实验3

task1.c

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#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：根据输入的分数转换为相应的等级字母
形参类型：int类型
返回值类型：char类型
问题2：A,B,C,D,E常量字符应该用单引号，缺少break语句

task2.c

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#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1：对n中的各个数字进行求和
问题2：能
区别：原方式为迭代的算法思维，新的方式为递归的算法思维

task3.c

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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("%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次方
问题2：是

task4.c

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#include <stdio.h>

int is_prime(int n) {
    if (n < 2) {
        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 = 2; 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.c

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#include <stdio.h>

int count = 0;

void hanoi(int n, char source, char target, char auxiliary) {
    if (n > 0) {
        hanoi(n - 1, source, auxiliary, target);
        printf("%d: %c --> %c\n", n, source, target);
        count++;
        hanoi(n - 1, auxiliary, target, source);
    }
}

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

task6.c

task7.c

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#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);  // 函数调用
        printf("最大公约数: %d\n\n", ans);
    }
    return 0;
}


int gcd(int a, int b, int c) {
    int min_num = a;
    // 找出a、b、c中的最小值
    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;
}