实验3

task 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：将分数转化为对应的等第；形参类型：整型 ；返回值类型：字符串

问题2：有问题 将break去除后 执行一个语句 输出对应等地时会接着输出之后的等第；双引号变为单引号 输出不了字符等第

task 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：求各个位数相加之和

问题2：可以有同等效果 第一种是迭代方式 第二种是递归方式

task 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：计算x的n次方

问题2：是递归函数

task 4

#include <stdio.h>

int is_prime(int n);

int main() {
    int count = 0;
    int n;

    printf("100 以内的孪生素数：\n");

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

int is_prime(int n) {
    int i;

    if (n <= 1) 
        return 0;
    
    for (i = 2; i * i <= n; i++) {
        if (n % i == 0) 
            return 0;
    }
    return 1;
}

task 5

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

int count=0;

void hanoi(unsigned int n,char from,char temp,char to);
void move(unsigned int n,char from,char to);

int main(){
    int n;

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

task 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 a,b, i;
    a = 1, b = 1;
    
    if (n < m)
        return 0;
    else if (n == m || m == 0)
        return 1;
    else {
        for (i = 1;i <= m;++i) 
            a *= i;
        for (i = n; i >= n - m + 1;--i)
            b *= i;
        
        return b / a;
        
    }

}

#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 a,b, i;
    a = 1, b = 1;
    
    if (n < m)
        return 0;

    else if (n == m || m == 0)
        return 1;

    else 
        return func(n-1,m)+func(n-1,m-1);

}

task 7

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

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

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

}