#include<stdio.h> #include<string.h> //#include<stdlib.h> void swap(char *x,char *y); char *insertChar(char ch,char *sent); int main() { char ch,sent[1000]; int i; printf("请输入给定字符ch:\n"); scanf("%c",&ch); printf("请输入给定字符串:\n"); for(i=0;i<=1000;i++) { scanf("%c",&sent[i]); } insertChar(ch,sent); } char *inserChar(char ch,char *sent) { int i; for(i=0;i<=1000;i++) { if(sent[i]==ch) { sent[i+1]='*'; for(i+2;i<=998;i++) { swap(&sent[i+2],&sent[i+3]); } } } for(i=0;sent[i]!='\0';i++) { printf("%c",sent[i]); } } void swap(char*x,char*y) { int temp; temp=*x; *x=*y; *y=temp; }
时间: 2024-01-18 14:03:09 浏览: 118
这是一个 C 语言程序,用于在给定字符串中插入指定字符并将其替换为星号。程序的主要功能是定义了一个名为 insertChar 的函数,该函数接受两个参数:一个是指定字符 ch,另一个是字符串 sent。该函数通过遍历字符串 sent 来查找指定字符 ch 并在其后面插入星号,同时使用 swap 函数交换字符以将星号移至正确的位置。最后,该函数将修改后的字符串 sent 打印到控制台中。程序中还包含了一个 swap 函数,用于交换两个字符的位置。
相关问题
#include<stdio.h> #include<stdlib.h> #include<string.h> #include<windows.h> struct stu { int id; char name[10]; float grade[11]; float sum; float average; float gpa; struct stu* next; };按照这个结构体来续写C语言代码。要求将学生信息保存在硬盘文件中,通过制定文件路径和文件名称来导入学生信息;按照学号对所有学生信息进行升序、降序排列,并输出到屏幕显示或输出到文件
好的,以下是续写的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
struct stu {
int id;
char name[10];
float grade[11];
float sum;
float average;
float gpa;
struct stu* next;
};
void saveToFile(struct stu* head, char* fileName) {
FILE* fp = fopen(fileName, "wb");
if (fp == NULL) {
printf("Failed to open file!\n");
return;
}
struct stu* p = head;
while (p != NULL) {
fwrite(p, sizeof(struct stu), 1, fp);
p = p->next;
}
fclose(fp);
printf("Successfully saved to file!\n");
}
struct stu* readFromFile(char* fileName) {
FILE* fp = fopen(fileName, "rb");
if (fp == NULL) {
printf("Failed to open file!\n");
return NULL;
}
struct stu* head = NULL, *p = NULL, *q = NULL;
while (!feof(fp)) {
p = (struct stu*)malloc(sizeof(struct stu));
if (p == NULL) {
printf("Memory allocation failed!\n");
return NULL;
}
fread(p, sizeof(struct stu), 1, fp);
if (p->id == 0) {
free(p);
break;
}
p->next = NULL;
if (head == NULL) {
head = p;
}
else {
q->next = p;
}
q = p;
}
fclose(fp);
printf("Successfully read from file!\n");
return head;
}
void printList(struct stu* head) {
if (head == NULL) {
printf("The list is empty!\n");
return;
}
printf("ID\tName\tGrade1\tGrade2\tGrade3\tGrade4\tGrade5\tGrade6\tGrade7\tGrade8\tGrade9\tGrade10\tSum\tAverage\tGPA\n");
struct stu* p = head;
while (p != NULL) {
printf("%d\t%s\t", p->id, p->name);
for (int i = 0; i < 10; i++) {
printf("%.2f\t", p->grade[i]);
}
printf("%.2f\t%.2f\t%.2f\n", p->sum, p->average, p->gpa);
p = p->next;
}
}
void swap(struct stu* a, struct stu* b) {
struct stu temp = *a;
*a = *b;
*b = temp;
}
void sortList(struct stu* head, int order) {
if (head == NULL) {
printf("The list is empty!\n");
return;
}
int count = 0;
struct stu* p = head;
while (p != NULL) {
count++;
p = p->next;
}
for (int i = 0; i < count - 1; i++) {
p = head;
for (int j = 0; j < count - i - 1; j++) {
if (order == 1) {
if (p->id > p->next->id) {
swap(p, p->next);
}
}
else {
if (p->id < p->next->id) {
swap(p, p->next);
}
}
p = p->next;
}
}
printf("Successfully sorted!\n");
}
int main() {
struct stu* head = NULL;
char fileName[100];
int choice, order;
while (1) {
system("cls");
printf("1. Import student information from file\n");
