程序稍作修改，如下：
#include<iostream>using namespace std;
#include<mpi.h>int main(int argc, char * argv[] ){ double start, stop; int *a, *b, *c, *buffer, *ans; int size = 1000; int rank, numprocs, line;
MPI_Init(NULL,NULL);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
line = size/numprocs;
b = new int [ size * size ];
ans = new int [ size * line ];
start = MPI_Wtime(); if( rank ==0 ){
a = new int [ size * size ];
c = new int [ size * size ]; for(int i=0;i<size; i++) for(int j=0;j<size; j++){
a[ i*size + j ] = i*j;
b[ i*size + j ] = i + j;
} for(int i=1;i<numprocs;i++){// send b
MPI_Send( b, size*size, MPI_INT, i, 0, MPI_COMM_WORLD );
} for(int i=1;i<numprocs;i++){// send part of a
MPI_Send( a + (i-1)*line*size, size*line, MPI_INT, i, 1, MPI_COMM_WORLD);
} for(int i = (numprocs-1)*line;i<size;i++){// calculate block 1
for(int j=0;j<size;j++){ int temp = 0; for(int k=0;k<size;k++)
temp += a[i*size+k]*b[k*size+j];
c[i*size+j] = temp;
}
} for(int k=1;k<numprocs;k++){// recieve ans
MPI_Recv( ans, line*size, MPI_INT, k, 3, MPI_COMM_WORLD, MPI_STATUS_IGNORE); for(int i=0;i<line;i++){ for(int j=0;j<size;j++){
c[ ((k-1)*line + i)*size + j] = ans[i*size+j];
}
}
}
FILE *fp = fopen("c.txt","w"); for(int i=0;i<size;i++){ for(int j=0;j<size;j++)
fprintf(fp,"%d\t",c[i*size+j]);
fputc('\n',fp);
}
fclose(fp);
stop = MPI_Wtime();
printf("rank:%d time:%lfs\n",rank,stop-start); delete [] a,c;
} else{
buffer = new int [ size * line ];
MPI_Recv(b, size*size, MPI_INT, 0, 0, MPI_COMM_WORLD, MPI_STATUS_IGNORE);
MPI_Recv(buffer, size*line, MPI_INT, 0, 1, MPI_COMM_WORLD, MPI_STATUS_IGNORE); for(int i=0;i<line;i++) for(int j=0;j<size;j++){ int temp=0; for(int k=0;k<size;k++)
temp += buffer[i*size+k]*b[k*size+j];
ans[i*size+j] = temp;
}
MPI_Send(ans, line*size, MPI_INT, 0, 3, MPI_COMM_WORLD); delete [] buffer; delete [] ans;
} delete [] b;
MPI_Finalize(); return 0;
}
线程 0 发送矩阵 b，以及 a 的分块矩阵给其他线程，然后自己做一部分矩阵乘法，然后接收其他线程的结果，然后输出最终答案。
2. 用 MPI_Scatter 和 MPI_Gather，代码同样参考自 https://blog.csdn.net/lcx543576178/article/details/45892839
稍作修改，如下：
#include<iostream>using namespace std;
#include<mpi.h>int main(){ int my_rank; int num_procs; int size = 1000; double start, finish;
MPI_Init(NULL,NULL);
MPI_Comm_rank(MPI_COMM_WORLD, &my_rank);
MPI_Comm_size(MPI_COMM_WORLD, &num_procs);
int line = size / num_procs;
cout<<" line = "<<line<<endl; int * local_a = new int [ line * size ]; int * b = new int [ size * size ]; int * ans = new int [ line * size ]; int * a = new int [ size * size ]; int * c = new int [ size * size ]; if( my_rank == 0 ){
start = MPI_Wtime(); for(int i=0;i<size;i++){ for(int j=0;j<size;j++){
a[ i*size + j ] = i*j;
b[ i*size + j ] = i + j;
}
}
MPI_Scatter(a, line * size, MPI_INT, local_a, line * size, MPI_INT, 0, MPI_COMM_WORLD );
MPI_Bcast(b, size*size, MPI_INT, 0, MPI_COMM_WORLD); for(int i= 0; i< line;i++){ for(int j=0;j<size;j++){ int temp = 0; for(int k=0;k<size;k++)
temp += a[i*size+k] * b[k*size + j];
ans[i*size + j ] = temp;
}
}
MPI_Gather( ans, line * size, MPI_INT, c, line * size, MPI_INT, 0, MPI_COMM_WORLD ); for(int i= num_procs *line; i< size;i++){ for(int j=0;j<size;j++){ int temp = 0; for(int k=0;k<size;k++)
temp += a[i*size+k] * b[k*size + j];
c[i*size + j ] = temp;
}
}
FILE *fp = fopen("c2.txt","w"); for(int i=0;i<size;i++){ for(int j=0;j<size;j++)
fprintf(fp,"%d\t",c[i*size+j]);
fputc('\n',fp);
}
fclose(fp);
finish = MPI_Wtime();
printf(" time: %lf s \n", finish - start );
} else{ int * buffer = new int [ size * line ];
MPI_Scatter(a, line * size, MPI_INT, buffer, line * size, MPI_INT, 0, MPI_COMM_WORLD );
MPI_Bcast( b, size * size, MPI_INT, 0, MPI_COMM_WORLD ); /*
cout<<" b:"<<endl;
for(int i=0;i<size;i++){
for(int j=0;j<size;j++){
cout<<b[i*size + j]<<",";
}
cout<<endl;
} */
for(int i=0;i<line;i++){ for(int j=0;j<size;j++){ int temp = 0; for(int k=0;k<size;k++)
temp += buffer[i*size+k] * b[k*size + j]; //cout<<"i = "<<i<<"\t j= "<<j<<"\t temp = "<<temp<<endl;
ans[i*size + j] = temp;
}
} /*
cout<<" ans:"<<endl;
for(int i=0;i<line;i++){
for(int j=0;j<size;j++){
cout<<ans[i*size + j]<<",";
}
cout<<endl;
} */
MPI_Gather(ans, line*size, MPI_INT, c, line*size, MPI_INT, 0, MPI_COMM_WORLD ); delete [] buffer;
} delete [] a, local_a, b, ans, c;
MPI_Finalize(); return 0;
}
进程 0 向所有进程（包括自己）分发（MPI_Scatter）a 矩阵的各个分块矩阵，并广播（MPI_Bcast）b 矩阵。其他进程接收 a 的分块矩阵与 b 矩阵，做乘法，并返回结果（MPI_Gather）。进程 0 收集所有结果（包括自己的）（MPI_Gather），如果 a 矩阵分块有剩余，进程 0 就做掉相关乘法，最后输出所有结果。
3. 测时间。下图为上面两种方法的耗时，
4. 总结：①确实进程越多，耗时越少。
②由于消息传递需要成本，而且不是每个进程都同时开始和结束，所以随着进程数的上升，平均每进程的效率下降。
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