实验3

task.1 代码：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
char score_to_grade(int x);
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'; break;
    }
    return ans;
}

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问题一：功能：将输入分数转化为相应等级返回，形参类型：int，返回值：char

问题二：有问题，当case进行结果时不会停止，导致ans值总以E返回

task.2

代码：

#include<stdio.h>
int sum_digits(int x);
int main()
{
    int x,y;
    while(printf("Enter n:"),scanf("%d",&x)!=EOF)
    {
        y=sum_digits(x);
        printf("x=%d,y=%d\n",x,y);
    }
    return 0;
}

int sum_digits(int x) {
    int ans=0;
    while(x!=0)
    {
        ans+=x%10;
        x/=10;
    }
    return ans;
}

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问题一：将输入数字每一分位上的数字相加

问题二：能，利用递归函数不断将数字每一分位数字剥离相加

task.3

代码：

#include<stdio.h>
int power(intx, inty);
int main()
{
    int x, n, 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(x, 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;
    }
}
    

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问题1：将x进行n次方相乘

问题2：是，y=x^n

task.4

代码：

#include<stdio.h>
#include<stdlib.h>
int is_prime(int x);
int main()
{
    printf("100以内的素数：\n");
    int x = 1, y = 3, ans1, ans2, s = 0;
    while (x <= 100)
    {
        ans1 = is_prime(x);
        ans2 = is_prime(y);
        if (ans1 == 1 && ans2 == 1)
        {
            printf("%d %d\n", x, y);
            s++;
        }
        x++;
        y++;
    }
            
    printf("100以内的孪生素数有：%d", s);
    return 0;
}
int is_prime(int x)
{
    int n = 2;
    if (x == 1)
        return 0;
    while (n <= x / 2)
    {
        if (x % n != 0)
            n++;
        else
        {
            return 0;
            break;
        }
    }
    return 1;
}

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task.5

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
void move(int n, char from, char to);
void hanoi(int n, char from, char temp, char to);
int s = 0;
int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        hanoi(n, 'A', 'B', 'C');
        printf("一共移动了%d次", s);
    }
    return 0;
}
void hanoi(int n, char from, char temp, char to)
{
    if (n == 1)
    {
        move(n, from, to);
    }
    else
    {
        hanoi(n - 1, from, to, temp);
        move(n, from, to);
        hanoi(n - 1, temp, from, to);
    }
    
}
void move(int n, char from, char to)
{
    printf("%d:%c->%c\n", n, from, to);
    s++;
}

task.6

代码

#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)
{
    int s;
    int t = n - m + 1;
    for (s = 1; n >= t; n--)
        s *= n;

    for (; m >= 1; m--)
        s /= m;
    return s;
}
/*int func(int n, int m)
{
    int s = 0;
    if (m > n)
        s = 0;
    else if (m == n)
        s = 1;    
    else if (m > 0 && n > 1)
        s = func(n - 1, m) + func(n - 1, m - 1);
    else
        s = 1;
    return s;
}
*/

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task.7

代码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
void print_charman(int n);
int a;
int main() {
    int n;
    printf("Enter n: ");
    scanf("%d", &n);
    a = n;
    print_charman(n); 
    return 0;
}
void print_charman(int n)
{
    if (n > 0)
    {
        int s, z;
        for (s = 2 * n - 1; s > 0; s--)
            printf(" O \t");
        printf("\n");
        for (z = a - n ; z > 0; z--)
            printf("\t");
        for (s = 2 * n - 1; s > 0; s--)
            printf("<H>\t");
        printf("\n");
        for (z = a - n; z > 0; z--)
            printf("\t");
        for (s = 2 * n - 1; s > 0; s--)
            printf("I I\t");
        printf("\n\n");
        for (z=a-n+1; z > 0; z--)
            printf("\t");
        print_charman(n - 1);
    }
}

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