实验三

Task 1:

#include <stdio.h>
char score_to_grade(int score); // 函数声明
int main() {
  int score;
  char grade;
  while(scanf("%d", &score) != EOF) {
  grade = score_to_grade(score); // 函数调用
    printf("分数: %d, 等级: %c\n\n", score, grade);
 }
  return 0;
}
// 函数定义
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;
}

 

问题一:score_to_grade的作用是将分数转换成等级；形参类型是整型，返回值类型是字符型

问题二:没有break终止，程序会继续执行后续代码，ans会被后续分支的赋值覆盖，最后输出的全是E

Task 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;
}

 

 

问题一:sum_digits是计算输入的n的所有位数字之和

问题二:能实现同样的输出。原：迭代的思维；现：递归的思维，拆分成计算去掉个位后剩余数字的和，以及个位数字

Task 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;
}
}

 

问题一:power的功能是实现x的n次方

问题二:是。

 

 

Task4:

 

#include <stdio.h>
int is_prime(int n);
int main() 
{
    int count = 0;
    int n;
    printf("100以内的孪生素数:\n");
    for(n = 2; n <= 98; n++) 
    {
        if (is_prime(n) && is_prime(n + 2)) 
        {
            printf("%d %d\n", n, n + 2);
            count++;
        }
    }
    printf("\n100以内的孪生素数共有%d个。\n", count);
    return 0;
}

int is_prime(int n)
{
    int i;
    if (n<2)
    {
        return 0;
    }
    for(i=2;i*i <= n;i++)
    {
        if(n%i == 0)
            {
                return 0;
            }
    }
    return 1;
}

 

Task 5:

 迭代实现：

#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) 
        return 0;
    if (m > n - m) 
        m = n - m;
    int a = 1, b = 1;    
    if (m == 0 || m == n) 
        return 1;
    int i;
    for (i = 1; i <= m; i++) 
    {
        a *= (n - m + i);
        b *= i;
    }
    return a / b;
}

 递归实现：

 

 

 

#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) return 0;
    if (m == 0 || m == n) return 1;
    return func(n - 1, m) + func(n - 1, m - 1);
}

 Task 6:

#include <stdio.h>

int gcd(int a,int b,int c);

int main() 
{
    int a, b, c;
    int ans;
    while(scanf("%d%d%d", &a, &b, &c) != EOF) 
    {
        ans = gcd(a, b, c); 
        printf("最大公约数: %d\n\n", ans);
    }
    return 0;
}
int gcd(int a, int b, int c) 
{
    int min = a;
    if (b < min) min = b;
    if (c < min) min = c;
    int i;
    for (i = min; i >= 1; i--) 
    {
        if (a % i == 0 && b % i == 0 && c % i == 0) 
        {
            return i; 
        }
    }
    return 1; 
}

 

Task 7:

#include <stdio.h>
#include <stdlib.h>

void print_charman(int n);
int main() 
{
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); 
    return 0;
}
void print_charman(int n) 
{
    int i,j;
    for (  i = n; i >= 1; i--) 
    {
        for (  j = 0; j < n - i; j++) 
        {
            printf("\t");
        }
        for (  j = 0; j < (2*i-1); j++) 
        {
            printf(" O ");
            if (j < (2*i-1) - 1) 
            {
                printf("\t");
            }
        }
        printf("\n");
        for (  j = 0; j < n - i; j++) 
        {
            printf("\t");
        }
        for (  j = 0; j < (2*i-1); j++) 
        {
            printf("<H>");
            if (j < (2*i-1) - 1) 
            {
                printf("\t");
            }
        }
        printf("\n");
        for (  j = 0; j < n - i; j++) 
        {
            printf("\t");
        }
        for (  j = 0; j < (2*i-1); j++) 
        {
            printf("I I");
            if (j < (2*i-1) - 1) 
            {
                printf("\t");
            }
        }
        printf("\n");
        if (i > 1) {
            printf("\n");
        }
    }
}