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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- [阿里通义] #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)256
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- [讯飞星火] 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)275
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- [讯飞星火] 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)182