实验3

 #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);  
         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;
 }

问题1把输入的百分之分数转化为对应登记的字符，形参类型：int整型；返回值char字符型

       2所有的 case 无break最终所有结果都是E；

二

#include <stdio.h>

int sum_digits(int n);  // 函数声明

int main() {
    int n;
    int ans;

    while(printf("Enter n: "), scanf("%d", &n) != EOF) {
        ans = sum_digits(n);    // 函数调用
        printf("n = %d, ans = %d\n\n", n, ans);
    }

    return 0;
}

// 函数定义
int sum_digits(int n) {
    int ans = 0;

    while(n != 0) {
        ans += n % 10;
        n /= 10;
    }

    return ans;
}

问题1 计算并返回一个整数n的各个位数之和

      2能，原版本是迭代，新版本是递归，

三

#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1 power的功能是递归计算并返回x的n次幂

      2 是递归

四

#include <stdio.h>
int classify_triangle(int a, int b, int c);
int main(){
    int a, b, c, x;
    while(scanf("%d%d%d", &a, &b, &c) != EOF){
        x = classify_triangle(a, b, c);
        switch(x){
        case 0: printf("不能构成三角形\n\n");
            break;

        case 1:printf("普通三角形\n\n");
            break;

        case 2:printf("等边三角形\n\n");
            break;

        case 3:printf("等腰三角形\n\n");
            break;

        case 4:printf("直角三角形\n\n");
            break;
        }
    }

        return 0;
    }

int classify_triangle(int a, int b, int c){
    if(a >= b + c || b >= a + c || c >= a + b)
        return 0;

    if(a == b && b == c)
        return 2;

    if(a == b || a == c || c == b)
        return 3;

    if(a*a == b*b + c*c || b*b == a*a +c*c || c*c == a*a + b*b)
        return 4;

    else
        return 1;
}

#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);
}

#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", ans);
    }
    return 0;
}

int gcd(int a, int b, int c) {
    int min = a;
    int i;

    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;
        }
    }
}

#include <stdio.h>
 #include <stdlib.h>
 void print_charman(int n);
 int main() {
         int n;
         printf("Enter n: ");
         while (scanf("%d", &n) != EOF) {
                 printf("input n: %d\n", n);
                 print_charman(n);
                 printf("\nEnter n: ");

    }
         return 0;

}

 void print_charman(int n) {
         for (int i = 1; i <= n; i++) {
                 for (int j = 1; j < i; j++) {
                         printf("\t");

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

        }
                 printf("\n");

                     for (int j = 1; j < i; j++) {
                         printf("\t");

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

        }
                 printf("\n");

                     for (int j = 1; j < i; j++) {
                         printf("\t");

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

        }
                 printf("\n");

    }

}