实验三

任务1：

#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：作为一个函数，根据分数划分等级。形参为整数型score，返回值为字符型ans。

问题2：有，少了break，程序不能及时终止；双引号应该为单引号。

任务2：

#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：分别求出每位的数并求和

问题2：能，原函数为迭代，分别算出每位的数并加起来，改函数为递归，通过调用自身来求出结果

任务3：

#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：算出x的n次方

问题2：是， 偶数先计算x的n/2次幂，再将结果平方。奇数减一转为偶数，计算n-1次幂，然后将结果乘以x

 

任务4：

#include <stdio.h>
#include <math.h>

int is_prime(int n);

int main()
{
    int n;
    int cnt = 0;
    printf("100以内的孪生素数:\n");
    for (n = 1; n <= 98; n++)
    {
        if (is_prime(n) && is_prime(n + 2))
        {
            printf("%d %d\n", n, n + 2);
            cnt++;
        }
    }
    printf("100以内的孪生素数共有%d个\n", cnt);
    return 0;
}

int is_prime(int n)
{
    int i;
    if (n == 1)
        return 0;
    if (n == 2)
    {
        return 1;
    }
    if (n % 2 == 0)
        return 0;
    for (i = 3; i <= sqrt(n); i += 2)
    {
        if (n % i == 0)
        {
            return 0;
        }
    }
    return 1;
}

任务5：

 

#include <stdio.h>

int cnt = 0;

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

 

任务6：

#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)
{
    int ans;
    int x = 1, y = 1;
    int i;
    for (i = n; i >= n - m + 1; i--)
    {
        x *= i;
    }
    for (i = m; i >= 1; i--)
    {
        y *= i;
    }
    ans = x / y;
    return ans;
}

 

#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 == 0 || n == m)
        return 1;
    else
        return func(n - 1, m) + func(n - 1, m - 1);
}

 任务7：

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

static int a = 0, b = 0, c = 0;
void print_charman();

int main()
{
    int n;

    printf("Enter n:");
    scanf("%d", &n);
    print_charman(n);

    return 0;
}

void print_charman(int n)
{
    int i;
    if (n == 0)
    {
        return;
    }
    for (i = 0; i < a; i++)
    {
        printf("   ");
        printf("\t");
    }
    a++;
    for (i = 0; i < 2 * n - 1; i++)
    {
        printf(" O ");
        printf("\t");
    }
    printf("\n");
    for (i = 0; i < b; i++)
    {
        printf("   ");
        printf("\t");
    }
    b++;
    for (i = 0; i < 2 * n - 1; i++)
    {
        printf("<H>");
        printf("\t");
    }
    printf("\n");
    for (i = 0; i < c; i++)
    {
        printf("   ");
        printf("\t");
    }
    c++;
    for (i = 0; i < 2 * n - 1; i++)
    {
        printf("I I");
        printf("\t");
    }
    printf("\n");
    print_charman(n - 1);
}