实验3

任务1

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#include <stdio.h>

char score_to_grade(int score);

int main(){
	int score;
	char grade;
	
	while (scanf_s("%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：将百分制整数分数转换为对应等级字符（A/B/C/D/E）；形参类型为 int ，返回值类型为 char 。
问题2：代码存在3个问题：① ans = "A"; 等写法类型不匹配，应使用单引号字符常量；② case 8 、 case 7 、 case 6 缺少 break ，会发生穿透导致逻辑错误；③缺乏对非法分数的前置校验，逻辑不严谨。

任务2

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#include <stdio.h>

int sum_digits(int n); 

int main() {
	int n;
	int ans;
	while (printf("Enter n: "), scanf_s("%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1：计算整数 n 的各位数字之和。
问题2：递归版本可以实现相同输出；循环版本采用迭代思想，空间复杂度O(1)，效率高；递归版本采用分治思想，代码简洁但空间复杂度O(log₁₀n)，存在函数调用开销。
任务3

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#include <stdio.h>
int power(int x, int n); 

int main() {
	int x, n;
	int ans;
	while (printf("Enter x and n: "), scanf_s("%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：计算整数 x 的 n 次方（ xⁿ ），采用快速幂递归算法实现。
问题2：x^n = {1, n=0; x*x^(n-1), n>0且n为奇数; [x^(n/2)]2, n>0且n为偶数}
任务4

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#include <stdio.h>


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

int main() {
	int a, b, c;
	while (scanf("%d%d%d", &a, &b, &c) != EOF)
		classify_triangle(a, b, c);

	return 0;
}

int classify_triangle(int a, int b, int c) {
	int i;
	if (a + b <= c || a + c <= b || b + c <= a)
		i = 1;
	else if (a == b && b == c)
		i = 2;
	else if (a * a + b * b == c * c || a * a + c * c == b * b || b * b + c * c == a * a)
		i = 3;
	else if (a == b || b == c || a == c)
		i = 4;
	else
		i = 5;

	switch (i) {
	case 1:
		printf("不能构成三角形\n\n");
		break;
	case 2:
		printf("等边三角形\n\n");
		break;
	case 3:
		printf("直角三角形\n\n");
		break;
	case 4:
		printf("等腰三角形\n\n");
		break;
	case 5:
		printf("普通三角形\n\n");
		break;
	}
	return 0;
}

任务5

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

int main() {
	int n, m;
	int ans;
	while (scanf_s("%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;

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

任务6

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#include <stdio.h>

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

int min(int a, int b, int c) {
    int min_val = a;
    if (b < min_val) min_val = b;
    if (c < min_val) min_val = c;
    return min_val;
}

int main() {
    int a, b, c;
    int ans;

    while (scanf_s("%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 i = min(a, b, c);

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

任务7

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#include <stdio.h>
#include <stdlib.h>
void print_charman(int n);
int main() {
    int n;
    printf("Enter n: ");
    while (scanf_s("%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");
    }
}