实验3

任务1

原代码

 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     
15     return 0;
16 }
17 
18 char score_to_grade(int score){
19     char ans;
20     
21     switch(score/10){
22     case 10:
23     case 9: ans='A';break;
24     case 8: ans='B';break;
25     case 7: ans='C';break;
26     default: ans='E';
27     }
28     
29     return ans;
30 }

图片

判断等级   int char

有，没有break语句会继续执行下面指令

任务2

源代码

#include<stdio.h>

int sum_digits(int n);

int main(){
    int n;
    int ans;
    
    while(printf("Enter n:"),scanf("%d",&n)!=EOF){
        ans=sum_digits(n);
        printf("n=%d,ans=%d\n\n",n,ans);
    }
    
    return 0;
}

int sum_digits(int n){
    int ans=0;
    while(n!=0){
        ans+=n%10;
        n/=10;
    }
    
    return ans;
}  //迭代 
/*int sum_digits(int n){
    if(n<10)
        return n;
        
return sum_digits(n/10)+n%10;
}//递归 */

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将各个位上的数相加求和

可以，迭代和递归的区别

任务3

源代码

 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 }

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x的n次方

任务4

源代码

 1 #include<stdio.h>
 2 int is_prime(int n);
 3 
 4 int main(){
 5     int n,i;
 6     i=0;
 7     for(n=2;n<=100;n++)
 8     {
 9         if(is_prime(n)==1&&is_prime(n+2)==1){
10             printf("%d %d\n",n,n+2);
11             i+=1;
12             }
13     }
14     printf("100以内的孪生素数共有%d个",i);
15     
16     return 0; 
17 }
18 
19 is_prime(int n){
20     int m,ans;
21     for(m=2;m<=n/2;m++)
22         if(n%m==0)
23         break;
24     if(m>n/2)
25         ans=1;
26     else
27         ans=0;
28         
29     return ans;
30     
31 }

图片

任务5

源代码

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

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任务6

递归

源代码

 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 /*int func(int n,int m){
16     int a,b,c,d,ans;
17     b=d=1;
18     for(a=0;a<=m-1;a++)
19         b=b*(n-a);
20     for(c=1;c<=m;c++)
21         d=d*c;
22     ans=b/d;
23     return ans;
24 }//迭代 */
25 int func(int n,int m){
26     if(n<m)
27         return 0;
28     if(m==0)
29         return 1;
30      else if(m==1)
31         return n;
32     else
33         return func(n-1,m)+func(n-1,m-1);
34 }//递归 

图片

迭代

源代码

 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 int func(int n,int m){
16     int a,b,c,d,ans;
17     b=d=1;
18     for(a=0;a<=m-1;a++)
19         b=b*(n-a);
20     for(c=1;c<=m;c++)
21         d=d*c;
22     ans=b/d;
23     return ans;
24 }//迭代 
25 /*int func(int n,int m){
26     if(n<m)
27         return 0;
28     if(m==0)
29         return 1;
30      else if(m==1)
31         return n;
32     else
33         return func(n-1,m)+func(n-1,m-1);
34 }//递归 */

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任务7

源代码

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 char print_charman(int n);
 5 
 6 int main() {
 7     int n;
 8 
 9     printf("Enter n: ");
10     scanf("%d", &n);
11     print_charman(n);
12 
13     return 0;
14 }
15 char print_charman(int n)
16 {
17     int i,a,b,c,d,e;
18     e=n;
19     for(i=1;i<=e;i++)
20     {
21          for(d=1;d<=e-n;d++)
22        printf("      ");
23        for(a=1;a<=(2*n-1);a++)
24        printf("  O   ");
25        printf("\n");
26         for(d=1;d<=e-n;d++)
27        printf("      ");
28        for(b=1;b<=(2*n-1);b++)
29        printf(" <H>  ");
30        printf("\n");
31         for(d=1;d<=e-n;d++)
32        printf("      ");
33        for(c=1;c<=(2*n-1);c++)
34        printf(" I I  ");
35        printf("\n");
36        n--;
37     }
38     
39 }

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