实验3 c语言函数应用编程

task1:

 

#include<stdio.h>

char score_to_grade(int score);

int main() {
    int score;
    char grade;

    while (scanf_s("%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：功能是将键盘输入的分数转化成对应的等级  形参类型是整型  返回值类型是字符型

问题2：1 改成双引号后无法输出对应的等级    2 语句后不加break,程序执行到对应语句后无法终止，还会输出后面语句的结果

task2:

 

#include<stdio.h>

int sum_digits(int n);

int main() {
    int n;
    int ans;

    while (printf("Enter n:"), scanf_s("%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：功能是得出输入数各位的加和

问题2：可以实现同等效果  原算法将数字的各位取出依次加和，改后通过递归实现

task3:

 

#include<stdio.h>

int power(int x, int n);

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

    while (printf("Enter x and n :"), scanf_s("%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：功能是计算x的n次方

问题2：是递归函数

 

 

 

 

task4:

#include<stdio.h>
#include<math.h>
int is_prime(int n);


int main() {
    int n,sum;
    n = 1;
    sum = 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);
            sum = sum + 1;
        }
    }
    printf("100以内的孪生素数共有%d个", sum);

    return 0;
        
    
    }









int is_prime(int n) {
    int i, m;
    m = sqrt(n);

    for (i = 2;i<=m;++i)
        if (n % i == 0)
            break;
    if (i > m&&n>1)
        return 1;
    else
        return 0;
}
    

 

 

 

task5:

 

#include<stdio.h>
void hanoi(int n, char from, char temp, char to);
void moveplate(int n, char from, char to);
auto sum = 0;
int main() {
    int n;
    while (scanf_s("%d", &n)!=EOF) {
        hanoi(n, 'A', 'B', 'C');
        printf("一共移动了%d次\n\n", sum);
        sum = 0;
    }
    return 0;
}

void hanoi(int n, char from, char temp, char to) {
    if (n == 1) {
        moveplate(n, from, to);
    }

    else {
        hanoi(n - 1, from, to,temp);
        moveplate(n, from, to);
        hanoi(n - 1, temp, from, to);
    }
}
void moveplate(int n, char from, char to){
    printf("%d:%c-->%c\n", n, from, to);
    sum += 1;
}

 

 

 

task6:

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




    }

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

 

 

 

 

 

task7:

#include<stdio.h>

int main() {
    int n;
    printf("Enter n:");
    scanf_s("%d", &n);
    print_charman(n);

    return 0;
}


int print_charman(int n) {
    int i,j;
    for (i = n;i > 0;--i) {
        for (int m = 0;m < n - i;m++)
            printf("\t");
        for (j = 1;j <=(2*i-1);j++) {
            printf(" o \t");
        }
        printf("\n");

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

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

    }