实验三

实验任务1 
 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <time.h>
 4 #include <windows.h>
 5 #define N 80
 6 
 7 void print_text(int line, int col, char text[]);  // 函数声明 
 8 void print_spaces(int n);  // 函数声明 
 9 void print_blank_lines(int n); // 函数声明 
10 
11 int main() {
12     int line, col, i;
13     char text[N] = "hi, April~";
14     
15     srand(time(0)); // 以当前系统时间作为随机种子
16     
17     for(i = 1; i <= 10; ++i) {
18         line = rand() % 25;
19         col =  rand() % 80;
20         print_text(line, col, text);
21         Sleep(1000);  // 暂停1000ms
22     }
23     
24     return 0; 
25 }
26 
27 // 打印n个空格 
28 void print_spaces(int n) {
29     int i;
30     
31     for(i = 1; i <= n; ++i)
32         printf(" ");
33 }
34 
35 // 打印n行空白行
36 void print_blank_lines(int n) {
37     int i;
38     
39     for(i = 1; i <= n; ++i)
40         printf("\n");
41  } 
42 
43 // 在第line行第col列打印一段文本 
44 void print_text(int line, int col, char text[]) {
45     print_blank_lines(line-1);      // 打印(line-1)行空行 
46     print_spaces(col-1);            // 打印(col-1)列空格
47     printf("%s", text);         // 在第line行、col列输出text中字符串
48 }

10秒内每秒在随机位置输出一个“hi，April~”

实验任务二

 1 // 利用局部static变量的特性，计算阶乘
 2 
 3 #include <stdio.h>
 4 long long fac(int n); // 函数声明
 5 
 6 int main() {
 7     int i, n;
 8 
 9     printf("Enter n: ");
10     scanf("%d", &n);
11 
12     for (i = 1; i <= n; ++i)
13         printf("%d! = %lld\n", i, fac(i));
14 
15     system("pause");
16     return 0;
17 }
18 
19 // 函数定义
20 long long fac(int n) {
21     static long long p = 1;
22     printf("p = %lld\n",p);//打印局部变量p的值 static变量保留上一个过程操作后的p（即为（n-1）!的值）作为初始p的值
23 
24     p = p * n;
25 
26     return p;
27 }

// 练习：局部static变量特性

#include <stdio.h>
int func(int, int);        // 函数声明

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

    p1 = func(k, m);    // 函数调用
    p2 = func(k, m);    // 函数调用
    printf("%d, %d\n", p1, p2);

    system("pause");
    return 0;
}

// 函数定义
//与分析结果一致
int func(int a, int b) {
    static int m = 0, i = 2;

    i += m + 1;
    m = i + a + b;

    return m;
}

一致

实验任务三

 1 #include <stdio.h>
 2 long long func(int n); // 函数声明
 3 
 4 int main() {
 5     int n;
 6     long long f;
 7 
 8     while (scanf("%d", &n) != EOF) {
 9         f = func(n); // 函数调用
10         printf("n = %d, f = %lld\n", n, f);
11     }
12 
13     return 0;
14 }
15 
16 // 函数定义
17 long long func(int n){
18     long long f;
19     if((n-1)>=0)
20         f = 2 * func(n-1) + 1;
21     if((n-1)==0)
22         f = 1;
23     if(n==0)
24         f = 0;
25 
26     return f;
27 }

实验任务四

 1 #include <stdio.h>
 2 int func(int n, int m);
 3 
 4 int main() {
 5     int n, m;
 6 
 7     while(scanf("%d%d", &n, &m) != EOF)
 8         printf("n = %d, m = %d, ans = %d\n", n, m, func(n, m));
 9     
10     return 0;
11 }
12 
13 // 函数定义
14 int func(int n, int m){
15     int ans;
16     if ( m > n )
17         ans = 0;
18     else if ( m == n )
19         ans = 1;
20     else if ( m < n ){
21         if ( m == 0 )
22             ans =1;
23         if ( m != 0 )
24             ans = func(n-1,m) + func(n-1,m-1);
25         }
26 
27     return ans;
28 }

实验任务五

#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 = 1.0;
    double k = 1.0 / x ; 
    if (y > 0){
        for(int i = 1; i <= y; i++)
            ans *= (double)x ;
        }
    if (y == 0)
        ans = 1.0 ;
    if (y < 0)
        for (int j = y;j < 0 ;j++)
            ans *= k ;
    return ans ;
}

实验任务六

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 void hanoi(unsigned int n, char from, char temp, char to);
 5 void moveplate(unsigned int n, char from, char to);
 6 
 7 int tot;
 8 
 9 int main() {
10     unsigned int n;
11     
12     while(scanf("%u", &n) != EOF) {
13         tot = 0;
14         
15         hanoi(n, 'A', 'B', 'C');
16         
17         printf("一共移动了%d次\n", tot);
18     }
19     
20     system("pause");
21     
22     return 0;
23 }
24 
25 void hanoi(unsigned int n, char from, char temp, char to) {
26     if(n == 1)
27         moveplate(n, from, to);
28     else {
29         hanoi(n - 1, from, to, temp);
30         moveplate(n, from, to);
31         hanoi(n - 1, temp, from, to);
32     }
33 }
34 
35 void moveplate(unsigned int n, char from, char to) {
36     tot ++;
37     printf("%u: %c --> %c\n", n, from, to);
38 }

实验任务七

 1 #include <stdio.h>
 2 #include <math.h>
 3 int is_prime(int n);
 4 int main(){
 5     int n = 20;
 6     int num1,num2,z;
 7     for (int f = 2;f <= n; f = f+2 ){
 8         for (int p = 2;p <= f/2;p++){
 9             if ( is_prime(p) == 1 ){
10                 num1 = p;
11                 z = f - num1;
12                 if( is_prime(z) == 1){
13                     num2 = z;
14                     printf("%d = %d + %d\n",f,num1,num2);
15                 }
16             }
17             if( is_prime(z) == 1){
18                 break;
19             }
20         }
21     }
22     return 0;
23 }
24 
25 int is_prime(int p){
26     int flag = 1;
27     for(int j = 2;j <= sqrt(p);j++){
28         if ( p % j == 0){
29             flag = 0;
30             break;
31         }
32     }
33     return flag;
34 }

实验任务八

#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) {
    long t = 0, t1 = 0;
    while(s > 0) {
        int tmp = s % 10;
        s /= 10;
        if(tmp % 2 == 1) t1 = t1 * 10 + tmp;
    }
    while(t1 > 0) {
        t = t * 10 + (t1 % 10);
        t1 /= 10;
    }
    return t;
}

 

posted @ 2023-04-02 22:51  银桥  阅读(63)  评论(0)    收藏  举报