实验3

实验任务1

任务1

#define _CRT_SECURE_NO_WARNINGS
#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
判断一个分数的等级，score，char
问题2
最后返回的ans均为E，没有break，ans会被一直赋值为对应case后面的值，最终被赋予为E

实验任务2

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int sum_digits(int n); // 函数声明
int main() {
	int n;
	int ans;
	while (printf("Enter n: "), scanf("%d", &n) != EOF) {
		ans = sum_digits(n); // 函数调用
			printf("n = %d, ans = %d\n\n", n, ans);
	}
	return 0;
}
// 函数定义
int sum_digits(int n) {
	int ans = 0;
	while (n != 0) {
		ans += n % 10;
		n /= 10;
	}
	return ans;
}

问题1
将输入的数字的每一位上的数字加起来
问题2
能，原来的代码是迭代的方式，通过循环的方式来实现运算，第二种代码是递归的方式，通过自身调用自身来实现运算

实验任务3

#define _CRT_SECURE_NO_WARNINGS
#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
通过递归来计算x的n次方
问题2
是，

实验任务4

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int is_prime(int n) {
	for (int i = 2; i <=n/2; i++) {
		if (n % i == 0)
			return 0;
	}
	return 1;
}
int main() {
	int s = 0;
	printf("100以内的孪生素数:\n");
	for (int n = 2; n <= 100; n++) {
		if (is_prime(n) && is_prime(n + 2)) {
			s += 1;
			printf("%d %d\n", n, n + 2);
		}
	}
	printf("100以内的孪生素数共有%d个",s);
}

实验任务5

递归

#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) {
		if (n == 1 && m == 1)
			return 1;
		else if (m == 0)
			return 1;
		else if (n < m)
			return 0;
		else
			return func(n - 1, m) + func(n - 1, m - 1);

	
	}

迭代

#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) {
		if (m > n)
			return 0;
		if (m == 0 || m == n)
			return 1;
		long long a=1, b=1;
		for (int i = 1; i <= m; i++) {
			a *= (n - i + 1);
			b *= i;
		}

		return a/b;
	
	}

实验任务6

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int gcd(int a, int b, int c);
int main() {
	int a, b, c;
	int ans;

	while (scanf("%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 x, int y, int z) {
	int m,n,i;
	m =(x > y ? x : y);
	i = (m > z ? m: z);
	for (; i > 0; i--) {
		if (x % i == 0 && y % i == 0 && z % i == 0) {
			n = i;
			break;
		}
	}
	return n;
}

实验任务7

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
void print_charman(int a);
int main() {
	int n;
	printf("Enter n: ");
	scanf("%d", &n);
	print_charman(n); 
	return 0;
}
void print_charman(int a) {
	int count = 0;
	int x = 0;
	
	for (int I = 2 * a - 1; I > 0; I = I - 2) {
		for(int i2 = 1; i2 <= x; i2++) {
			printf(" ");
		}
		for (int i1 = 1; i1 <= I; i1++) {
			printf("  O  ");
		}
		printf("\n");

		for (int i2 = 1; i2 <= x; i2++) {
			printf(" ");
		}
		for (int i1 = 1; i1 <= I; i1++) {
			printf(" <H> ");
		}
		printf("\n");

		for (int i2 = 1; i2 <= x; i2++) {
			printf(" ");
		}
		for (int i1 = 1; i1 <= I; i1++) {
			printf(" I I ");
		}
		printf("\n");
	
		count++;
		x += 5;
	}
	
}