/* Going to select the highest power object that we are compatible with */ for (i = 0; i < port->SrcCapsHeaderReceived.NumDataObjects; i++) { switch (port->SrcCapsReceived[i].PDO.SupplyType) { case pdoTypeFixed: /* pdoTypeFixed */ objVoltage = port->SrcCapsReceived[i].FPDOSupply.Voltage * 50; if (objVoltage > port->PortConfig.SinkRequestMaxVoltage) { /* If the voltage is greater than our limit... */ continue; } else { /* Calculate the power for comparison */ objCurrent = port->SrcCapsReceived[i].FPDOSupply.MaxCurrent * 10; objPower = (objVoltage * objCurrent) / 1000; } break; case pdoTypeVariable: case pdoTypeBattery: case pdoTypeAugmented: PPSAPDO = i + 1; objVoltage = port->SrcCapsReceived[i].PPSAPDO.MaxVoltage * 100; if (objVoltage > port->PortConfig.SinkRequestMaxVoltage) { /* If the voltage is greater than our limit... */ continue; } else { /* Calculate the power for comparison */ objCurrent = port->SrcCapsReceived[i].PPSAPDO.MaxCurrent * 50; objPower = (objVoltage * objCurrent) / 1000; } break; default: /* Ignore other supply types for now */ objPower = 0; break; } /* Look for highest power */ if (objPower >= MaxPower) { MaxPower = objPower; SelVoltage = objVoltage; reqPos = i + 1; } }代码分析

专业认知 - 用 Python 制作 13 个小游戏，边玩边学！(含源码).pdf

if score > highest_score: highest_score = score # 更新屏幕 food_sprites_group.update() food_sprites_group.draw(screen) hero.draw(screen) pygame.display.flip() clock.tick(60) #### ...

python2-kombu-4.6.3-3.el7.noarch.rpm

官方离线安装包，测试可用。使用rpm -ivh [rpm完整包名] 进行安装

<%@page pageEncoding="utf-8" %> 用户注册 <button type="button" class="close" data-dismiss="modal" aria-label="Close"> × </button> <form method="post" action="${pageContext.request.contextPath }/model/register"> <label class="col-sm-2 col-form-label">用户名</label> <input class="form-control" id="username" name="un" type="text" required /> <label class="col-sm-2 col-form-label">密码</label> <input class="form-control" name="pw" type="password" required /> <label class="col-sm-2 col-form-label">电话</label> <input class="form-control" name="tel" type="number" required /> <label class="col-sm-2 col-form-label">地址</label> <input class="form-control" name="addr" type="text" required /> <button type="button" class="btn btn-secondary" data-dismiss="modal">关闭</button> <button type="submit" class="btn btn-primary" id="regButton">确定</button> </form> 解析每一行代码

print(f'数学成绩最高分：{highest_score}') print(f'数学成绩最低分：{lowest_score}') print(f'数学成绩平均分：{average_score}') 1. 第一行是指定页面编码为 UTF-8。 2. 注释行，提示该模态框是用 elif sub...

List<int> aryList1 = new List<int>(); aryList1.AddRange(new int[] { 85, 92, 67, 94, 94 }); List<int> aryList2 = new List<int>(); aryList2.AddRange(new int[] { 50, 60, 57, 95 }); List<int> aryList3 = new List<int>(); aryList3.AddRange(new int[] { 95 }); List> classGrade = new List>(); classGrade.Add(aryList1); classGrade.Add(aryList2); classGrade.Add(aryList3); List<int> highestGrade(List> p2) { List<int> result = new List<int>(); for (int i = 0; i < p2.Count; i++) { int max; List<int> temp = p2[i]; for (int j = 1; j < temp.Count; j++) { if (temp[0] < temp[j]) { max = temp[j]; temp[j] = temp[0]; temp[0] = max; } } result.Add(temp[0]); } return result; }; Console.WriteLine($"The highest grade is {highestGrade(classGrade).Max()}"); 改进代码

List<List<int>> classGrade = new List<List<int>> { new List<int> { 85, 92, 67, 94, 94 }, new List<int> { 50, 60, 57, 95 }, new List<int> { 95 } }; Console.WriteLine($"The highest grade is {...

