Find the logic behind the output and code for it. Input: Number of elements in set 1: 4 Elements are: 9, 9, 9, 9 Number of elements in set 2: 3 Elements are: 1,1,1 Output: 1, 0, 1, 1, 0 Input: Number of elements in set 1: 11 Elements are: 7,2,3,4,5,3,1,2,7,2,8 Number of elements in set 2: 3 Elements are: 1,2,3 Output: 7,2,3,4,5,3,1,2,8,5,1

#include<stdio.h>

#include<conio.h>

void sump(int n1,int n2,int s1[],int s2[]){

int temp,s[100],c=0,sum=0,i,j,k;

for(i=n1-1,j=n2-1;j>=0;i--,j--){

sum=s1[i]+s2[j]+c;

if(sum>9){

sum=sum-10;

c=1;

}

else{

c=0;

}

s[i]=sum;

//printf("%d ",sum);

}

if(i>=0){

for(k=i;k>=0;k--){

sum=s1[k]+c;

if(sum>9){

sum=sum-10;

c=1;

}

else{

c=0;

}

s[k]=sum;

if(k==0 && c==1){

for(i=n1;i>0;i--){

temp=s[i-1];

s[i]=temp;

}

n1=n1+1;

s[i]=c;

//printf("c=%d ",c);

}

}

}

printf("%d",s[0]);

for(i=1;i<n1;i++)

printf(",%d",s[i]);

//printf("%d",s[n1]);

}

void main(){

int s11[100],s22[100],n11,n22,i1,j1;

clrscr();

scanf("%d",&n11);

for(i1=0;i1<n11;i1++)

scanf("%d,",&s11[i1]);

scanf("%d",&n22);

for(j1=0;j1<n22;j1++)

scanf("%d,",&s22[j1]);

if(n11>n22 || n11==n22)

sump(n11,n22,s11,s22);

else

sump(n22,n11,s22,s11);

getch();

}