更新分数后跳升学生排名

给定三个数组名称[] ， marks []和updates [] ，其中：

1. names []包含学生的姓名。
2. mark []包含同一学生的分数。
3. updates []包含将用来更新这些学生的分数的整数。

任务是找到更新后分数最高的学生姓名，以及该学生的排名(即先前的排名-当前的排名)的跳跃。

注意：学生的详细信息按分数的降序排列，如果有两个以上的学生获得相等的分数(也是最高分)，则选择以前排名较低的那个。

例子：

Input: names[] = {“sam”, “ram”, “geek”, “sonu”}, 
marks[] = {99, 75, 70, 60}, 
updates[] = {-10, 5, 9, 39} 
Output: Name: sonu, Jump: 3 
Updated marks are {89, 80, 79, 99}, its clear that sonu has the highest marks with jump of 3

Input: names[] = {“sam”, “ram”, “geek”}, 
marks[] = {80, 79, 75}, 
updates[] = {0, 5, -9} 
Output: Name: ram, Jump: 1 

方法：创建一个结构struct Student来存储每个学生的信息，每个学生将具有3个学生的属性名称，学生的标记，学生的prev_rank 。
现在，根据updates []的内容更新分数，然后对数组进行一次遍历，找到更新后分数最高的学生。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Structure to store the information of
// students
struct Student {
    string name;
    int marks = 0;
    int prev_rank = 0;
};
 
// Function to print the name of student who
// stood first after updation in rank
void nameRank(string names[], int marks[],
                      int updates[], int n)
{
    // Array of students
    Student x[n];
 
    for (int i = 0; i < n; i++) {
 
        // Store the name of the student
        x[i].name = names[i];
 
        // Update the marks of the student
        x[i].marks = marks[i] + updates[i];
 
        // Store the current rank of the student
        x[i].prev_rank = i + 1;
    }
 
    Student highest = x[0];
    for (int j = 1; j < n; j++) {
        if (x[j].marks >= highest.marks)
            highest = x[j];
    }
 
    // Print the name and jump in rank
    cout << "Name: " << highest.name
         << ", Jump: "
         << abs(highest.prev_rank - 1)
         << endl;
}
 
// Driver code
int main()
{
    // Names of the students
    string names[] = { "sam", "ram", "geek" };
 
    // Marks of the students
    int marks[] = { 80, 79, 75 };
 
    // Updates that are to be done
    int updates[] = { 0, 5, -9 };
 
    // Number of students
    int n = sizeof(marks) / sizeof(marks[0]);
 
    nameRank(names, marks, updates, n);
}

Java

// Java implementation of the approach
import java.util.*;
   
class GFG{
  
static class Student
{
    String name;
    int marks, prev_rank;
      
    Student()
    {
        this.marks = 0;
        this.prev_rank = 0;
    }
}
      
// Function to print the name of student who
// stood first after updation in rank
static void nameRank(String []names, int []marks,
                      int []updates, int n)
{
     
    // Array of students
    Student []x = new Student[n];
   
    for(int i = 0; i < n; i++)
    {
        x[i] = new Student();
         
        // Store the name of the student
        x[i].name = names[i];
   
        // Update the marks of the student
        x[i].marks = marks[i] + updates[i];
   
        // Store the current rank of the student
        x[i].prev_rank = i + 1;
    }
   
    Student highest = x[0];
      
    for(int j = 1; j < n; j++)
    {
        if (x[j].marks >= highest.marks)
            highest = x[j];
    }
   
    // Print the name and jump in rank
    System.out.print("Name: " + highest.name +
                     ", Jump: " +
                     Math.abs(highest.prev_rank - 1));
}
  
// Driver code 
public static void main(String[] args)
{
     
    // Names of the students
    String []names = { "sam", "ram", "geek" };
      
    // Marks of the students
    int []marks = { 80, 79, 75 };
      
    // Updates that are to be done
    int []updates = { 0, 5, -9 };
      
