计算特殊矩阵中等于x的条目数量
原文： https://www.geeksforgeeks.org/count-entries-equal-to-x-in-a-special-matrix/
您将得到n阶方阵（matrix[][]），其中matrix[i][j] = i * j。 查找具有等于给定数字x的条目的单元格数。
注意：矩阵的索引从 1 开始，即1 <= i, j < = n。
示例：
Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 4
Output : 3
Input : matrix[][] = {1, 2, 3, 4,
2, 4, 6, 8,
3, 6, 9, 12,
4, 8, 12, 16}
x = 12
Output : 2
简单方法是遍历整个矩阵并检查单元格值是否等于给定的x，然后相应地增加计数值。 这种方法的时间复杂度很高，等于O(n ^ 2)。
// traverse and check whole matrix
for (int i=1; i<=n ; i++)
{
for (int j=1; j<=n; j++)
if (i * j == x)
count++;
}
// return count
return count;
有效方法仅查找给定x的因子数量在 0 到x之间，并且那些因子（包括除数和商）必须小于或等于n（矩阵的阶数）。 在这种情况下，时间复杂度将为O(n)。
// traverse and find the factors
for (int i=1; i<=n && i<=x ; i++)
{
// x%i == 0 means i is factor of x
// x/i <= n means i and j are <= n (for i*j=x)
if ( x/i <= n && x%i ==0)
count++;
}
// return count
return count;
C++
// CPP program for counting number of cell
// equals to given x
#include<bits/stdc++.h>
using namespace std;
// function to count factors as number of cell
int count (int n, int x)
{
int count == 0;
// traverse and find the factors
for (int i=1; i<=n && i<=x ; i++)
{
// x%i == 0 means i is factor of x
// x/i <= n means i and j are <= n (for i*j=x)
if ( x/i <= n && x%i ==0)
count++;
}
// return count
return count;
}
// driver program
int main()
{
int n = 8;
// we can manually assume matrix of order 8*8
// where mat[i][j] = i*j , 0<i,j<=n
int x = 24;
cout << count(n, x);
return 0;
}
Java
// Java program for counting number of
// cell equals to given x
class GFG
{
// function to count factors as
// number of cell
static int count (int n, int x)
{
int count = 0;
// traverse and find the factors
for (int i = 1; i <= n && i <= x ;
i++)
{
// x%i == 0 means i is factor
// of x. x/i <= n means i and
// j are <= n (for i*j=x)
if ( x / i <= n && x % i == 0)
count++;
}
// return count
return count;
}
// driver program
public static void main(String args[])
{
int n = 8;
// we can manually assume matrix
// of order 8*8 where
// mat[i][j] = i*j , 0<i,j<=n
int x = 24;
System.out.println(count(n, x));
}
}
/*This code is contributed by Danish kaleem*/
Python3
# Python 3 program for counting
# number of cell equals to given x
# function to count factors
# as number of cell
def count(n, x):
cnt = 0
# traverse and find the factors
for i in range(1, n + 1):
# // x%i == 0 means i is factor of x
# x/i <= n means i and j are <= n (for i*j=x)
if i <= x:
if x // i <= n and x % i == 0:
cnt += 1
return cnt
# Driver code
n = 8
x = 24
print(count(n, x))
# This code is contributed by Shrikant13
C#
// C# program for counting number
// of cell equals to given x
using System;
class GFG {
// function to count factors as
// number of cell
static int count (int n, int x) {
int count = 0;
// traverse and find the factors
for (int i = 1; i <= n &&
i <= x ; i++)
{
// x%i == 0 means i is factor
// of x. x/i <= n means i and
// j are <= n (for i*j=x)
if ( x / i <= n && x % i == 0)
count++;
}
// return count
return count;
}
// Driver Code
public static void Main()
{
int n = 8;
// we can manually assume matrix
// of order 8*8 where
// mat[i][j] = i*j , 0<i,j<=n
int x = 24;
Console.Write(count(n, x));
}
}
// This code is contributed by Nitin Mittal.
PHP
<?php
// PHP program for counting
// number of cell equals to
// given x
// function to count factors
// as number of cell
function c_ount ( $n, $x)
{
$Count = 0;
// traverse and find the factors
for ( $i = 1; $i <= $n and
$i <= $x ; $i++)
{
// x%i == 0 means i is
// factor of x x/i <= n
// means i and j are
// <= n (for i*j=x)
if ( $x / $i <= $n and
$x % $i == 0)
$Count++;
}
// return count
return $Count;
}
// Driver Code
$n = 8;
// we can manually assume
// matrix of order 8*8
// where mat[i][j] = i*j ,
// 0<i,j<=n
$x = 24;
echo c_ount($n, $x);
// This code is contributed by anuj_67\.
?>
输出：
4
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