work6

1.学生信息处理问题

• char name[25];要写成字符数组
• scanf_s(%s", stu_arr[i].name, 25, );数组名的本质是 “指向数组首元素的指针”，所以不需要加&符号；并且这个25必须要加（如果是scanf_s）
• 冒泡排序
  • for (int i = 0; i < len-1; i++) 外
  • for (int j = 0; j < len-1-i; j++)内
• %.2f保留两位小数

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

//学生信息
typedef struct student_s {
	int stu_id;
	char name[25];// 要写成字符数组
	float chinese_score;
	float math_score;
	float english_score;
}student_t;

void info_input(student_t* stu_arr, int len); //输入学生信息
void sort_by_id(student_t* stu_arr,int len);//按照学号排序
void print_arr(student_t* stu_arr, int len); // 数组输出
void find_chinese_top(student_t *stu_arr,int len);//语文最高分的学生信息
void find_math_top(student_t* stu_arr, int len);//数学最高分的学生信息
void find_english_top(student_t* stu_arr, int len);//英语最高分的学生信息
void course_average_score(student_t* stu_arr, int len);//输出每门课程的平均分
void sort_sum_score(student_t* stu_arr, int len);//按照总分排序
int main() {
	student_t stu_arr[5];// 结构体数组
	info_input(stu_arr, 5);//输入
	sort_by_id(stu_arr, 5);
	print_arr(stu_arr, 5);
	find_chinese_top(stu_arr, 5);
	find_math_top(stu_arr, 5);
	find_english_top(stu_arr, 5);
	course_average_score(stu_arr, 5);
	sort_sum_score(stu_arr, 5);
	print_arr(stu_arr, 5);
	return 0;
}
//输入
void info_input(student_t* stu_arr, int len) {
	for (int i = 0; i < len; i++) {
		scanf_s("%d%s%f%f%f",
			&stu_arr[i].stu_id,
			stu_arr[i].name, 25,
			&stu_arr[i].chinese_score,
			&stu_arr[i].math_score,
			&stu_arr[i].english_score);
		printf("-----------------------\n");
	}
}
//输出
void print_arr(student_t* stu_arr, int len) {
	for (int i = 0; i < len; i++) {
		printf("学号：%d 姓名：%s 语文：%.1f 数学：%.1f 英语：%.1f\n",
			stu_arr[i].stu_id,
			stu_arr[i].name, 
			stu_arr[i].chinese_score,
			stu_arr[i].math_score,
			stu_arr[i].english_score);
		printf("-----------------------\n");
	}
}
// 按照学号排序（冒泡）
void sort_by_id(student_t* stu_arr,int len) {
	for (int i = 0; i < len-1; i++) {
		for (int j = 0; j < len-1-i; j++) {
			if (stu_arr[j].stu_id > stu_arr[j+1].stu_id) {
				student_t temp = stu_arr[j];
				stu_arr[j] = stu_arr[j+1];
				stu_arr[j+1] = temp;
			}
		}
	}
}
// 最高分的学生信息
void find_chinese_top(student_t* stu_arr, int len) {
	int max_score = 0;
	int max_i = 0;
	for (int i = 0; i < len; i++) {
		if (stu_arr[i].chinese_score > max_score) {
			 max_score = stu_arr[i].chinese_score;
			 max_i = i;
		}
	}
	printf("语文状元是：学号：%d 姓名：%s 语文：%.1f 数学：%.1f 英语：%.1f\n",
		stu_arr[max_i].stu_id,
		stu_arr[max_i].name,
		stu_arr[max_i].chinese_score,
		stu_arr[max_i].math_score,
		stu_arr[max_i].english_score);
}
void find_math_top(student_t* stu_arr, int len) {
	int max_score = 0;
	int max_i = 0;
	for (int i = 0; i < len; i++) {
		if (stu_arr[i].math_score > max_score) {
			max_score = stu_arr[i].math_score;
			max_i = i;
		}
	}
	printf("数学状元是：学号：%d 姓名：%s 语文：%.1f 数学：%.1f 英语：%.1f\n",
		stu_arr[max_i].stu_id,
		stu_arr[max_i].name,
		stu_arr[max_i].chinese_score,
		stu_arr[max_i].math_score,
		stu_arr[max_i].english_score);
}
void find_english_top(student_t* stu_arr, int len) {
	int max_score = 0;
	int max_i = 0;
	for (int i = 0; i < len; i++) {
		if (stu_arr[i].english_score > max_score) {
			max_score = stu_arr[i].english_score;
			max_i = i;
		}
	}
	printf("英语状元是：学号：%d 姓名：%s 语文：%.1f 数学：%.1f 英语：%.1f\n",
		stu_arr[max_i].stu_id,
		stu_arr[max_i].name,
		stu_arr[max_i].chinese_score,
		stu_arr[max_i].math_score,
		stu_arr[max_i].english_score);
}
//输出每门课程的平均分
void course_average_score(student_t* stu_arr, int len) {
	float sum_chinese = 0;
	float sum_math = 0;
	float sum_english = 0;
	for (int i = 0; i < len; i++) {
		sum_chinese += stu_arr[i].chinese_score;
		sum_math += stu_arr[i].math_score;
		sum_english += stu_arr[i].english_score;
	}
	printf("avg chinese = %.2f, avg math = %.2f, avg english = %.2f\n", sum_chinese / len, sum_math / len, sum_english / len);
}
// 总分排序
void sort_sum_score(student_t* stu_arr, int len) {
	for (int i = 0; i < len - 1; i++) {
		for (int j = 0; j < len - 1 - i; j++) {
			if (stu_arr[j].chinese_score+ stu_arr[j].math_score+ stu_arr[j].english_score< stu_arr[j+1].chinese_score + stu_arr[j+1].math_score + stu_arr[j+1].english_score) {
				student_t temp = stu_arr[j];
				stu_arr[j] = stu_arr[j + 1];
				stu_arr[j + 1] = temp;
			}
		}
	}
}

2. 合并有序链表

假设存在两个链表，链表的各个结点数据域都是有序的。要求合并这两个链表，并且合并之后也是有序的。本题中不能重新申请堆空间。