实验3 C语言控制语句应用编程

#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) {
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); 
}

利用随机数改变下个字符串出现的位置

实验二

#include <stdio.h>
#include <stdlib.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));
    system("pause"); 
    return 0;
}
long long fac(int n) {
static long long p = 1;
printf("p = %lld\n", p);
p = p * n;
return p;
}

#include <stdio.h>
#include <stdlib.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);
system("pause");
return 0;
}
int func(int a, int b) {
static int m = 0, i = 2;
i += m + 1;
m = i + a + b;
return m;
}

预期8 17

用于声明一个内存不会被收回的地址了,内存中的数据在程序运行时都不改变。

实验三

#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 fun(int n);
long long func(int n)
{
    return fun(n)-1;

}
long long fun(int n)
{
    if(n==0)
        return 1;
    return 2*fun(n-1);
}

 实验四

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

实验五

#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)
{
if (n==0)
return 0;
return m+mul(n-1,m);
}

 

实验六

#include <stdio.h>
#include <stdlib.h>
int m=0;
void move(int n,char a,char c)
{
    printf("第%d个盘子,从%c移动到%c\n",n,a,c);
    m++;
}
void hanoi(int n,char a,char b,char c)
{
    if (n==1)
        move(n,a,c);
    else
    {
    hanoi(n-1,a,c,b);
    move(n,a,c);
    hanoi(n-1,b,a,c);
    }
}
int main()
{
    int n;
    char a='A',b='B',c='C';
    while(scanf("%d",&n)!=EOF)
    {
        hanoi(n,a,b,c);
    printf("共移动了%d次\n",m);
    }
    system("pause");
    return 0;
}

 

 

实验七

#include <stdio.h>
#include <stdlib.h>
int is_prime(int);
int main() 
{
     int sum,a,b;
     for (sum=2;sum<=20;sum=sum+2)
     {
         for(a=1;a<=17;a++)
         {
             if(is_prime(a))
             {
                 for(b=1;b<=17;b++)
                 {
                     if(is_prime(b)&&a<=b)
                     {
                         if(sum==a+b)
                         {
                             printf("%d =%d + %d",sum,a,b);
                             printf("\n");
                         }

                     }
                 }
             }
         }
     }
     system("pause");
}
int is_prime(int n)
{
    int i;
    if (n <= 1)
        return 0;
    for (i = 2; i < n; i++)
        if (n % i == 0)
            return 0;
    return 1;
}

实验八

#include <stdio.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 x[100];
    long a;
    int i=0,b;
    for(;s!=0;s=s/10)
    {
        a=s%10;
        if(a%2!=0)
        {
            x[i]=a;
            i++;
        }
    }
    
    for(b=i-1;b>=0;b--)
        s=s*10+x[b];
    return s;
}

posted @ 2022-11-05 19:03    阅读(59)  评论(0)    收藏  举报