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struct uclamp_min_limit_param { char *name; u64 min_limit; }; static struct uclamp_min_limit_param uclamp_min_limit[] = { {"top-app", 20 * pow10(uclamp_percent_shift)}, {"foreground", 10 * pow10(uclamp_percent_shift)}, }; static ssize_t cpu_uclamp_min_limit_write(struct kernfs_open_file *of, char *buf, size_t nbytes, loff_t off) { int ret, i; for (i = 0; i < array_size(uclamp_min_limit); i++) { if (!strcmp(of_css(of)->cgroup->kn->name, uclamp_min_limit[i].name)) { b
请问您关于这段代码是希望我为您解释它的功能、优化建议,还是回答某些与它相关的特定问题呢? [2024-10-21 22:58:48 | AI写代码神器 | 30点数解答]
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AI写代码神器
struct uclamp_min_limit_param { char *name; u64 min_limit; }; static struct uclamp_min_limit_param uclamp_min_limit[] = { {"top-app", 20 * pow10(uclamp_percent_shift)}, {"foreground", 10 * pow10(uclamp_percent_shift)}, }; static ssize_t cpu_uclamp_min_limit_write(struct kernfs_open_file *of, char *buf, size_t nbytes, loff_t off) { int ret, i; for (i = 0; i < array_size(uclamp_min_limit); i++) { if (!strcmp(of_css(of)->cgroup->kn->name, uclamp_min_limit[i].name)) { b(30点数解答 | 2024-10-21 22:58:48)
103
AI写代码神器
#include<stdio.h> #include<stdlib.h> #include<time.h> int producerand(int remainder); void initprocess(); void chosedisplace(); struct linknode* fifo(struct linknode* head, int randcount); void optimal(struct linknode* head, int randprocess); struct linknode* lru(struct linknode* head, int randprocess); struct linknode* initlink(); void choicestey(); int allotment(struct linknode* head); int checkfifooptimal(struct linknode* head, int checkpage); void recover(struct linknode* head, int randproc(60点数解答 | 2024-12-13 20:02:21)
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AI写代码神器
#include<stdio.h> #include<stdlib.h> #include<time.h> int producerand(int remainder); void initprocess(); void chosedisplace(); struct linknode* fifo(struct linknode* head, int randcount); void optimal(struct linknode* head, int randprocess); struct linknode* lru(struct linknode* head, int randprocess); struct linknode* initlink(); void choicestey(); int allotment(struct linknode* head); int checkfifooptimal(struct linknode* head, int checkpage); void recover(struct linknode* head, int randproc(858点数解答 | 2024-12-13 20:03:47)
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AI写代码神器
#include <stdio.h> #include <stdlib.h> #include <stdbool.h> #define maxsize 100 typedef char elemtype; typedef struct node { elemtype data; struct node* lchild; struct node* rchild; } btnode; typedef struct { btnode* data[maxsize]; int top; } stacktype; void initstack(stacktype* st) { st->top = -1; } bool stackempty(stacktype* st) { return st->top == -1; } bool push(stacktype* st, btnode* e) { if (st->top < maxsize - 1) { st->data[++st->top] = e;(95点数解答 | 2024-12-10 13:17:25)
72
AI写代码神器
采用c++语言,实现如下功能: 实现古典密码中的移位密码和仿射密码,具体实现如下接口: c++接口: <<<<<移位密码>>>>> bool is_valid_s(unsinged char k) { 判断k是否为合理的密钥 } int encrypt_s(unsigned char* p, unsigned char k) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_s(unsigned char* c, unsigned char k) { 密钥合法则返回1,且明文覆盖密文: 密钥不合法则返回0. } <<<<<仿射密码>>>>> bool in_valid_a(unsinged char a, unsigned char b) { 判断a,b是否为合理的密钥 } int encrypt_a(unsigned char* p, unsigned char a, unsigned char b) { 密钥合法则返回1,且密文覆盖明文: 密钥不合法则返回0. } int decrypt_a(unsigned char* c, unsign(812点数解答 | 2024-12-18 16:02:36)
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AI写代码神器
快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(710点数解答 | 2024-11-14 10:44:25)
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AI写代码神器
快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(951点数解答 | 2024-11-14 10:46:29)
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AI写代码神器
快速排序:# include< iostream>using namespace std;# include< stdlib. h># include< time. h>const int n =10;const int min= 10;const int max=99;int getrand( int min, int max){return ( rand()%( max-min+1)+ min);}void set( int r[], int n){int i;for(i=0;i<n;i++)r[i]=getrand(min, max);}void show( int r[], int n){int i;for(i=0;i<n;i++)cout << r[i]<<" ";cout << endl;}int partition( int r[], int i, int j){//由你完成}void quicksort( int r[], int i, int j){//由你完成}int main(){srand( time(0));int a[n];set(a, n);cout<<"(472点数解答 | 2024-11-14 10:51:44)
92
AI写代码神器
c语言现在有n枚硬币,其中有一枚是假的,假的硬币比真币轻,但也只轻一点点,而**和真币的外观是一模一样的,从外观上无法辨别硬币的真假,请写程序找出**。 程序的要求: 必须使用递归调用函数实现 递归函数原型必须为:int findfakecoin(int *p, int low, int high) 程序中必须包含函数:int getsum(int *p, int start, int end) main函数已经写好,请编写程序剩余部分的代码并提交。 main函数如下: int main() { int coins[n]; int m; int index; int getsum(int *p, int start, int end); int findfakecoin(int *p, int low, int high); scanf("%d", &m); for (int i = 0; i < m; ++i(497点数解答 | 2024-12-02 22:03:42)
116
AI写代码神器
#include <stdio.h> int i; struct stu { char num; char name; int grade[3] }; void input(struct stu stu) { scanf("%s%s", &stu.num, &stu.name); for (int i = 0; i < 3; i++) { scanf("%d", &stu.grade); } } void print(struct stu stu) { printf("%s,%s,", stu.num, stu.name); for (i = 0; i < 2; i++) { printf("%d,", a.grade[i]); }printf("%d\n",stu.grade[2]); } int main() { int n; struct stu a; scanf("%d", &n); for ( i = 0; i < n; i++) { input(a); } for (i = 0; i < n; i++) { print((371点数解答 | 2025-01-03 23:31:22)
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AI写代码神器
请把下面c语言代码转化成python语言: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> #define max_processes 4 // 进程结构体 struct process { char name[20]; int arrivaltime; int executiontime; int remainingtime; }; // 找到最短作业 struct process* findshortestjob(struct process processes[], int n, int currenttime) { struct process* shortestjob = null; for (int i = 0; i < n; i++) { if (processes[i].arrivaltime <= currenttime && processes[i].remainingtime > 0)(397点数解答 | 2024-11-14 16:08:06)
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AI写代码神器
检查代码 #include <ctype.h> #include <string.h> void verse(char* str) { *str = tolower(*str); int lens = strlen(str); printf("字符串长度为:%d\n", lens); for (int i = 0; i < lens / 2; i++) { char temp = str[i]; str[i] = str[lens - 1 - i]; str[lens - 1 - i] = temp; } } int pali(char *str) { int lens=strlen(str); char *reverse[lens+1]; strcpy(reverse, str); verse(reverse); return strcmp(str, reverse)==0; } int main() { char str[101];(655点数解答 | 2024-11-08 21:54:45)
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