#include<stdio.h>
#include<conio.h>
#include<math.h>
main()
{
int i,j,k,Qi[100],Nj[100],Aa[100][100],a1[10],lv1,lv2,sumQ,x[100][100],counter,istar,seq[100],prasad;
double D[100],d[100],temp,min;
FILE *fptr;
clrscr();
printf("________________________________________________________________________________");
printf("\t\t     TOYOTA MIXED MODEL ASSEMBLY LINE PROBLEM");
printf("\nAshutosh Nandeshwar\t\t  			       Devdatta Yogi");
printf("\n________________________________________________________________________________");
printf("\n Notes:");
printf("\n Maximum Number of Models: 10");
printf("\n Maximum Number of Parts : 100");
printf("\x7\n\n\nEnter Number of models: ");
scanf("%d",&i);

	for(lv1=0;lv1<i;lv1++)
	{
	printf("\nEnter Planned Production Quantity Qi%d ",lv1+1);
	printf("\nEnter Number of models for A%d : ",lv1+1);
	scanf("%d", &Qi[lv1]);
	a1[lv1]=Qi[lv1];
	}

printf("\nEnter Number of Parts: ");
scanf("%d",&j);

	for(lv1=0;lv1<i;lv1++)
	{
		for(lv2=0;lv2<j;lv2++)
		{
			printf("Enter number of parts of a%d for model A%d:",lv2+1,lv1+1);
			scanf("%d",&Aa[lv1][lv2]);
		}
	}
	for(lv2=0;lv2<j;lv2++)
	{
		for(lv1=0,Nj[lv2]=0;lv1<i;lv1++)
		{
			Nj[lv2]=Qi[lv1]*Aa[lv1][lv2]+Nj[lv2];
		}
	}

	for(lv1=0,sumQ=0;lv1<i;lv1++)
	{
		sumQ=sumQ+Qi[lv1];
	}

	for(counter=0,istar=0;counter<=sumQ;counter++)
	{

		if(counter==0)
		{
			for(lv1=0;lv1<j;lv1++)
			{
				x[lv1][counter]=0;
			}

		}

		else
		{
			for(lv1=0;lv1<j;lv1++)
			{
				x[lv1][counter]=x[lv1][counter-1]+Aa[istar][lv1];
			}
		}

		for(lv2=0;lv2<i;lv2++)
		{
			d[lv2]=0;

			for(lv1=0;lv1<j;lv1++)
			{
				temp =(double)((counter+1) * (double)Nj[lv1]/sumQ) - x[lv1][counter] - Aa[lv2][lv1];
				d[lv2] = d[lv2] + pow(temp,2.0);
			}
			D[lv2]=sqrt(d[lv2]);
		}


			if (a1[0]==0)
			{
				D[0]=10000.0;
			}

			if (a1[1]==0)
			{
				D[1]=10000.0;
			}

			if (a1[2]==0)
			{
				D[2]=10000.0;
			}


			if (a1[3]==0)
			{
				D[3]=10000.0;
			}

			if (a1[4]==0)
			{
				D[4]=10000.0;
			}

			if (a1[5]==0)
			{
				D[5]=10000.0;
			}


			if (a1[6]==0)
			{
				D[6]=10000.0;
			}

			if (a1[7]==0)
			{
				D[7]=10000.0;
			}

			if (a1[8]==0)
			{
				D[8]=10000.0;
			}

			if (a1[9]==0)
			{
				D[9]=10000.0;
			}

		for(lv1=1,min=D[0];lv1<i;lv1++)
		{
			if(D[lv1]<=min)
			{
				min=D[lv1];
			}
		}

		for(lv1=0;lv1<i;lv1++)
		{
			if(min==D[lv1])
			{
			istar=lv1;

			if (istar==0)
			{
				a1[0]=a1[0]-1;
			}
			if(istar==1)
			{
				a1[1]=a1[1]-1;
			}

			if (istar==2)

			{
				a1[2]=a1[2]-1;

			}
			if(istar==3)
			{
				a1[3]=a1[3]-1;
			}
			if(istar==4)
			{
				a1[4]=a1[4]-1;
			}

			if(istar==5)
			{
				a1[5]=a1[5]-1;
			}

			if(istar==6)
			{
				a1[6]=a1[6]-1;
			}

			if (istar==7)

			{
				a1[7]=a1[7]-1;

			}
			if(istar==8)
			{
				a1[8]=a1[8]-1;
			}
			if(istar==9)
			{
				a1[9]=a1[9]-1;
			}

			seq[counter]=istar;
			break;
			}
		}

	}
	clrscr();
	fptr=fopen("TOYOTA.TXT","w+");
	printf("\n The Sequence of this Mixed-Model Assembly Line Problem is:\n");
	for(lv2=0,prasad=1;lv2<sumQ;lv2++,prasad++)
	{
	printf("\n");
	printf("Sequence: %d ",lv2+1);
	printf("\t A%d ",seq[lv2]+1);
	fprintf(fptr,"\nSequence: %d \t A%d",lv2+1,seq[lv2]+1);
	if((prasad%20)==0)
	{       printf("\nPress any key to continue........");
		getch();
		clrscr();
	}
	if(prasad==sumQ)
	{
		printf("\x7\n***************************You have reached the Solution!**********************");
		printf("\nCheck TOYOTA.TXT for the current solution");
	}
	}
	fclose(fptr);
	getch();
}