Source code of Bigmultiplication problem in c

#include #include int main() { int n,q,f=0,s,m,d,a,i=-1,j=-1,a1[101],a2[101],sum[201],mul[101]; char ch; clrscr(); printf("\nEnter frist number : "); while((ch=getchar())!='\n') { switch(ch) { case '0':a=0;break; case '1':a=1;break; case '2':a=2;break; case '3':a=3;break; case '4':a=4;break; case '5':a=5;break; case '6':a=6;break; case '7':a=7;break; case '8':a=8;break; case '9':a=9;break; } i++; a1[i]=a; } printf("Enter second number : "); while((ch=getchar())!='\n') { switch(ch) { case '0':a=0;break; case '1':a=1;break; case '2':a=2;break; case '3':a=3;break; case '4':a=4;break; case '5':a=5;break; case '6':a=6;break; case '7':a=7;break; case '8':a=8;break; case '9':a=9;break; } j++; a2[j]=a; } for(a=0;a<=200;a++) sum[a]=0; printf("\nMultiplication = "); for(q=j;q>=0;q--) { m=101;d=0; for(a=i;a>=0;a--) { m--; n=((a1[a]*a2[q])+d)%10; d=((a1[a]*a2[q])+d)/10; mul[m]=n; } if(d>0) { m--; mul[m]=d; } d=0;n=200-f; for(a=100;a>=m;a--) { s=(sum[n]+mul[a]+d)%10; d=(sum[n]+mul[a]+d)/10; sum[n]=s; n--; } f++; if(d>0) sum[n]=d; } n=199-i-j; if(sum[n]==0) n=200-i-j; for(a=n;a<=200;a++) printf("%d",sum[a]); getch(); return 0; }