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[]); 
void print_spaces(int n); 
void print_blank_lines(int n); 
int main() {
    int line, col, i;
    char text[N] = "hi, November~";

    srand(time(0));
    for(i = 1; i <= 10; ++i) {
     ine = 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);                                                                           
}     

通过系统时间生成随机数，决定打印文本的几行几列

task2
#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;

}

task2
#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;
}

int func(int a, int b) {
    static int m = 0, i = 2;                                                                                    static是静态变量，再次进入函数时，使用上次的结果，变量始终存在。
    i += m + 1;
    m = i + a + b;
return m;
}

#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 f;
    
    if(n==0)
        f=0;
    else
    f=2*func(n-1)+1;
    return f;
} 

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

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

int mul(int n,int m)
{
    int result;
    
    if(n*m==0)
    result=0;
    else
    result=mul(n-1,m)+m;
return result;
    
}

#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()
{
    unsigned int n;
    int count;
    while(scanf("%u",&n)!=EOF){
        hanoi(n,'A','B','C');
        count=pow(2,n)-1;
    printf("一共移动了%d次",count);
    printf("\n");
}
    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);
}

#include<stdlib.h>
#include<math.h>
int is_prime(int a);
int main()
{
    int n=0;
    int m=0;
    int a;
    for(n=4;n<=20;n+=2)
    {
        for(a=2;a<=n/2;a++){
            if(is_prime(a)&&is_prime(n-a))
            {
                printf("%d=%d+%d",n,a,n-a);
                printf("\n");
            }
        }
    }

    
    return 0;
}



int is_prime(int a)
{
if(a==2)
{
    return 1;
}
int j=0;
    for(j=2;j<=sqrt(a);j++)
    {
        if(a%j==0)
        {
            return 0;
        }
    }
    return 1;
}

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

long fun(long s)
{
  int j;
    int i = 0;
    int count = 0;
    while (s)
    {
        j= s % 10;//得到一个尾数 
        if (j% 2 != 0)
        {
           i += j * pow(10, count) ;
            count++;
        }
        s = s / 10;
    }
    return i;
    
}