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 :
- 将h:m格式的给定时间转换为分钟数。它只是60*h+m。
- 计算两个计算时间(将K小时添加到初始时间)。
- 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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