次方：%g\n\n", x, y, ans); 
    }
    
    return 0;
 } 
 
 //函数定义
 double mypow(int x, int y)
 {
     double ans;
     
     if(x==1||y==0)
         ans = 1;
     if(y>0)
         ans = x*mypow(x, y-1);
     if(y<0)
         ans = mypow(x, y+1)/x;
         
     return ans;
  } 

程序运行截图

 

 

task6

程序源码

#include <stdio.h>
void hanoi(unsigned int n, char form, char temp, char to);
void moveplate(unsigned int n, char from, char to);

int count = 0; 

int main()
{
    unsigned int n;
    int count1=0;
    
    while(scanf("%u", &n)!=EOF){
        count = count - count1;
        hanoi(n, 'A', 'B', 'C');
        
        printf("\n");
        printf("一共移动了%d次.\n\n", count); 
        
        count1 = count;  
    }
    
    return 0;
}

//递归函数
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);
            
        }
} 

//移动函数
void moveplate(unsigned int n, char from, char to){
    printf("%u: %c --> %c\n", n, from, to);
    count++;    
}

程序运行截图

 

 

task7

程序源码

#include<stdio.h>
#include<math.h>

int is_prime(int a);
int main()
{
    int s, i;
    while(scanf("%d", &s) != EOF){
        for(i=2; i<=1.0*s/2; i++){
            if(is_prime(i)&&is_prime(s-i))
            {printf("%d = %d + %d\n", s, i, s-i);
            break;
            }
        }
    }
    return 0;
}

int is_prime(int a){
    int i, m;
    
    m = sqrt(1.0*a);
    
    for(i=2; i<=m; i++)
    if(a%i == 0)
        return 0;
    if(i>a)
        return 1;
}

程序运行截图

 

 

task8

程序源码

#include <stdio.h>
#include <math.h>
long func(long s);  //函数声明

int main()
{
    long s, t;
    
    printf("Enter a number: ");
    while(scanf("%ld", &s) != EOF)
    {
        t = func(s);  //函数调用
        printf("new number is: %ld\n\n", t);
        printf("Enter a number: "); 
    }
    
    return 0;
 } 
 
 //函数定义
 long func(long s)
 {
     int a=0, b=0, m, n;
    while(s != 0)
    {
        m = s%10;
        if(m%2 != 0)
            a = a*10 + m;
            s = s/10;
    }
    
    while(a != 0)
    {
        n = a%10;
        b = b*10 + n;
        a = a/10;
    }
    
    return b;
  } 

程序运行截图