实验3

task1

#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;
}

判断输出对应的等级，形参是整数，返回值是字符

符合一个case后ans会被赋值为D

task2

#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;
}

计算输入n的各位数之和

可以，原代码为迭代，改代码为递归，自己调用自己

task3

#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;
    }
}

 定义一个函数

计算x^n

task4

#include <stdio.h>

int is_prime(int n);

int main() {
    int i, j, s=0;
    
    printf("100以内的孪生素数：\n");
    for (i = 0; i < 100; ++i) {
        j = i + 2;
        if (is_prime(i) == 1 && is_prime(i) == is_prime(j)) {
            printf("%d %d\n", i, j);
            s += 1;
        }
    }
    printf("100以内的孪生素数有%d个", s);

    return 0;
}

int is_prime(int n) {
    int i, x=1;
    if (n != 1 && n != 2) {
        for (i = 3; i < n; ++i) {
            if (n % i == 0) {
                x = 0;
                break;
            }
        }
    }
    else
        x = 0;

    return x;
}

task5

#include <stdio.h>

int count = 0;
void hanoi(int n, char from, char temp, char to);

int main() {
    int n;
    while (scanf_s("%d", &n) != EOF) {
        count = 0;
        hanoi(n, 'A', 'B', 'C');
        printf("一共移动了%d次.\n", count);
    }
    return 0;
}
void hanoi(int n, char from, char temp, char to) {
    if (n == 1) {
        count++;
        printf("%d: %c --> %c\n", n, from, to);
    }
    else {
        hanoi(n - 1, from, to, temp);
        count++;
        printf("%d: %c --> %c\n", n, from, to);
        hanoi(n - 1, temp, from, to);
    }
}

task6

1

#include <stdio.h>
int func1(int n, int m);   // 函数声明

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

    while(scanf_s("%d%d", &n, &m) != EOF) {
        ans = func(n, m);   // 函数调用
        printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
    }
        
    return 0;
}

int func1(int n, int m) {
    int x, y;
    for (x = 1; n > n - m; --n)
        x *= n;
    for (y = 1; m > 1; --m)
        y *= m;
    return x / y;
}

 

2

#include <stdio.h>
int func(int n, int m);   // 函数声明

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

    while (scanf_s("%d%d", &n, &m) != EOF) {
        ans = func(n, m);   // 函数调用
        printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
    }

    return 0;
}

int func(int n, int m) {
    if (n == m || m == 0)
        return 1;
    if (m >n)
        return 0;
    else
        return func(n - 1, m) + func(n - 1, m - 1);
}

 

task7

#include <stdio.h>

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

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

    while(scanf_s("%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 t;
    t = a;
    if (t > b)
        t = b;
    if (t > c)
        t = c;
    while (t > 0) {
        if (a % t == 0 && b % t == 0 && c % t == 0)
            break;
        --t;
    }
    return t;
}