实验三

task1

#include <stdio.h>

char score_to_grade(int score);  // 函数声明

int main() {
    int score;
    char grade;

    while (scanf_s("%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;
}

A1 将成绩转换成等级；形参类型为整形，返回值为字符型。

A2 字符应该用单引号引起；每个case最后没有break，导致依次输出。

 

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;
}

A1 实现各个数位上数字求和

A2 能；原代码是用迭代方法，此代码是用递归的方法，以个位数为出口，实现自身调用。

 

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); 
        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;
    }
}

A1 实现幂函数运算

A2​ 1,x=0

      x*power(x,n-1),n是奇数

      [power(x,n/2)]的平方，n是偶数且n>0

 

 

task4

#define  _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
int is_prime(int n);
int main()
{
    int m=100, n=0;
    printf("100以内的孪生素数：\n");

    for (int i = 1; i <= (m - 2); i++)
    {
        if (is_prime(i) && is_prime(i + 2))
        {
            printf("%d %d\n", i, i + 2);
            n++;
        }

    }
    printf("100以内的孪生素数共%d个.", n);
    return 0;
}
int is_prime(int n)
{
    int p = 2;

    for (p; p <= n; p++)
    {
        if (n % p == 0)
            break;
    }
    if (n == p)
        return 1;
    else return 0;
}

 

task5

#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)
{
    double ans = 1;
    for (int i = n; i > n - m; i--)
    {
        ans *= i;
    }
    for (int j = 1; j <= m; j++)
        ans /= j;
    return ans;
}

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

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

    while (scanf("%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 (m > n) {
        return 0;
    }
    if (m == 0 || m == n) {
        return 1;
    }
    if (m == 1) {
        return n;
    }
    else if (m == 2) {
        return n * (n - 1) / 2;
    }
    else return func(n - 1, m) + func(n - 1, m - 1);
}

 

task6

#include <stdio.h>

int judge(int a, int b);
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 judge(int a, int b)
{
    int t;
    while (b != 0) {
        t = a % b;
        a = b;
        b = t;
    }

    return a;
}
int gcd(int a, int b, int c)
{
    int ans;

    ans = judge(a, b);
    ans = judge(ans, c);

    return ans;

}

 

task7

#include <stdio.h>
#include <stdlib.h>

int print_charman(n);

int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); // 函数调用

    return 0;
}

int print_charman(n)
{

    for (int i = 1; i <= n; i++)
    {
        for (int j=i-1;j>0;j--)
        {
            printf("\t");
        }
        for (int k = 1; k <= 2 * (n - i) + 1; k++)
        {
            printf(" O \t");
        }
        printf("\n");
        for (int j = i - 1; j > 0; j--)
        {
            printf("\t");
        }
        for (int k = 1; k <= 2 * (n - i) + 1; k++)
        {
            printf("<H>\t");
        }
        printf("\n");
        for (int j = i - 1; j > 0; j--)
        {
            printf("\t");
        }
        for (int k = 1; k <= 2 * (n - i) + 1; k++)
        {
            printf("I I\t");
        }
        printf("\n");
    }
}