计算待收罚款总额
给定一个日期和包含在该日期行驶的汽车数量的整数数组(一个整数)，任务是根据以下规则计算罚款总额:
- 奇数编号的汽车只能在奇数日期行驶。
- 偶数日期的偶数车。
- 否则一辆车将被罚款 250 卢比。
例:
Input: car_num[] = {3, 4, 1, 2}, date = 15
Output: 500
Car with numbers '4' and '2' will be fined
250 each.
Input: car_num[] = {1, 2, 3} , date = 16
Output: 500
Car with numbers '1' and '3' will be fined
250 each.
进场:
- 开始遍历给定的数组。
- 检查当前车号和日期是否不匹配，即一个是偶数，另一个是奇数，反之亦然。
- 如果不匹配，就对那个车号罚款。否则不行。
- 打印罚款总额。
以下是上述方法的实现:
C++
// C++ implementation to calculate
// the total fine collected
#include <bits/stdc++.h>
using namespace std;
// function to calculate the total fine collected
int totFine(int car_num[], int n, int date, int fine)
{
int tot_fine = 0;
// traverse the array elements
for (int i = 0; i < n; i++)
// if both car no and date are odd or
// both are even, then statement
// evaluates to true
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
// required total fine
return tot_fine;
}
// Driver program to test above
int main()
{
int car_num[] = { 3, 4, 1, 2 };
int n = sizeof(car_num) / sizeof(car_num[0]);
int date = 15, fine = 250;
cout << totFine(car_num, n, date, fine);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to calculate
// the total fine collected
class GFG
{
// function to calculate
// the total fine collected
static int totFine(int car_num[], int n,
int date, int fine)
{
int tot_fine = 0;
// traverse the array elements
for (int i = 0; i < n; i++)
// if both car no and date
// are odd or both are even,
// then statement evaluates to true
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
// required total fine
return tot_fine;
}
// Driver Code
public static void main(String[] args)
{
int car_num[] = { 3, 4, 1, 2 };
int n = car_num.length;
int date = 15, fine = 250;
System.out.println(totFine(car_num, n,
date, fine));
}
}
// This code is contributed
// by ChitraNayal
Python 3
# Python 3 program to calculate
# the total fine collected
# function to calculate the total fine collected
def totFine(car_num, n, date, fine) :
tot_fine = 0
# traverse the array elements
for i in range(n) :
# if both car no and date are odd or
# both are even, then statement
# evaluates to true
if (((car_num[i] ^ date) & 1) == 1 ):
tot_fine += fine
# required total fine
return tot_fine
# Driver Program
if __name__ == "__main__" :
car_num = [ 3, 4, 1, 2 ]
n = len(car_num)
date, fine = 15, 250
# function calling
print(totFine(car_num, n, date, fine))
# This code is contributed by ANKITRAI1
C
// C# implementation to calculate
// the total fine collected
using System;
class GFG
{
// function to calculate the
// total fine collected
static int totFine(int[] car_num, int n,
int date, int fine)
{
int tot_fine = 0;
// traverse the array elements
for (int i = 0; i < n; i++)
// if both car no and date
// are odd or both are even,
// then statement evaluates to true
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
// required total fine
return tot_fine;
}
// Driver Code
public static void Main()
{
int[] car_num = { 3, 4, 1, 2 };
int n = car_num.Length;
int date = 15, fine = 250;
Console.Write(totFine(car_num, n,
date, fine));
}
}
// This code is contributed
// by ChitraNayal
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation to calculate
// the total fine collected
// function to calculate the
// total fine collected
function totFine(&$car_num, $n,
$date, $fine)
{
$tot_fine = 0;
// traverse the array elements
for ($i = 0; $i < $n; $i++)
// if both car no and date
// are odd or both are even,
// then statement evaluates
// to true
if ((($car_num[$i] ^
$date) & 1) == 1)
$tot_fine += $fine;
// required total fine
return $tot_fine;
}
// Driver Code
$car_num = array(3, 4, 1, 2 );
$n = sizeof($car_num);
$date = 15;
$fine = 250;
echo totFine($car_num, $n,
$date, $fine);
// This code is contributed
// by ChitraNayal
?>
java 描述语言
<script>
// Javascript implementation to calculate
// the total fine collected
// function to calculate the total fine collected
function totFine(car_num, n, date, fine)
{
let tot_fine = 0;
// traverse the array elements
for (let i = 0; i < n; i++)
// if both car no and date are odd or
// both are even, then statement
// evaluates to true
if (((car_num[i] ^ date) & 1) == 1)
tot_fine += fine;
// required total fine
return tot_fine;
}
// Driver program to test above
let car_num = [ 3, 4, 1, 2 ];
let n = car_num.length;
let date = 15, fine = 250;
document.write(totFine(car_num, n, date, fine));
//This code is contributed by Mayank Tyagi
</script>
Output:
500
来源:https://www . geeksforgeeks . org/Microsoft-面试-实习经验/
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