printf("2. Save student information to file\n");
printf("3. Print student information\n");
printf("4. Sort student information by ID in ascending order\n");
printf("5. Sort student information by ID in descending order\n");
printf("6. Exit\n");
printf("Please enter your choice: ");
scanf_s("%d", &choice);
switch (choice) {
case 1:
printf("Please enter the file path and name: ");
scanf_s("%s", fileName, 100);
head = readFromFile(fileName);
Sleep(2000);
break;
case 2:
printf("Please enter the file path and name: ");
scanf_s("%s", fileName, 100);
saveToFile(head, fileName);
Sleep(2000);
break;
case 3:
printList(head);
system("pause");
break;
case 4:
sortList(head, 1);
system("pause");
break;
case 5:
sortList(head, 2);
system("pause");
break;
case 6:
printf("Goodbye!\n");
return 0;
default:
printf("Invalid choice!\n");
Sleep(1000);
break;
}
}
return 0;
}
```
这段代码实现了以下功能:
1. 将学生信息保存到硬盘文件中,并通过指定文件路径和文件名称来导入学生信息。
2. 按照学号对所有学生信息进行升序、降序排列,并输出到屏幕显示或输出到文件。
3. 通过菜单的方式来选择要执行的操作。
希望对你有所帮助!
优化这段代码:#include <stdio.h> typedef struct inf1 { char name; int grade; }inf; void sort(int s[],int r[],int n); void swap(int r[],int i,int j); int main() { int n,i; scanf("%d",&n); inf chart[n]; //inf *chart=(inf *)malloc(n * sizeof(inf)); int rank[n]; int mark[n]; for(i=0;i<n;i++) { scanf("%s,%d",&chart[i].name,&chart[i].grade); mark[i]=i; } for(i=0;i<n;i++) rank[i]=chart[i].grade; sort(rank,mark,n); for(i=0;i<n;i++) printf("%s,%d",chart[mark[i]].name,chart[mark[i]].grade); return 0; } void sort(int s[],int r[],int n) { int i,j; for(i=0;i<n;i++) { for(j=0;j<n;j++) { if(s[r[i]]<s[r[j]]) swap(r,i,j); } } } void swap(int r[],int i,int j) { int temp; temp=r[i]; r[i]=r[j]; r[j]=temp; }
这段代码存在一些需要优化的问题:
1. 结构体 `inf1` 中的 `name` 只能存储单个字符,应该改为 `char name[10]` 等可以存储字符串的类型。
2. 在 `scanf` 中使用 `%s` 格式化字符串时,需要指定字符串的长度,否则可能会发生缓冲区溢出的问题。
3. 在排序中使用了冒泡排序,时间复杂度为 $O(n^2)$,在数据较大时会导致性能问题。
以下是对代码的优化建议:
```
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct {
char name[10];
int grade;
} Student;
void sort(Student *students, int n);
void merge(Student *students, int left, int mid, int right);
int main() {
int n, i;
scanf("%d", &n);
Student *students = (Student *)malloc(n * sizeof(Student));
for (i = 0; i < n; i++) {
scanf("%s%d", students[i].name, &(students[i].grade));
}
sort(students, n);
for (i = 0; i < n; i++) {
printf("%s %d\n", students[i].name, students[i].grade);
}
free(students);
return 0;
}
void sort(Student *students, int n) {
int step, left, mid, right;
for (step = 1; step < n; step *= 2) {
for (left = 0; left < n - step; left = right) {
right = left + step * 2;
if (right > n) {
right = n;
}
mid = left + step;
merge(students, left, mid, right);
}
}
}
void merge(Student *students, int left, int mid, int right) {
int i = left, j = mid, k = 0;
Student *tmp = (Student *)malloc((right - left) * sizeof(Student));
while (i < mid && j < right) {
if (students[i].grade < students[j].grade) {
tmp[k++] = students[i++];
} else {
tmp[k++] = students[j++];
}
}
while (i < mid) {
tmp[k++] = students[i++];
}
while (j < right) {
tmp[k++] = students[j++];
}
memcpy(students + left, tmp, k * sizeof(Student));
free(tmp);
}
```
优化后的代码使用归并排序,时间复杂度为 $O(n\log n)$,可以更好地处理大量数据。同时,结构体中的 `name` 变为了字符串类型,避免了缓冲区溢出的问题。
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