实验3

试验任务1

程序源码

#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; 
}


void print_spaces(int n) {
    int i;
    
    for(i = 1; i <= n; ++i)
        printf(" ");
}


void print_blank_lines(int n) {
    int i;
    
    for(i = 1; i <= n; ++i)
        printf("\n");
 } 


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：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

试验任务2

程序源码1.1

#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;

}
long long fac(int n){
 static long long p=1;
 p=p*n;
 return p;
}

 

程序源码1.2

#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;

}
long long fac(int n){
    static long long p=1;
    printf("p=%lld\n",p);
    p=p*n;
    return p;
}

   

程序源码2

#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",p1,p2);
    return 0;
    
}
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2

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

实验任务3

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

运行结果

实验任务4

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

运行结果

 

实验任务5

程序源码（不适用递归）

#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;
} 
double mypow(int x,int y)
{
    int i;
    double s=1;
    while(y>0){
    s=s*x;
    y=y-1;}
    while(y<0){
    s=s/x;
    y=y+1;}
    
    return (s);
}

运行结果

 

使用递归

#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;
} 
double mypow(int x,int y)
{
    double s=1;
    if(y==0)
    {s=1;}
    else
    if(y>0)
    s=x*mypow(x,y-1);
    if(y<0)
    s=mypow(x,y+1)/x;
    return (s);
}

 

运行结果

 实验任务6

#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 cishu(unsigned int n);
int main(){
    unsigned int n,s;
    while((scanf("%u",&n)!=EOF)){
    hanoi(n,'A','B','C');
    s=pow(2,n)-1;
    printf("一共移动了%d次\n",s);}
    return 0;
}
void hanoi(unsigned int n,char from, char temp, char to)
{
    if(n==1)
      {moveplate(n,from,to);
      n==1;}
    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);
    
}

运行结果

 实验任务7

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;
}

运行结果

 

实验任务8

程序源码

#include<stdio.h>
#include<math.h>
long func(long s);
int main()
{
    long s, t;
    printf("Enter a number:");
    while (scanf("%ld",&s)!=EOF){
    t=func(s);
    printf("new number is:%ld\n\n",t);
    printf("Enter a number:");
    }
    return 0;
}
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;
}

运行结果