给定两个整数 N 和 M .问题是找到最接近的数字 N 并且可以被 M .如果有多个这样的数字，则输出具有最大绝对值的数字。如果 N 完全可以被 M ，然后输出 N 只有要求时间复杂度为O（1）。 限制条件： m！=0 例如：
null
Input : n = 13, m = 4Output : 12Input : n = -15, m = 6Output : -18Both -12 and -18 are closest to -15, but-18 has the maximum absolute value.
资料来源： 微软面试经验|设置125 .
我们找到n/m的值。让这个值为q。然后我们找到两种可能性中最接近的。一个是q*m，另一个是（m*（q+1））或（m*（q-1）），这取决于给定的两个数字中是否有一个是负数。 算法：
closestNumber(n, m) Declare q, n1, n2 q = n / m n1 = m * q if (n * m) > 0 n2 = m * (q + 1) else n2 = m * (q - 1) if abs(n-n1) < abs(n-n2) return n1 return n2
C++
// C++ implementation to find the number closest to n // and divisible by m #include <bits/stdc++.h> using namespace std; // function to find the number closest to n // and divisible by m int closestNumber( int n, int m) { // find the quotient int q = n / m; // 1st possible closest number int n1 = m * q; // 2nd possible closest number int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); // if true, then n1 is the required closest number if ( abs (n - n1) < abs (n - n2)) return n1; // else n2 is the required closest number return n2; } // Driver program to test above int main() { int n = 13, m = 4; cout << closestNumber(n, m) << endl; n = -15; m = 6; cout << closestNumber(n, m) << endl; n = 0; m = 8; cout << closestNumber(n, m) << endl; n = 18; m = -7; cout << closestNumber(n, m) << endl; return 0; } |
JAVA
// Java implementation to find the number closest to n // and divisible by m public class close_to_n_divisible_m { // function to find the number closest to n // and divisible by m static int closestNumber( int n, int m) { // find the quotient int q = n / m; // 1st possible closest number int n1 = m * q; // 2nd possible closest number int n2 = (n * m) > 0 ? (m * (q + 1 )) : (m * (q - 1 )); // if true, then n1 is the required closest number if (Math.abs(n - n1) < Math.abs(n - n2)) return n1; // else n2 is the required closest number return n2; } // Driver program to test above public static void main(String args[]) { int n = 13 , m = 4 ; System.out.println(closestNumber(n, m)); n = - 15 ; m = 6 ; System.out.println(closestNumber(n, m)); n = 0 ; m = 8 ; System.out.println(closestNumber(n, m)); n = 18 ; m = - 7 ; System.out.println(closestNumber(n, m)); } } // This code is contributed by Sumit Ghosh |
Python3
# Python 3 implementation to find # the number closest to n # Function to find the number closest # to n and divisible by m def closestNumber(n, m) : # Find the quotient q = int (n / m) # 1st possible closest number n1 = m * q # 2nd possible closest number if ((n * m) > 0 ) : n2 = (m * (q + 1 )) else : n2 = (m * (q - 1 )) # if true, then n1 is the required closest number if ( abs (n - n1) < abs (n - n2)) : return n1 # else n2 is the required closest number return n2 # Driver program to test above n = 13 ; m = 4 print (closestNumber(n, m)) n = - 15 ; m = 6 print (closestNumber(n, m)) n = 0 ; m = 8 print (closestNumber(n, m)) n = 18 ; m = - 7 print (closestNumber(n, m)) # This code is contributed by Nikita tiwari. |
C#
// C# implementation to find the // number closest to n and divisible by m using System; class GFG { // function to find the number closest to n // and divisible by m static int closestNumber( int n, int m) { // find the quotient int q = n / m; // 1st possible closest number int n1 = m * q; // 2nd possible closest number int n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); // if true, then n1 is the required closest number if (Math.Abs(n - n1) < Math.Abs(n - n2)) return n1; // else n2 is the required closest number return n2; } // Driver program to test above public static void Main() { int n = 13, m = 4; Console.WriteLine(closestNumber(n, m)); n = -15; m = 6; Console.WriteLine(closestNumber(n, m)); n = 0; m = 8; Console.WriteLine(closestNumber(n, m)); n = 18; m = -7; Console.WriteLine(closestNumber(n, m)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP implementation to find // the number closest to n and // divisible by m // function to find the number // closest to n and divisible by m function closestNumber( $n , $m ) { // find the quotient $q = (int) ( $n / $m ); // 1st possible closest number $n1 = $m * $q ; // 2nd possible closest number $n2 = ( $n * $m ) > 0 ? ( $m * ( $q + 1)) : ( $m * ( $q - 1)); // if true, then n1 is the // required closest number if ( abs ( $n - $n1 ) < abs ( $n - $n2 )) return $n1 ; // else n2 is the required // closest number return $n2 ; } // Driver Code $n = 13; $m = 4; echo closestNumber( $n , $m ), "" ; $n = -15; $m = 6; echo closestNumber( $n , $m ), "" ; $n = 0; $m = 8; echo closestNumber( $n , $m ), "" ; $n = 18; $m = -7; echo closestNumber( $n , $m ), "" ; // This code is contributed by jit_t ?> |
Javascript
<script> // Javascript implementation to find // the number closest to n and // divisible by m // function to find the number // closest to n and divisible by m function closestNumber(n, m) { // find the quotient let q = parseInt(n / m); // 1st possible closest number let n1 = m * q; // 2nd possible closest number let n2 = (n * m) > 0 ? (m * (q + 1)) : (m * (q - 1)); // if true, then n1 is the // required closest number if (Math.abs(n - n1) < Math.abs(n - n2)) return n1; // else n2 is the required // closest number return n2; } // Driver Code let n = 13; let m = 4; document.write(closestNumber(n, m) + "<br>" ); n = -15; m = 6; document.write(closestNumber(n, m) + "<br>" ); n = 0; m = 8; document.write(closestNumber(n, m) + "<br>" ); n = 18; m = -7; document.write(closestNumber(n, m) + "<br>" ); // This code is contributed by gfgking </script> |
输出：
12-18021
时间复杂性： O（1） 本文由 阿尤什·焦哈里 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
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