实验三

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;
}#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;
}

ans1：为分数评等级；形参类型为整型，返回值类型字符型。
ans2：应用单引号表示字符；未写break所以后面的选项会依次执行。
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;
}#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;
}

ans1:将一个数上的所有数相加。
ans2：能，修改前是迭代思想，修改后为递归思想。
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;
    }
}

ans1：求x的n次方。
ans2：是，

task4

#include <stdio.h>
#include <math.h>
int main() {
	int cnt = 0,c;
	printf("100以内的孪生素数：");
	for (int i =2; i <= 100; i++) {
		c = i + 2;
		if (is_prime(i)&&is_prime(c)) {
			printf("%d\t%d\n",i,c);
			cnt++;
		}
	}
	printf("100以内的孪生素数有%d对", cnt);
	return 0;
}
int is_prime(int a) {
	int t=1;
	for (int i = 2; i <= sqrt(a); i++) {
		if (a % i == 0) {
			t = 0;
			break;
		}
	}
	return t;
}

task5.1迭代

#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 n1 = 1, m1 = 1, t=1, ans;
		if (n < m) {
			ans = 0;
		}
		else if(n>=m){
			for (int i = 1; i <= n; i++) {
				n1 = n1 * i;
			}
			for (int i = 1; i <= m; i++) {
				m1 = m1 * i;
			}
			for (int i = 1; i <= n - m; i++) {
				t = t * i;
			}
			ans = n1 / (m1 * t);
		}
		return ans;
	}

task5.2递归

#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 ans;
		if (n < m) {
			ans = 0;
		}
		else if (n!=0&&m == 1) {
			ans = n;
		}
		else if (n!=0&&m == 0) {
			ans = 1;
		}
		else {

			ans = func(n - 1, m) + func(n - 1, m - 1);
		}
		return ans;
	}

task6

#include <stdio.h>
int gcd(int a, int b, int c);
int main() {
    int a, b, c;
    int ans;
    while (scanf_s("%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 gcd(int a, int b, int c) {
    int min, ans;
    if (a <= b && a <= c) {
        min = a;
    }
    else if(b <= c && b <= a) {
        min = b;
    }
    else {
        min = c;
    }
    if (a % min == 0 && b % min == 0 && c % min == 0) {
        ans = min;
    }
    else {
        for (int i = min; i >= 1; i--) {
            if (a % i == 0 && b % i == 0 && c % i == 0) {
                ans = i;
                break;
            }
        }
    }
    return ans;
}

task7

#include <stdio.h>
#include <stdlib.h>
void printf_charman(int n);
int main() {
	int n;
	printf("input:");
	scanf_s("%d", &n);
	printf_charman(n);
	return 0;
}
void printf_charman(int n) {
	for (int i = 0; i / 2 <= n; i+=2) {
		for (int j = 0; j < i / 2; j++) {
			printf("\t");
		}
		int m;
		m = i;
		for (; m < 2 * n - 1; m++) {
			printf(" o\t");
		}
		printf("\n");
		m = i;
		for (int j = 0; j < i / 2; j++) {
			printf("\t");
		}
		for (; m < 2 * n - 1; m++) {
			printf("<H>\t");
		}
		printf("\n");
		m = i;
		for (int j = 0; j < i / 2; j++) {
			printf("\t");
		}
		for (; m < 2 * n - 1; m++) {
			printf("I I\t");
		}
		printf("\n");
	}

}