实验3

1

#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);  // 函数调用
        printf("分数: %d, 等级: %c\n\n", score, grade);
    }

    return 0;
}

// 函数定义
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;
}

1.将分数转化为对应的等级字符

2.缺少break，程序无法执行后续的case代码

 

2

#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;
}

1.计算输入n的各位数字之和，通过循环不断取出整数的个位数字累加到ans中

2.能等同

原方法是用迭代循环的方式用while循环，新方法用的是递归当n大于等于10时去掉个位数

3

#include <stdio.h>

int power(int x, int n);    // 函数声明

int main() {
    int x, n;
    int ans;

    while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {
        ans = power(x, n);  // 函数调用
        printf("n = %d, ans = %d\n\n", n, ans);
    }
    
    return 0;
}

// 函数定义
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;
    }
}

1.实现x的n次方的计算

2.是

当n为基数，x=x*x的n-1次方，当n为偶数x=x的二分之n次方的平方，当n=0,x=1

4

#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 = 2; i <= 98; i++) {
        if (is_prime(i) && is_prime(i + 2)) {
            printf("%d %d\n", i, i + 2);
            count++;
        }
    }
    printf("总数: %d\n", count);
    return 0;
}

5

 

#include <stdio.h>
int moveCount = 0;
void hanoi(int n, char from, char to, char aux) {
    if (n == 1) {
        moveCount++;
        printf("%d: %c --> %c\n", moveCount, from, to);
    }
    else {
        hanoi(n - 1, from, aux, to);
        moveCount++;
        printf("%d: %c --> %c\n", moveCount, from, to);
        hanoi(n - 1, aux, to, from);
    }
}

int main() {
    int n;
    while (1) {
        printf("请输入盘子数量n（输入非数字可结束程序）: ");
        if (scanf_s("%d", &n) != 1) {
            break;
        }
        moveCount = 0;
        printf("\n");
        hanoi(n, 'A', 'C', 'B');
        printf("一共移动了%d次.\n\n", moveCount);
    }
    return 0;
}

 

6.1

#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 > n) return 0;  
    if (m == 0 || m == n) return 1;  
    int i;
    double result = 1.0;
    if (m > n - m) {
        m = n - m; 
    }
    for (i = 0; i < m; i++) {
        result *= (n - i);
        result /= (i + 1);
    }
    return (int)result;
}

6.2

#define _CRT_SECURE_NO_WARNINGS
#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);  
}

7

#define _CRT_SECURE_NO_WARNINGS
#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_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;
}