    // Number of students
    int n = marks.length;
      
    nameRank(names, marks, updates, n);
}
}
 
// This code is contributed by pratham76

Python3

# Python3 implementation of the approach
 
# Function to prthe name of student who
# stood first after updation in rank
def nameRank(names, marks, updates, n):
     
    # Array of students
    x = [[0 for j in range(3)] for i in range(n)]
    for i in range(n):
         
        # Store the name of the student
        x[i][0] = names[i]
         
        # Update the marks of the student
        x[i][1]= marks[i] + updates[i]
         
        # Store the current rank of the student
        x[i][2] = i + 1
         
    highest = x[0]
    for j in range(1, n):
        if (x[j][1] >= highest[1]):
            highest = x[j]
             
    # Print the name and jump in rank
    print("Name: ", highest[0], ", Jump: ",
            abs(highest[2] - 1), sep="")
 
# Driver code
 
# Names of the students
names= ["sam", "ram", "geek"]
 
# Marks of the students
marks = [80, 79, 75]
 
# Updates that are to be done
updates = [0, 5, -9]
 
# Number of students
n = len(marks)
 
nameRank(names, marks, updates, n)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;
  
class GFG{
 
public class Student
{
    public string name;
    public int marks, prev_rank;
     
    public Student()
    {
        this.marks = 0;
        this.prev_rank = 0;
    }
}
     
// Function to print the name of student who
// stood first after updation in rank
static void nameRank(string []names, int []marks,
                      int []updates, int n)
{
     
    // Array of students
    Student []x = new Student[n];
  
    for(int i = 0; i < n; i++)
    {
        x[i] = new Student();
         
        // Store the name of the student
        x[i].name = names[i];
  
        // Update the marks of the student
        x[i].marks = marks[i] + updates[i];
  
        // Store the current rank of the student
        x[i].prev_rank = i + 1;
    }
  
    Student highest = x[0];
     
    for(int j = 1; j < n; j++)
    {
        if (x[j].marks >= highest.marks)
            highest = x[j];
    }
  
    // Print the name and jump in rank
    Console.Write("Name: " + highest.name +
                  ", Jump: " +
                  Math.Abs(highest.prev_rank - 1));
}
 
// Driver code 
public static void Main(string[] args)
{
     
    // Names of the students
    string []names = { "sam", "ram", "geek" };
     
    // Marks of the students
    int []marks = { 80, 79, 75 };
     
    // Updates that are to be done
    int []updates = { 0, 5, -9 };
     
    // Number of students
    int n = marks.Length;
     
    nameRank(names, marks, updates, n);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript implementation of the approach
     
// Function to print the name of student who
// stood first after updation in rank
function nameRank(names, marks, updates, n)
{
     
    // Array of students
    let x = new Array(n);
    for(let i = 0; i < n; i++)
    {
        x[i] = new Array(3);
        for(let j = 0; j < 3; j++)
        {
            x[i][j] = 0;
        }
    }
     
    for(let i = 0; i < n; i++)
    {
         
        // Store the name of the student
        x[i][0] = names[i]
          
        // Update the marks of the student
        x[i][1] = marks[i] + updates[i]
          
        // Store the current rank of the student
        x[i][2] = i + 1
    }
     
    let highest = x[0];
     
    for(let j = 1; j < n; j++)
    {
        if (x[j][1] >= highest[1])
            highest = x[j]
    }
     
    // Print the name and jump in rank
    document.write("Name: ", highest[0], ", Jump: ",
                   Math.abs(highest[2] - 1), sep = "")
}
 
// Driver code
 
// Names of the students
let names = [ "sam", "ram", "geek" ];
 
// Marks of the students
let marks = [ 80, 79, 75 ];
 
// Updates that are to be done
let updates = [ 0, 5, -9 ];
 
// Number of students
let n = marks.length;
 
nameRank(names, marks, updates, n)
 
// This code is contributed by unknown2108
 

输出：

Name: ram, Jump: 1