计算特殊矩阵中等于 x 的条目 · GeeksForGeeks 中文系列教程

# 计算特殊矩阵中等于 x 的条目 > 原文： [https://www.geeksforgeeks.org/count-entries-equal-to-x-in-a-special-matrix/](https://www.geeksforgeeks.org/count-entries-equal-to-x-in-a-special-matrix/) 您将得到 n 阶方矩阵（matrix [] []），其中 matrix [i] [j] = i * j。查找具有等于给定数字 x 的条目的像元数。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像元值是否等于给定的 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 // 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 // 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 ```py # 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# ```cs // 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 // 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 ``` 本文由 [**Shivam Pradhan（anuj_charm）**](http://www.facebook.com/ma5ter6it) 提供。如果您喜欢 GeeksforGeeks 并希望做出贡献，则还可以使用 [tribution.geeksforgeeks.org](http://www.contribute.geeksforgeeks.org) 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。查看您的文章出现在 GeeksforGeeks 主页上，并帮助其他 Geeks。