实验3

task1

#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;
}

问题1：将输入的成绩转换为等级

问题2：缺少break语句

task2

#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：计算输入整数n的各位数字之和

问题2：能实现。原本使用while循环方式，改后使用递归方式

task3

#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x的n次幂

问题2：是，

task4

#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 i = 3; i + 2 <= 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;
}

task5

#include <stdio.h>

void hanoi(int n, char a, char b, char c, int* count) {
    if (n == 1) {
        printf("%d: %c --> %c\n", n, a, c);
        (*count)++;
    }
    else {
        hanoi(n - 1, a, c, b, count);
        printf("%d: %c --> %c\n", n, a, c);
        (*count)++;
        hanoi(n - 1, b, a, c, count);
    }
}

int main() {
    int n;
    while (scanf_s("%d", &n) != EOF) {
        int count = 0;
        hanoi(n, 'A', 'B', 'C', &count);
        printf("一共移动了%d次.\n", count);
    }
    return 0;
}

task6

#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) {
    int result = 1;
    for (int i = 1; i <= m; i++) {
        result *= (n - (m - i));
        result /= i;
    }
    return result;
}

 

task7

#include <stdio.h>

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

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;
}