Accum TrustedAccum::TEEaccum(Stats &stats, Nodes nodes, Vote<Void, Cert> votes[MAX_NUM_SIGNATURES]) { View v = votes[0].getCData().getView(); View highest = 0; Hash hash = Hash(); std::set signers; for(int i = 0; i < MAX_NUM_SIGNATURES && i < this->qsize; i++) { Vote<Void, Cert> vote = votes[i]; CData<Void, Cert> data = vote.getCData(); Sign sign = vote.getSign(); PID signer = sign.getSigner(); Cert cert = data.getCert(); bool vd = verifyCData(stats, nodes, data, sign); bool vc = verifyCert(stats, nodes, cert); if(data.getPhase() == PH1_NEWVIEW && data.getView() == v && signers.find(signer) == signers.end() && vd && vc) { if(DEBUG1) { std::cout << KMAG << "[" << this->id << "]" << "inserting signer" << KNRM << std::endl; } signers.insert(signer); if(cert.getView() >= highest) { highest = cert.getView(); hash = cert.getHash(); } } else { if(DEBUG1) { std::cout << KMAG << "[" << this->id << "]" << "vote:" << vote.prettyPrint() << KNRM << std::endl; } if(DEBUG1) { std::cout << KMAG << "[" << this->id << "]" << "not inserting signer (" << signer << ") because:" << "check-phase=" << std::to_string(data.getPhase() == PH1_NEWVIEW) << "(" << data.getPhase() << "," << PH1_NEWVIEW << ")" << ";check-view=" << std::to_string(data.getView() == v) << ";check-notin=" << std::to_string(signers.find(signer) == signers.end()) << ";verif-data=" << std::to_string(vd) << ";verif-cert=" << std::to_string(vc) << KNRM << std::endl; } } } bool set = true; unsigned int size = signers.size(); std::string text = std::to_string(set) + std::to_string(v) + std::to_string(highest) + hash.toString() + std::to_string(size); Sign sign(this->priv,this->id,text); return Accum(v, highest, hash, size, sign); }

- Vote<Void, Cert> votes[MAX_NUM_SIGNATURES]：一个最大长度为MAX_NUM_SIGNATURES的Vote<Void, Cert>数组。函数的主要功能是根据给定的投票数组，计算并返回一个Accum对象。函数内部的操作如下： - ...

改进代码 list<int> arylist1 = new list<int>(); arylist1.addrange(new int[] { 85, 92, 67, 94, 94 }); list<int> arylist2 = new list<int>(); arylist2.addrange(new int[] { 50, 60, 57, 95 }); list<int> arylist3 = new list<int>(); arylist3.addrange(new int[] { 95 }); list> classgrade = new list>(); classgrade.add(arylist1); classgrade.add(arylist2); classgrade.add(arylist3); list<int> highestgrade(list> p2) { list<int> result = new list<int>(); for (int i = 0; i < p2.count; i++) { int max; list<int> temp = p2[i]; for (int j = 1; j < temp.count; j++) { if (temp[0] < temp[j]) { max = temp[j]; temp[j] = temp[0]; temp[0] = max; } } result.add(temp[0]); } return result; }; console.writeline($"the highest grade is {highestgrade(classgrade).max()}");

list<int> highestgrade(list<list<int>> p2) { list<int> result = new list<int>(); for (int i = 0; i < p2.count; i++) { int max = p2[i][0]; list<int> temp = p2[i]; for (int j = 1; j < temp.count; ...

for (int i = sz - 1; i >= 0; --i, n >>= 1) { str[i] = (n & 1) + '0'; } str[sz] = '\0';

The loop starts from the most significant bit (i.e., the highest power of 2), and for each bit, it checks whether it is 0 or 1. If it is 1, the corresponding character in the string is set to '1', ...

