Experiment 12:problem G: A powerful matrix operation comes

1. This topic is mainly the overloading of multiplication operators, card me for a long time, matrix multiplication with 3 nested for loop, to divide the matrix of the multiplication result is the first matrix of the row, the second matrix of the column composed of the Matrix.
2. When you overload the +,* operator, you can pass two matrix references in the argument list, representing the matrix before and after the operation, or the matrix reference after the only operator , using the implied matrix that the this pointer points to. I'm using the latter.

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Problem G: A powerful matrix operation comes Problem G: Powerful matrix operations come time Limit: 1 Sec Memory Limit: MB
Submit: 171 Solved: 98
[Submit] [Status] [Web Board] Description

Defines a matrix class that is used to store an array. Overloading its +, * operators, respectively, for calculating the sum and product of two matrices, and overloading its << and >> operators for outputting and entering a matrix. Requires that when two matrices cannot be additive or multiplicative, error should be output.

Input

Enter line 1th n>0, which indicates that there are n sets of test cases with a total of 2N matrices.

Each set of test cases consists of 2 matrices. Each matrix first enters the number of rows, columns, and then all the elements of the matrix.

Output

Each test case produces a set of outputs. The specific format is shown in the sample. Note: Error should be output when addition or multiplication is not possible.

Sample Input2 of 111 in the same.Sample OutputCase 1:3 4Case 2:error2 2Case 3:errorerrorHINT Append codeappend.cc, [Submit] [Status] [Web Board]

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#include <iostream>#defineMAX 102using namespacestd;classmatrix{ Public:    intR,c,error; intM[max][max]; Matrix (): Error (0) {} Matrixoperator+(ConstMatrix &M) {Matrix tmp; TMP.R=M.R; TMP.C=M.C; /*for (int i=0;i<tmp.r;i++) for (int j=0;j<tmp.c;j++) tmp.m[i][j]=0;*/        if(r!=m.r| | c!=m.c) {tmp.error=1; returntmp; }         for(intI=0; i<r; i++)             for(intj=0; j<c; J + +) {Tmp.m[i][j]=m[i][j]+M.m[i][j]; }        returntmp; } Matrixoperator*(ConstMatrix &M) {Matrix tmp; TMP.R=R; TMP.C=M.C;  for(intI=0; i<tmp.r;i++)             for(intj=0; j<tmp.c;j++) Tmp.m[i][j]=0; if(c!=M.R) {tmp.error=1; returntmp; }         for(intI=0; i<r; i++)        {             for(intj=0; j<m.c; J + +)            {                intsum =0;  for(intk=0; k<m.r; k++) {sum+ = m[i][k] *M.m[k][j]; } Tmp.m[i][j]=sum; }        }        returntmp; } Friend IStream&operator>> (IStream & is, Matrix &l); Friend Ostream&operator<< (ostream &os,matrix &M);}; IStream&operator>> (IStream & is, Matrix &M) {     is>>M.r>>M.C;  for(intI=0; i<m.r; i++)         for(intj=0; j<m.c; J + +)        {             is>>M.m[i][j]; }    return  is;} Ostream&operator<< (ostream &os,matrix &M) {    if(m.error==1) {OS<<"Error"<<Endl; returnOS; }     for(intI=0; i<m.r; i++)         for(intj=0; j<m.c; J + +)        {            if(j!=m.c-1) OS<<M.m[i][j]<<" "; ElseOS<<M.m[i][j]<<Endl; }    ///OS<<endl;return OS;}intMain () {intcases, I; CIN>>cases;  for(i =0; i < cases; i++) {Matrix A, B, C, D; CIN>>A>>C; C= A +B; D= A *B; cout<<" Case"<<i +1<<":"<<Endl; cout<<C<<Endl; cout<<D; }    return 0;}

Experiment 12:problem G: A powerful matrix operation comes