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\nInput:<\/strong><\/p>\n
\n 3\n 10 20\n 20 30\n 30 10\n <\/pre>\n<\/div>\n
\nOutput:<\/strong><\/p>\n
\n 3000\n <\/pre>\n<\/div>\n
Solution Process<\/h2>\n
1. Understanding the Basic Concept<\/h3>\n
\n Matrix multiplication is determined by the dimensions of the two matrices. Multiplying matrix A (p\u00d7q) and matrix B (q\u00d7r) results in matrix C (p\u00d7r).
\n The number of operations needed is p * q * r<\/code>. If multiple matrices are provided, it is important to find the optimal multiplication order.\n<\/p>\n
2. Utilizing Dynamic Programming<\/h3>\n\n This problem can be solved using dynamic programming. We can construct a DP table to determine the optimal order of matrix multiplication. \n Let dp[i][j]<\/code> represent the minimum number of operations required to multiply from the i<\/code>th matrix to the j<\/code>th matrix.\n<\/p>\n 3. Filling the DP Table<\/h3>\n\n 1. dp[i][i] = 0<\/code> (No operations are needed when multiplying a matrix with itself.) \n 2. When multiplying two or more matrices, divide the calculation based on the intermediate matrix. \n 3. The order of matrix multiplication is determined as follows.\n<\/p>\n \nfor length from 2 to n:\n for i from 1 to n-length+1:\n j = i + length - 1\n dp[i][j] = \u221e\n for k from i to j-1:\n cost = dp[i][k] + dp[k+1][j] + dimensions[i-1]*dimensions[k]*dimensions[j]\n dp[i][j] = min(dp[i][j], cost)\n<\/pre>\n4. Python Implementation and C++ Code<\/h3>\n\n Let’s implement the above algorithm in C++.\n<\/p>\n \n#include \n#include \n#include \n\nusing namespace std;\n\nint main() {\n int n;\n cin >> n;\n vector dimensions(n + 1);\n for (int i = 0; i < n; i++) {\n int p, q;\n cin >> p >> q;\n dimensions[i] = p;\n if (i == n - 1) dimensions[i + 1] = q; \/\/ last matrix's number of columns\n }\n\n vector> dp(n, vector(n, 0));\n\n for (int length = 2; length <= n; length++) {\n for (int i = 0; i < n - length + 1; i++) {\n int j = i + length - 1;\n dp[i][j] = INT_MAX;\n for (int k = i; k < j; k++) {\n int cost = dp[i][k] + dp[k + 1][j] + dimensions[i] * dimensions[k + 1] * dimensions[j + 1];\n dp[i][j] = min(dp[i][j], cost);\n }\n }\n }\n\n cout << dp[0][n - 1] << endl;\n return 0;\n}\n<\/int><\/vector<\/int><\/limits.h><\/vector><\/iostream><\/pre>\n5. Time Complexity<\/h2>\n\n The time complexity of this algorithm is O(n^3). This occurs because we need to consider all possible multiplication orders when there are n matrices.\n<\/p>\n 6. Conclusion<\/h2>\n\n The problem of minimizing matrix multiplication operations can be efficiently solved using dynamic programming. \n This problem is very useful in developing algorithm problem-solving skills and is an important topic applicable in various fields. \n By implementing this problem in C++, you can gain a deeper understanding of the concepts of algorithms and enhance your abilities to apply them in actual coding tests.\n<\/p>\n 7. References<\/h2>\n\nMatrix Chain Multiplication – Wikipedia<\/a><\/li>\n Matrix Chain Multiplication – GeeksforGeeks<\/a><\/li>\n HackerRank – Matrix Multiplication<\/a><\/li>\n<\/ul>\n<\/body><\/p>\n","protected":false},"excerpt":{"rendered":"Problem Description There is a problem of finding the minimum number of operations required to multiply a given set of matrices. If matrix A has a size of pq and matrix B has a size of qr, the number of operations required to multiply these two matrices is p * q * r. 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Output<\/h3>\n
\n Print the minimum number of operations required to multiply the matrices.\n<\/p>\n

Solution Process<\/h2>\n

5. Time Complexity<\/h2>\n
\n The time complexity of this algorithm is O(n^3). This occurs because we need to consider all possible multiplication orders when there are n matrices.\n<\/p>\n

Output<\/h3>\n\n Print the minimum number of operations required to multiply the matrices.\n<\/p>\n

Solution Process<\/h2>\n

5. Time Complexity<\/h2>\n\n The time complexity of this algorithm is O(n^3). This occurs because we need to consider all possible multiplication orders when there are n matrices.\n<\/p>\n

Output<\/h3>\n
\n Print the minimum number of operations required to multiply the matrices.\n<\/p>\n

5. Time Complexity<\/h2>\n
\n The time complexity of this algorithm is O(n^3). This occurs because we need to consider all possible multiplication orders when there are n matrices.\n<\/p>\n