实验3

task1

实验源码

#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
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);

}

 程序功能，做出字符小人跑动的样子

task2.1

程序源码

//利用局部statuc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;
    
    p =p*n;
    printf("p=%lld\n",p);
    return p;
}

运行结果

task2.2

程序源码

//联系：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;
}

运行结果

task3

#include <stdio.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
//函数补足
long long func(int n){
    long long s;
    
    if(n==1)return 1;
      else
        return (2*func(n-1)+1);
        
    
    
    
        
}

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;
}

int func(int n, int m)
{
    if(n==m||m==0)
      return 1;
    else if(m>n)
      return 0;
    else if(m==1)
      return n;
    else 
      return func(n-1,m)+func(n-1,m-1);
}

运行结果

task5

程序源码

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

int main(){
    int x,y;
    double ans;
    
    while(scanf("%d%d",&x,&y)!=EOF){
        ans=mypow(x,y);
        printf("%d的%d次方；%g\n\n",x,y,ans);
        
    }
    
    return 0;
}

//函数定义
//函数补足
double mypow(int x,int y){
    double s=1.0;
    int i;
    if(y>0)
      s=x*mypow(x,y-1);
    if(y==0)
      
      s=1.0;
    
    if(y<0)
      s=mypow(x,y+1)/x;
    return s;  
}

运行结果

task6

程序源码

#include<stdio.h>
#include<stdlib.h>
void hanoi(unsigned int n, char from, char temp, char to);
void moveplate(unsigned int n, char from, char to);
int i=0;
int main()
{
    unsigned int n;
    while(scanf("%u",&n)!=EOF)
    {
    i=0;
    hanoi(n,'A','B','C');
    printf("一共移动%d次\n",i);
    }   
    return 0;
}
void hanoi(unsigned int n, char from, char temp, char to)
{
    if (n==1)
       moveplate(n,from,temp);
    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);
    i++;
}

运行结果

 task7

程序源码

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

运行结果

task8

程序源码

#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
// 待补足。。。
long fun(long s)
{
    long a,count=0,b=0;
    
    do
    {
        a = s % 10;
        if (a % 2 != 0)
        {
            b = b + a*pow(10,count);
            count++;
        }
        s = s / 10;
    } while (s != 0);
    return b;
}

运行结果