实验3

任务1

#include <stdio.h>

char score_to_grade(int score);  

int main() {
    int score;
    char grade;

    while(scanf("%d", &score) != EOF) {
        grade = score_to_grade(score);  
        printf("分数: %d, 等级: %c\n\n", score, grade);
    }

    return 0;
}


char score_to_grade(int score) {
    char ans;

    switch(score/10) {
    case 10:
    case 9:   ans = 'A'; break;
    case 8:   ans = 'B'; break;
    case 7:   ans = 'C'; break;
    case 6:   ans = 'D'; break;
    default:  ans = 'E';
    }

    return ans;
}

 函数score_to_grade的功能是根据输入的分数返回对应的等级。形参是整形，返回值是字符型。

有问题，没有break，满足条件的代码和后续代码都会执行。

 

任务2

#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;
}

函数sum_digits的功能是计算输入值的各个位数之和。

能实现等同的效果。第一种算法是迭代，第二种算法是递归。

 

任务3

#include <stdio.h>

int power(int x, int n);    

int main() {
    int x, n;
    int ans;

    while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {
        ans = power(x, n);  
        printf("n = %d, ans = %d\n\n", n, ans);
    }
    
    return 0;
}


int power(int x, int n) {
    int t;

    if(n == 0)
        return 1;
    else if(n % 2)
        return x * power(x, n-1);
    else {
        t = power(x, n/2);
        return t*t;
    }
}

函数power的功能是计算x的n次方。

是递归函数，

 

任务4

#include<stdio.h>
#include<math.h>
int is_prime(int n);
int main(){
    int i,count=0;
    printf("100以内的孪生素数:\n");
    for(i=2;i<=100;++i){
        if(is_prime(i)&&is_prime(i+2)){
            printf("%d %d\n",i,i+2);
            count++;
        }
    }
        printf("100以内的孪生素数共有%d个\n",count);
        return 0;
    
}
int is_prime(int n){
    int j;
    for(j=2;j<=sqrt(n);++j){
        if(n%j==0)
        return 0;
    }
    return 1;
}

 

任务5

#include<stdio.h>
 int cnt=0;
 void hanoi(unsigned int n,char from,char temp,char to);
 void moveplate(unsigned int n,char from,char to);
  
  int main()
  {
     unsigned int n;
     while(scanf("%d",&n) !=EOF){
     
        cnt=0;
        hanoi(n,'A','B','C');
        printf("\n");
        printf("一共移动了%d次\n",cnt);
    }
     
     return 0;
 }
 
 void hanoi(unsigned int n,char from,char temp,char to)
 {
    if(n==1)
       moveplate(n,from ,to);
    else
    {
       hanoi(n-1,from,to,temp);
       moveplate(n,from,to);
       hanoi(n-1,temp,from,to);
 }
 }
 
 void moveplate(unsigned int n,char from,char to)
 {
     printf("%u:%c-->%c\n",n,from,to);
          cnt++;
 }

 

任务6

#include <stdio.h>
int func(int n, int m); 

int main() {
    int n, m;
    int ans;

    while(scanf("%d%d", &n, &m) != EOF) {
        ans = func(n, m);   
        printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
    }
        
    return 0;
}
int func(int n,int m){
    int a,b=1,c,d=1,ans;
    for(a=n;a>=n-m+1;--a)
    b=b*a;
    for(c=m;c>=1;--c)
    d=d*c;
    ans=b/d;
    return ans;
}

#include <stdio.h>
int func(int n, int m); 

int main() {
    int n, m;
    int ans;

    while(scanf("%d%d", &n, &m) != EOF) {
        ans = func(n, m);   
        printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
    }
        
    return 0;
}
int func(int n,int m){
    if(m==n||m==0)
     return 1;
    if(m>n)
      return 0;
    if(n!=0&&m!=0)
      return func(n-1,m)+func(n-1,m-1);
}

 

任务7

int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); 
       
    return 0;
}

void print_charman(int n){

    int i;
    int j;
     for(i = n;i >= 1;i --)
     {
        int a;
        for(a = 1;a <= n-i;a ++)
        printf("\t");
        
        for(j = 1; j <= 2*i-1;j ++)
        printf(" O \t");
        printf("\n");
        for(a = 1;a <= n-i;a ++)
        printf("\t");
        
        for(j = 1; j <= 2*i-1;j ++)
        printf("<H>\t");
        printf("\n");
        for(a = 1;a <= n-i;a ++)
        printf("\t");
        
        for(j = 1; j <= 2*i-1;j ++)
        printf("I I\t");
        printf("\n");
     }
 }