实验3

任务1

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

 功能是将成绩转化为ABCDE,形参是整型，返回值是字符型

不能修改，原因是没有break，程序会一直运行到default

任务2

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

 问题1：函数的功能是计算一个数字每位数字之和

问题2：可以实现同等效果，一个是迭代，一个是递归

任务3

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

 问题1：函数功能是实现次方的运算

问题2：是递归函数 数学模型是：当n为奇数x^n=x*x^(n-1)   当n为偶数时，x^n=[x^(n/2)]^2

任务4

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

任务5

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

任务6

迭代：

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

 递归：

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

任务7

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 void print_charman(int n)
 4 {
 5     int i,j,a;
 6     for(i=n;i>=1;i--)
 7     {
 8         for(a=n-i;a>0;a--)
 9             printf("\t");
10         for(j=2*i-1;j>=1;j--)
11             printf(" O\t");
12         printf("\n");
13         for(a=n-i;a>0;a--)
14             printf("\t");
15         for(j=2*i-1;j>=1;j--)
16             printf("<H>\t");
17         printf("\n");
18         for(a=n-i;a>0;a--)
19             printf("\t");
20         for(j=2*i-1;j>=1;j--)
21             printf("I I\t");
22         printf("\n");
23         printf("\n");
24         printf("\n");
25     }
26 }
27 
28 int main(){
29     int n;
30     printf("Enter n:");
31     scanf("%d",&n);
32     print_charman(n);
33     system("pause");
34     return 0;
35 }