116.复矩阵乘法
【摘要】
#include "stdio.h"#define MAX 255 void CMatrixMul(ar,ai,br,bi,m,n,k,cr,ci) /*复矩阵相乘*/int m,n,k; /*m是矩阵A的行数，n是矩阵B的行数，k是矩阵B的列数*//*ar,br,cr分别是矩阵A,B,C的实部，ai,bi,ci分别是矩阵A,B, C的...
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#include "stdio.h"
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#define MAX 255
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void CMatrixMul(ar,ai,br,bi,m,n,k,cr,ci) /*复矩阵相乘*/
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int m,n,k; /*m是矩阵A的行数，n是矩阵B的行数，k是矩阵B的列数*/
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/*ar,br,cr分别是矩阵A,B,C的实部，ai,bi,ci分别是矩阵A,B, C的虚部*/
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double ar[],ai[],br[],bi[],cr[],ci[];
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{
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int i,j,l,u,v,w;
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double p,q,s;
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for (i=0; i<=m-1; i++)
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for (j=0; j<=k-1; j++)
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{
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u=i*k+j;
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cr[u]=0.0; ci[u]=0.0;
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for (l=0; l<=n-1; l++)
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{
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v=i*n+l; w=l*k+j;
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p=ar[v]*br[w];
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q=ai[v]*bi[w];
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s=(ar[v]+ai[v])*(br[w]+bi[w]);
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cr[u]=cr[u]+p-q;
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ci[u]=ci[u]+s-p-q;
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}
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}
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return;
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}
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print_matrix(A,m,n)/*打印的矩阵A(m*n)的元素*/
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int m,n; /*矩阵的阶数*/
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double A[]; /*矩阵A*/
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{
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int i,j;
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for (i=0; i<m; i++)
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{
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for (j=0; j<n; j++)
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printf("%13.7f\t",A[i*n+j]);
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printf("\n");
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}
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}
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main()
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{
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int i,j,n,m,k;
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double Ar[MAX],Br[MAX],Cr[MAX],Ai[MAX],Bi[MAX],Ci[MAX];
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static double cr[3][4],ci[3][4];
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static double ar[3][4]={ {1.0,2.0,3.0,-2.0}, /*矩阵A的实部*/
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{1.0,5.0,1.0,3.0},
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{0.0,4.0,2.0,-1.0}};
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static double ai[3][4]={ {1.0,-1.0,2.0,1.0}, /*矩阵A的虚部*/
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{-1.0,-1.0,2.0,0.0},
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{-3.0,-1.0,2.0,2.0}};
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static double br[4][4]={ {1.0,4.0,5.0,-2.0}, /*矩阵B的实部*/
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{3.0,0.0,2.0,-1.0},
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{6.0,3.0,1.0,2.0},
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{2.0,-3.0,-2.0,1.0}};
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static double bi[4][4]={ {-1.0,-1.0,1.0,1.0}, /*矩阵B的虚部*/
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{2.0,1.0,0.0,5.0},
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{-3.0,2.0,1.0,-1.0},
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{-1.0,-2.0,1.0,-2.0}};
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clrscr();
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puts("**********************************************************");
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puts("* This is a complex-matrix-multiplication program. *");
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puts("* It calculate the two matrixes C(m*k)=A(m*n)B(n*k). *");
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puts("**********************************************************");
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printf(" >> Please input the number of rows in A, m= ");
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scanf("%d",&m);
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printf(" >> Please input the number of cols in A, n= ");
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scanf("%d",&n);
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printf(" >> Please input the number of cols in B, k= ");
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scanf("%d",&k);
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printf(" >> Please input the %d elements in Ar one by one:\n >> ",m*n);
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for(i=0;i<m*n;i++)
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scanf("%lf",&Ar[i]);
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printf(" >> Please input the %d elements in Ai one by one:\n >> ",m*n);
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for(i=0;i<m*n;i++)
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scanf("%lf",&Ai[i]);
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printf(" >> Please input the %d elements in Br one by one:\n >> ",n*k);
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for(i=0;i<n*k;i++)
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scanf("%lf",&Br[i]);
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printf(" >> Please input the %d elements in Bi one by one:\n >> ",n*k);
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for(i=0;i<n*k;i++)
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scanf("%lf",&Bi[i]);
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CMatrixMul(Ar,Ai,Br,Bi,m,n,k,Cr,Ci); /*进行计算*/
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/*输出乘积结果的实部*/
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printf(" Real part of C(%d*%d)=A(%d*%d)*B(%d*%d):\n",m,k,m,n,n,k);
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print_matrix(Cr,m,k);
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/*输出乘积结果的虚部*/
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printf(" Complex part of C(%d*%d)=A(%d*%d)*B(%d*%d):\n",m,k,m,n,n,k);
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print_matrix(Ci,m,k);
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printf(" Press any key to quit...");
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getch();
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}
文章来源: blog.csdn.net，作者：程序员编程指南，版权归原作者所有，如需转载，请联系作者。
原文链接：blog.csdn.net/weixin_41055260/article/details/124655254
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