请翻译这段注释 /** * psched_ratecfg_precompute__() - Pre-compute values for reciprocal division * @rate: Rate to compute reciprocal division values of * @mult: Multiplier for reciprocal division * @shift: Shift for reciprocal division * * The multiplier and shift for reciprocal division by rate are stored * in mult and shift. * * The deal here is to replace a divide by a reciprocal one * in fast path (a reciprocal divide is a multiply and a shift) * * Normal formula would be : * time_in_ns = (NSEC_PER_SEC * len) / rate_bps * * We compute mult/shift to use instead : * time_in_ns = (len * mult) >> shift; * * We try to get the highest possible mult value for accuracy, * but have to make sure no overflows will ever happen. * * reciprocal_value() is not used here it doesn't handle 64-bit values. */

* time_in_ns = (len * mult) >> shift; * * 我们尽量获取最高可能的mult值以提高精度，但必须确保不会发生溢出。 * * 这里没有使用reciprocal_value()函数，因为它无法处理64位的值。 */

for (int i = 0; i < triangleCount * 3; ++i) { unsigned int code = indices[i]; indices[i] = highest - code; if (code == 0) { ++highest; } }如何反写回去

for (int i = 0; i < triangleCount * 3; ++i) { unsigned int code = indices[i]; indices[i] = highest - code; if (code == 0) { ++highest; } } // 反写回去 for (int i = 0; i < triangleCount * 3; ++i)...

/用结构体编程:某班有5名学生，每名学生的数据包括学号、姓名、3 门课程的成绩; 从键盘输入5名学生数据，输出5名同学平均成绩，以及平均分最高的学生的数据（包括学号、姓名、 3 门课成绩、平均成绩)/ #include<stdio.h> struct student{ long int no; char name[20];//用英文输入姓拼音 int score[3]; double ave_score; }stu[5]; struct student p; int main(){ int i,j; int sum; printf("Please enter students' numbers first, then name, finally their scores\n"); for(i=0;i<5;i++) scanf("%ld",&stu[i].no); for(i=0;i<5;i++) gets(stu[i].name); for(i=0;i<5;i++) for(j=0;j<3;j++) scanf("%d",&stu[i].score[j]); i=0,j=0; while(i<5){ sum = 0; while(j<3){ sum += stu[i].score[j]; j++; } stu[i].ave_score = 1.0 sum / 3; i++; } for(i=0;i<5;i++) printf("%lf ",&stu[i].ave_score); p = stu; for(i=1;i<5;i++) if(stu[i].ave_score > p->ave_score) p = &stu[i]; printf("%lf",p->ave_score); }

printf("The student with the highest average score is:\n"); printf("name:%s\n",p->name); printf("no:%ld\n",p->no); printf("score:%d %d %d\n",p->score[0],p->score[1],p->score[2]); printf("ave_...

Apply the NMF to electrical brain signals data clustering. If possible, please give the visualization results (or any other analytical results) for clustering, where the dataset file can be downloaded from the ***\*‘./dataset/ebs/waveform-5000.csv’\****.

The color of each data point represents the cluster membership, which is determined by the component with the highest weight for that data point. Please note that the above code is just an example ...

