更新分数后学生排名跃升
给定三个数组名称【】、标记【】和更新【】其中:
- 人名[] 包含学生的姓名。
- 分数[] 包含同一学生的分数。
- 更新[] 包含更新这些学生分数的整数。
任务是在学生排名上升和跳跃后找到分数最高的学生的名字，即以前的排名-当前的排名。
注意:学生的详细信息按分数降序排列，如果两个以上的学生得分相同(也是最高的)，那么选择之前排名较低的一个。
示例:
输入:Name[]= {“Sam”、“ram”、“geek”、“sonu”}， marks[] = {99、75、70、60}， updates[] = {-10、5、9、39 } T5】输出: Name: sonu，Jump: 3 Updated marks 为{89、80、79、99}，很明显 sonu 的最高分是跳 3
输入:名称[]= {“Sam”、“ram”、“geek”}， 标记[] = {80，79，75}， 更新[] = {0，5，-9 } T5】输出:名称:ram，跳转:1
方法:创建一个结构结构学生存储每个学生的信息，每个学生将有 3 个属性学生的名字、学生的标记、学生的 prev_rank 。 现在，根据更新的内容更新分数【】然后用数组的单次遍历，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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# Python3 implementation of the approach
# Function to print the 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
java 描述语言
<script>
// 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
</script>
Output:
Name: ram, Jump: 1
时间复杂度:O(N) T3】辅助空间: O(N)
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