实验3

task1.c

#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;
}

answer1:将分数转换为对应的等级，int   ，  char

answer2:会报错，ans是char类型的字符型，字母应该用单引号包裹，而双引号表示字符串；

同时后续没有break； 这使得若已经执行该语句，后续语句仍然会执行。

 

task2.c

#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;
}

answer1:计算某个整数的个位数字之和

answer2:可以

原方式为迭代，通过while循环锥刺提取个位数字并累加，

问中为递归，将其分解为数字最后一位与剩余各部分数字之和，进行分解与累加。

 

task3.c

#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;
    }
}

answer1:将x变为x的n次方

answer2：是，通过递归模式将幂运算的规模逐渐缩小。

 

 

task4.c

#include <stdio.h>

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

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

 

 

task5.c

diedai:

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

digui:

#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 < 0 || m > n)
      {
         return 0;
     }
     if (m == 0 || m == n)
      {
         return 1;
     }

     return func(n - 1, m) + func(n - 1, m - 1);
 }

 

 

 

 

task6.c

 

#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;

     for (int i = min; i >= 1; i--) 
     {
         if (a % i == 0 && b % i == 0 && c % i == 0)
          {
             return i;
         }
     }
     return 1;
 }

 

 

task7.c

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

void print_charman(int n);

int main() 
{
    int n;

    printf("input n: ");
    scanf("%d", &n);
    print_charman(n);

    return 0;
}


void print_charman(int n)
 {
    int i, j, k;
    

    for (i = n; i >= 1; i--) 
    {

        int charman_count = 2 * i - 1;

        int tab_count = (n - i);
        
        for (k = 0; k < tab_count; k++) 
        {
            printf("\t");
        }
        for (j = 0; j < charman_count; j++) 
        {
            printf(" 0 "); 
            if (j < charman_count - 1) {
                printf("\t");
            }
        }
        printf("\n");

        for (k = 0; k < tab_count; k++)
         {
            printf("\t");
        }
        for (j = 0; j < charman_count; j++) 
        {
            printf("<H>");
            if (j < charman_count - 1) 
            {
                printf("\t");
            }
        }
        printf("\n");

        for (k = 0; k < tab_count; k++) {
            printf("\t");
        }
        for (j = 0; j < charman_count; j++) {
            printf("I I");
            if (j < charman_count - 1) {
                printf("\t");
            }
        }
        printf("\n");
        
        if (i > 1) {
            printf("\n");
        }
    }
}

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