#include <iostream> #include <iomanip> #include <string.h> #include <cmath> #define M 3 // 课程门数 #define N 4 //学生数组中的学生个数 using namespace std; class Student { public: Student() {} Student(const Student&); void input_info() { cin>>id; cin>>name; for(int i=0; i<3; i++) cin>>score[i]; isFail=false; for(int i=0; i<3; i++) if(score[i]<60) isFail=true; } void input_lesson_ids() { for(int i=0; i<M; i++) cin>>lesson_id[i]; } void show_info() { cout<<"Student id:"<<id<<endl; cout<<"Student name:"<<name<<endl; cout<<setw(10)<<"lesson_id "; for(int i=0; i<M; i++) cout<<setw(10)<<lesson_id[i]; cout<<setw(10)<<"Average"; cout<<endl; cout<<setw(10)<<"scores "; for(int i=0; i<M; i++) cout<<setw(10)<<score[i]; cout<<setw(10)<<ave(); cout<<endl; if(isFail) cout<<"The student failed."<<endl; else cout<<"The student didn\'t fail."<<endl; cout<<"------华丽的分割线--------"<<endl; } float ave()//求平均成绩 { //实现求平均成绩并返回 ...... } string get_name() { return name; } private: int id; string name; bool isFail; static int lesson_id[M]; float score[M]; }; int Student::lesson_id[M]; Student::Student(const Student& s) { //完成拷贝构造函数的实现，拷贝出的对象和原对象一致 ...... } int main() { Student cs[N]; cs[0].input_lesson_ids();// 用一个学生对象对静态数组成员赋值 for(int i=0; i<N; i++) cs[i].input_info(); //求出最高平均成绩并按要求格式输出相关语句 ...... //求出最低平均成绩并按要求格式输出相关语句 ...... //按照平均成绩的高低对学生数组进行排序 ...... for(int i=0; i<N; i++)//输出排序后的结果 cs[i].show_info(); return 0; }补全c++

cout << "The student with the highest average score:" << endl; cs[max_index].show_info(); // 求出最低平均成绩并按要求格式输出相关语句 float min_ave = 100; int min_index = 0; for(int i = 0; i < ...

输入: 201 202 203 1 zhang 67 88 92 2 li 44 77 99 3 wang 76 82 95 4 zhao 63 82 86 输出: Student wang got the highest average score as 84.3333 Student li got the lowest average score as 73.3333 Student id:3 Student name:wang lesson_id 201 202 203 Average scores 76 82 95 84.3333 The student didn't fail. ------华丽的分割线-------- Student id:1 Student name:zhang lesson_id 201 202 203 Average scores 67 88 92 82.3333 The student didn't fail. ------华丽的分割线-------- Student id:4 Student name:zhao lesson_id 201 202 203 Average scores 63 82 86 77 The student didn't fail. ------华丽的分割线-------- Student id:2 Student name:li lesson_id 201 202 203 Average scores 44 77 99 73.3333 The student failed. ------华丽的分割线-------- 程序部分代码如下，请补充完整。 #include <iostream> #include <iomanip> #include <string.h> #include <cmath> #define M 3 // 课程门数 #define N 4 //学生数组中的学生个数 using namespace std; class Student { public: Student() {} Student(const Student&); void input_info() { cin>>id; cin>>name; for(int i=0; i<3; i++) cin>>score[i]; isFail=false; for(int i=0; i<3; i++) if(score[i]<60) isFail=true; } void input_lesson_ids() { for(int i=0; i<M; i++) cin>>lesson_id[i]; } void show_info() { cout<<"Student id:"<<id<<endl; cout<<"Student name:"<<name<<endl; cout<<setw(10)<<"lesson_id "; for(int i=0; i<M; i++) cout<<setw(10)<<lesson_id[i]; cout<<setw(10)<<"Average"; cout<<endl; cout<<setw(10)<<"scores "; for(int i=0; i<M; i++) cout<<setw(10)<<score[i]; cout<<setw(10)<<ave(); cout<<endl; if(isFail) cout<<"The student failed."<<endl; else cout<<"The student didn\'t fail."<<endl; cout<<"------华丽的分割线--------"<<endl; } float ave()//求平均成绩 { //实现求平均成绩并返回 ...... } string get_name() { return name; } private: int id; string name; bool isFail; static int lesson_id[M]; float score[M]; }; int Student::lesson_id[M]; Student::Student(const Student& s) { //完成拷贝构造函数的实现，拷贝出的对象和原对象一致 ...... } int main() { Student cs[N]; cs[0].input_lesson_ids();// 用一个学生对象对静态数组成员赋值 for(int i=0; i<N; i++) cs[i].input_info(); //求出最高平均成绩并按要求格式输出相关语句 ...... //求出最低平均成绩并按要求格式输出相关语句 ...... //按照平均成绩的高低对学生数组进行排序 ...... for(int i=0; i<N; i++)//输出排序后的结果 cs[i].show_info(); return 0; }

cout << "Student " << cs[max_ave_index].get_name() << " got the highest average score as " << fixed << setprecision(4) << max_ave << endl; //求出最低平均成绩并按要求格式输出相关语句 int min_ave_...

