# 計算大小為 n 的矩陣中 k 的頻率，其中 matrix（i，j）= i + j > 原文： [https://www.geeksforgeeks.org/count-frequency-k-matrix-size-n-matrixi-j-ij/](https://www.geeksforgeeks.org/count-frequency-k-matrix-size-n-matrixi-j-ij/) 給定大小為 n * n 的矩陣。 計算該矩陣中給定元素 k 的頻率。 **這里的基本索引是 1。** 例子： ``` Input : n = 4, k = 7 Output : 2 Explanation The matrix will be 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8 in the given matrix where M(i, j) = i+j, frequency of 7 is 2 Input : n = 5, k = 4 Output : 3 Explanation The matrix will be 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9 6 7 8 9 10 Explanation In the given matrix where M(i, j) = i+j, frequency of 4 is 3 ``` **第一種方法** 1）構造大小為 n * n 的矩陣。 2）用 M（i，j）= i + j 填充值。（回想這里的基本索引為 1）。 3）迭代遍歷矩陣并計算給定元素的頻率。 這種方法效率不高，因為如果矩陣大小很大，將導致超過時間限制。時間復雜度將為 O（n ^ 2）。 **有效方法** 在此方法中，我們避免創建大小為 n * n 的矩陣。 例如 如果 n = 10，則矩陣為 ``` 2 3 4 5 6 7 8 9 10 11 3 4 5 6 7 8 9 10 11 12 4 5 6 7 8 9 10 11 12 13 5 6 7 8 9 10 11 12 13 14 6 7 8 9 10 11 12 13 14 15 7 8 9 10 11 12 13 14 15 16 8 9 10 11 12 13 14 15 16 17 9 10 11 12 13 14 15 16 17 18 10 11 12 13 14 15 16 17 18 19 11 12 13 14 15 16 17 18 19 20 ``` 現在，請注意第二對角線中的值如何相同，我們還可以發現計數增加的模式，例如 1、2、3、4 ，在這里我們可以看到 如果（n + 1 ）> = k，則 k 的頻率為 k-1 ，否則頻率將為 2 * n + 1-k ## CPP ``` // CPP program to find the frequency of k? // in matrix where m(i, j)=i+j #include <bits/stdc++.h> using namespace std; int find(int n, int k) { ????if (n + 1 >= k) ????????return (k - 1); ????else ????????return (2 * n + 1 - k); } // Driver Code int main() { ????int n = 4, k = 7; ????int freq = find(n, k); ????if (freq < 0) ????????cout << " element not exist \n "; ????else ????????cout << " Frequency of " << k? ?????????????<< " is " << freq << "\n"; ????return 0; } ``` ## Java ```java // Java program to find the? // frequency of k in matrix // in matrix where m(i, j)=i+j import java.util.*; import java.lang.*; public class GfG{ ????public static int find(int n, int k) ????{ ????????if (n + 1 >= k) ????????????return (k - 1); ????????else ????????????return (2 * n + 1 - k); ????} ????// Driver function? ????public static void main(String argc[]) ????{ ????????int n = 4, k = 7; ????????int freq = find(n, k); ????????if (freq < 0) ????????????System.out.print(" element" ????????????+ " not exist \n "); ????????else ????????????System.out.print(" Frequency" ????????????+ " of " + k + " is " + ????????????freq + "\n"); ????} } // This code is contributed by Sagar Shukla ``` ## Python3 ```py # Python program to find # the frequency of k? # in matrix where # m(i, j)=i+j import math def find( n,? k): ????if (n + 1 >= k): ????????return (k - 1) ????else: ????????return (2 * n + 1 - k) # Driver Code n = 4 k = 7 freq = find(n, k) if (freq < 0): ????print ( " element not exist") else: ????print(" Frequency of " , k ," is " , freq ) # This code is contributed # by Gitanjali. ``` ## C# ```cs // C# program to find the? // frequency of k in matrix // in matrix where m(i, j)=i+j using System; public class GfG{ ????public static int find(int n, int k) ????{ ????????if (n + 1 >= k) ????????????return (k - 1); ????????else ????????????return (2 * n + 1 - k); ????} ????// Driver function? ????public static void Main() ????{ ????????int n = 4, k = 7; ????????int freq = find(n, k); ????????if (freq < 0) ????????????Console.WriteLine(" element" ????????????+ " not exist "); ????????else ????????????Console.WriteLine(" Frequency" ????????????+ " of " + k + " is " + ????????????freq ); ????} } // This code is contributed by vt_m ``` ## PHP ```php <?php // PHP program to find the frequency of k? // in matrix where m(i, j)=i+j function find($n, $k) { ????if ($n + 1 >= $k) ????????return ($k - 1); ????else ????????return (2 * $n + 1 - $k); } ????// Driver Code ????$n = 4;? ????$k = 7; ????$freq = find($n, $k); ????if ($freq < 0) ????????echo " element not exist \n "; ????else ????????echo " Frequency of " , $k ????????????, " is " , $freq , "\n"; // This code is contributed by anuj_67\. ?> ``` Output: ``` Frequency of 7 is 2 ``` * * * * * *