实验3

任务1

 1 #include <stdio.h>
 2 
 3 char score_to_grade(int score);  
 4 
 5 int main() {
 6     int score;
 7     char grade;
 8 
 9     while(scanf("%d", &score) != EOF) {
10         grade = score_to_grade(score);  
11         printf("分数: %d, 等级: %c\n\n", score, grade);
12     }
13 
14     return 0;
15 }
16 
17 char score_to_grade(int score) {
18     char ans;
19 
20     switch(score/10) {
21     case 10:
22     case 9:   ans = 'A'; break;
23     case 8:   ans = 'B'; break;
24     case 7:   ans = 'C'; break;
25     case 6:   ans = 'D'; break;
26     default:  ans = 'E';
27     }
28 
29     return ans;
30 }

1 判断分数等级  整形 字符形

2 有问题 A到D为字符串型  E为字符型 没有break会不断执行

任务2

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

1 让一个数的各个位数相加

2 能     第一种通过对一个数取余10后得到的数进行相加 第二种先判断是否为个位数，再通过递归

任务3

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

1 计算x的n次幂

2 是   

任务4

 1 #include <stdio.h>
 2 #include <math.h>
 3 
 4 int is_prime(int num);
 5 
 6 int main() {
 7     printf("100以内的孪生素数：\n"); 
 8     int count = 0; 
 9     for (int i = 2; i < 100; i++) {
10         if (is_prime(i) && is_prime(i + 2)) {
11             printf("%d  %d \n", i, i + 2);
12             count++;
13         }
14     }
15     printf("100以内的李生素数共有: %d 对\n", count);
16     return 0;
17 }
18 
19 int is_prime(int num) {
20     if (num <= 1) return 0; 
21     if (num == 2) return 1; 
22     if (num % 2 == 0) return 0; 
23     for (int i = 3; i <= sqrt(num); i += 2) {
24         if (num % i == 0) {
25             return 0;
26         }
27     }
28     return 1;
29 }

任务5

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 void hanoi(unsigned int n,char from,char temp,char to);
 4 void moveplate(unsigned int n,char from,char to);
 5 int moves=0;
 6 int main(){
 7     
 8     unsigned int n;
 9     while(scanf("%u",&n)!=EOF){
10         moves=0;
11         hanoi(n,'A','B','C');
12         printf("一共移动了%d次\n\n",moves);
13         
14         
15     }
16     
17     return 0;
18 }
19     void hanoi(unsigned int n,char from,char temp,char to){
20         
21         if(n==1){
22             moves++;
23             moveplate(n,from,to);
24         }
25         
26         
27         else
28         {
29             
30             hanoi(n-1,from,to,temp);
31             moves++;
32             moveplate(n,from,to);
33             hanoi(n-1,temp,from,to);
34             
35         }
36         
37     }
38     void moveplate(unsigned int n,char from,char to){
39         printf("%u:%c-->%c\n",n,from,to);
40         
41     }

任务6 

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

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

任务7

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 void print_charman(int n);
 5 
 6 
 7 int main() {
 8     int n;
 9 
10     printf("Enter n: ");
11     scanf("%d", &n);
12     print_charman(n); 
13        
14     return 0;
15 }
16 
17 void print_charman(int n){
18      int m = 0,i,j;
19     for (i = n; i >= 1; i--)
20      {
21          for (j = 0; j < m; j++)
22          {
23              printf("\t");
24          }
25          for (j = 0; j < 2 * i - 1; j++)
26          {
27              printf(" O\t");
28          }
29          printf("\n");
30          for (j = 0; j < m; j++)
31          {
32              printf("\t");
33          }
34          for (j = 0; j < 2 * i - 1; j++)
35          {
36              printf("<H>\t");
37          }
38          printf("\n");
39          for (j = 0; j < m; j++)
40          {
41              printf("\t");
42          }
43          for (j = 0; j < 2 * i - 1; j++)
44          {
45              printf("I I\t");
46          }
47          printf("\n"); 
48          m++;
49      }
50              
51          }