sign int i = 0x81111111; i = (i >> 31) | (-i >>> 31); System.out.println( i ); reserve highest 1 int i = 0x88000088; i |= (i >> 1); i |= (i >> 2); i |= (i >> 4); i |= (i >> 8); i |= (i >> 16); i -= (i >>> 1); System.out.printf( "0x%X%n", i ); reserve lowest 1 int i = 0x88000088; i &= -i; System.out.printf( "0x%X%n", i ); number of leading 0 int i = 0x88000088; int n = 0; if (i >>> 16 == 0) { n += 16; i <<= 16; } if (i >>> 24 == 0) { n += 8; i <<= 8; } if (i >>> 28 == 0) { n += 4; i <<= 4; } if (i >>> 30 == 0) { n += 2; i <<= 2; } if (i >>> 31 == 0) { n += 1; i <<= 1; } if (i == 0) n++; // n -= i >>> 31; System.out.println( n ); number of 1 int i = 0x81111111; i = ((i&0xAAAAAAAA) >>> 1) + (i&0x55555555); i = ((i&0xCCCCCCCC) >>> 2) + (i&0x33333333); i = ((i&0xF0F0F0F0) >>> 4) + (i&0x0F0F0F0F); i = ((i&0xFF00FF00) >>> 8) + (i&0x00FF00FF); i = ((i&0xFFFF0000) >>> 16) + (i&0x0000FFFF); System.out.println( i ); reverse order int i = 0x11111111; i = ((i&0xAAAAAAAA)>>>1) | ((i&0x55555555)<<1); i = ((i&0xCCCCCCCC)>>>2) | ((i&0x33333333)<<2); i = ((i&0xF0F0F0F0)>>>4) | ((i&0x0F0F0F0F)<<4); i = ((i&0xFF00FF00)>>>8) | ((i&0x00FF00FF)<<8); i = ((i&0xFFFF0000)>>>16) | ((i&0x0000FFFF)<<16); System.out.printf( "0x%X%n", i );翻译代码

1.将i初始化为0x81111111，然后对i进行位运算，将i的最高位和最低位进行保留，并输出i的值。 2.将i初始化为0x88000088，然后对i进行位运算，将i的最高位进行保留，并输出i的值。 3.将i初始化为0x88000088，然后对i...

#include<stdio.h> #include<stdlib.h> #include<math.h> #include<time.h> #define N 50 typedef struct process

for (int i = 0; i < N; i++) { processes[i].id = i + 1; processes[i].priority = rand() / (float)RAND_MAX; // 随机优先级 processes[i].timestamp = time(NULL); // 设置当前时间 } } // 示例：找到具有...

int curr; /* highest queued rt task prio */

curr 是实时调度器中的一个重要变量，用于记录当前就绪队列中优先级最高的实时进程的优先级。在 Linux 内核中，实时进程分为两种类型：SCHED_FIFO 和 SCHED_RR。SCHED_FIFO 表示先进先出调度策略，即按照优先级...

