预期时间和给定时间之间的时差

考虑到最初的时钟时间 h1:m1 现在的时钟时间呢 h2:m2 ,表示时间和分钟 24小时 时钟格式。现在的时钟 h2:m2 可能正确,也可能不正确。也给出了一个变量K,表示经过的小时数。任务是以秒为单位计算延迟,即。 时间 预期时间和给定时间之间的差异。 例如:

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输入 :h1=10,m1=12,h2=10,m2=17,k=2 输出 :115分钟 时钟最初显示10:12。2小时后,它必须显示12:12。但此时,时钟显示为10:17。因此,时钟必须滞后115分钟。答案是115。 输入 :h1=12,m1=00,h2=12,m2=58,k=1 输出 :2分钟 时钟最初显示12:00。1小时后必须显示13:00。但此时,时钟显示12:58。因此,时钟必须滞后2分钟。所以答案是2。

方法 :

  1. 将h:m格式的给定时间转换为分钟数。它只是60*h+m。
  2. 计算两个计算时间(将K小时添加到初始时间)。
  3. 以分钟为单位找出差异,这将是答案。

以下是上述方法的实施情况。

C++

// C++ program to calculate clock delay
#include <bits/stdc++.h>
// Function definition with logic
int lagDuration( int h1, int m1, int h2, int m2, int k)
{
int lag, t1, t2;
// Conversion to minutes
t1 = (h1 + k) * 60 + m1;
// Conversion to minutes
t2 = h2 * 60 + m2;
// Calculating difference
lag = t1 - t2;
return lag;
}
// Driver Code
int main()
{
int h1 = 12, m1 = 0;
int h2 = 12, m2 = 58;
int k = 1;
int lag = lagDuration(h1, m1, h2, m2, k);
printf ( "Lag = %d minutes" , lag);
return 0;
}


JAVA

// Java program to
// calculate clock delay
class GFG
{
// Function definition
// with logic
static int lagDuration( int h1, int m1,
int h2, int m2,
int k)
{
int lag, t1, t2;
// Conversion to minutes
t1 = (h1 + k) * 60 + m1;
// Conversion to minutes
t2 = h2 * 60 + m2;
// Calculating difference
lag = t1 - t2;
return lag;
}
// Driver Code
public static void main(String args[])
{
int h1 = 12 , m1 = 0 ;
int h2 = 12 , m2 = 58 ;
int k = 1 ;
int lag = lagDuration(h1, m1, h2, m2, k);
System.out.println( "Lag = " + lag +
" minutes" );
}
}
// This code is contributed
// by Kirti_Mangal


Python3

# Python3 program to calculate clock delay
# Function definition with logic
def lagDuration(h1, m1, h2, m2, k):
lag, t1, t2 = 0 , 0 , 0
# Conversion to minutes
t1 = (h1 + k) * 60 + m1
# Conversion to minutes
t2 = h2 * 60 + m2
# Calculating difference
lag = t1 - t2
return lag
# Driver Code
h1, m1 = 12 , 0
h2, m2 = 12 , 58
k = 1
lag = lagDuration(h1, m1, h2, m2, k)
print ( "Lag =" , lag, "minutes" )
# This code has been contributed
# by 29AjayKumar


C#

// C# program to
// calculate clock delay
using System;
class GFG
{
// Function definition
// with logic
static int lagDuration( int h1, int m1,
int h2, int m2,
int k)
{
int lag, t1, t2;
// Conversion to minutes
t1 = (h1 + k) * 60 + m1;
// Conversion to minutes
t2 = h2 * 60 + m2;
// Calculating difference
lag = t1 - t2;
return lag;
}
// Driver Code
public static void Main()
{
int h1 = 12, m1 = 0;
int h2 = 12, m2 = 58;
int k = 1;
int lag = lagDuration(h1, m1, h2, m2, k);
Console.WriteLine( "Lag = " + lag +
" minutes" );
}
}
// This code is contributed
// by Akanksha Rai(Abby_akku)


PHP

<?php
// PHP program to
// calculate clock delay
function lagDuration( $h1 , $m1 ,
$h2 , $m2 , $k )
{
// Conversion to minutes
$t1 = ( $h1 + $k ) * 60 + $m1 ;
// Conversion to minutes
$t2 = $h2 * 60 + $m2 ;
// Calculating difference
$lag = $t1 - $t2 ;
return $lag ;
}
// Driver Code
$h1 = 12;
$m1 = 0;
$h2 = 12;
$m2 = 58;
$k = 1;
$lag = lagDuration( $h1 , $m1 , $h2 ,
$m2 , $k );
echo "Lag = $lag minutes" ;
// This code is contributed
// by Kirti_Mangal


Javascript

<script>
// javascript program to
// calculate clock delay
// Function definition
// with logic
function lagDuration(h1 , m1 , h2 , m2 , k) {
var lag, t1, t2;
// Conversion to minutes
t1 = (h1 + k) * 60 + m1;
// Conversion to minutes
t2 = h2 * 60 + m2;
// Calculating difference
lag = t1 - t2;
return lag;
}
// Driver Code
var h1 = 12, m1 = 0;
var h2 = 12, m2 = 58;
var k = 1;
var lag = lagDuration(h1, m1, h2, m2, k);
document.write( "Lag = " + lag + " minutes" );
// This code is contributed by aashish1995
</script>


输出:

Lag = 2 minutes

时间复杂性 :O(1)

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