4) Translate the following C code into MIPS (assume i, j, a are stored in $S1,...

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4) Translate the following C code into MIPS (assume i, j, a are stored in $S1, $S2, $S3, and the base addresses of A and B are stored in $S4, $S5 respectively). NOTE: Arrays A and B hold single precision floating point values. int a = 500; for (int i=0; i<a; i++){ } for(j=0; j<a; j++){ } if((C[j] / a) < 20){ } A[i+a*j] = A[i+a*j] * B[i]; 4) Translate the following C code into MIPS (assume i, j, a are stored in $S1, $S2, $S3, and the base addresses of A and B are stored in $S4, $S5 respectively). NOTE: Arrays A and B hold single precision floating point values. int a = 500; for (int i=0; i<a; i++){ } for(j=0; j<a; j++){ } if((C[j] / a) < 20){ } A[i+a*j] = A[i+a*j] * B[i];