#include<iostream> #include<iomanip> using namespace std; const int n=3; struct student { char num[6]; char name[8]; int score[4]; float avr; }stu[n]; int main() { int i,j,max,maxi,sum; float average; for(i=0;i<n;i++) { cout<<"input scores of student"<<i+1<<endl; cout<<"NO.:"; cin>>stu[i].num; cout<<"name:"; cin>>stu[i].name; for(j=0;j<3;j++) {cout<<"score"<<j+1<<":"; cin>>stu[i].score[j];} cout<<endl; } average=0; max=0; maxi=0; for(i=0;i<n;i++) { sum=0; for(j=0;j<3;j++) sum+=stu[i].score[j]; stu[i].avr=sum/3.0; average+=stu[i].avr; if(sum>max) {max=sum; maxi=i;} } average/=n; cout<<" No. name score1 score2 score3 average"<<endl; for(i=0;i<n;i++) {cout<<setw(8)<<stu[i].num<<" "<<setw(10)<<stu[i].name<<" "; for(j=0;j<3;j++) cout<<setw(3)<<stu[i].score[i]<<" "; cout<<stu[i].avr<<endl;} cout<<"average="<<average<<endl; cout<<"The highest score is :"<<stu[maxi].name<<", score total:"<<max<<endl; return 0; }

这段代码实现了一个简单的学生成绩管理系统，可以输入学生的学号、姓名和三门课的成绩，计算出每个学生的平均成绩以及全班的平均成绩和最高分。在输入学生信息时，使用了一个循环，依次输入每个学生的信息。...

Apply the power method to find the highest eigenvalue for the inverse of Matrix A = [2 -3 1; -2 5 -2; 3 -2 -5] in Matlab. Set the stopping criterion as 1% (Estimated Eigenvector need to be normalized). Estimated Eigenvector (normalized) and Estimated High

Here's how you can apply the power method to find the highest eigenvalue for the inverse of Matrix A in Matlab: matlab % Define the matrix A A = [2 -3 1; -2 5 -2; 3 -2 -5]; % Set the stopping ...

Hizli-Satis-ve-Stok-Takip-Otomasyonu-master_highest4w7_C#_asp.ne

《高效销售与库存跟踪自动化系统——C#与ASP.NET实现详解》在IT行业中，高效地管理和跟踪销售及库存是企业运营的关键环节。本项目"高效销售与库存跟踪自动化系统"（Hizli-Satis-ve-Stok-Takip-Otomasyonu-master）...

相关推荐

专业认知 - 用 Python 制作 13 个小游戏，边玩边学！(含源码).pdf

python2-kombu-4.6.3-3.el7.noarch.rpm

for (int i = sz - 1; i >= 0; --i, n >>= 1) { str[i] = (n & 1) + '0'; } str[sz] = '\0';

for (int i = 0; i < triangleCount * 3; ++i) { unsigned int code = indices[i]; indices[i] = highest - code; if (code == 0) { ++highest; } }如何反写回去

Apply the NMF to electrical brain signals data clustering. If possible, please give the visualization results (or any other analytical results) for clustering, where the dataset file can be downloaded from the ***\*‘./dataset/ebs/waveform-5000.csv’\****.

#include<stdio.h> #include<stdlib.h> #include<math.h> #include<time.h> #define N 50 typedef struct process

int curr; /* highest queued rt task prio */

Apply the power method to find the highest eigenvalue for the inverse of Matrix A = [2 -3 1; -2 5 -2; 3 -2 -5] in Matlab. Set the stopping criterion as 1% (Estimated Eigenvector need to be normalized). Estimated Eigenvector (normalized) and Estimated High

Hizli-Satis-ve-Stok-Takip-Otomasyonu-master_highest4w7_C#_asp.ne

最新推荐

16-17 数据挖掘算法基础 - 分类与回归1(1).ipynb

精选微信小程序源码：停车场管理小程序（含源码+源码导入视频教程&文档教程，亲测可用）

最新闪客网盘系统源码支持限速+按时收费+文件分享+可对接易支付

C语言数组操作：高度检查器编程实践

管理建模和仿真的文件

【KUKA系统变量进阶】：揭秘从理论到实践的5大关键技巧

如何使用Python编程语言创建一个具有动态爱心图案作为背景并添加文字'天天开心（高级版）'的图形界面？

基于Swift开发的嘉定单车LBS iOS应用项目解析

"互动学习：行动中的多样性与论文攻读经历"

PROTEUS符号定制指南：个性化元件创建与修改的全面攻略