#define BURG 206
#define CAN 145
#define combo 351
#include stdio.h
void main(){
int burgp,canp,combop,n1,n2,n3;
combop(n2){
printf("enter no of combo")
scanf("%d",n2);
for(combop=3;combop>=2000;combop++)
{printf("%d",combop*n2);}
}
sodap(n3,combop){
printf("enter no of combo")
scanf("%d",n3);
for(sodap=3;sodapp>=combop;sodap++)
{printf("%d",sodapp*n2);}
}

burgp(n1,sodap){
printf("enter no of combo")
scanf("%d",n1);
for(burgp=2;burgp>=sodap;burgp++)
{printf("%d",burgp*n1);}
}

profit(burgp,sodap,combop,n1,n2,n3){
int profit=(burgp*n1)+(sodap*n2)+(combop*n2);
}


//scenario(t) says 100 cases which means i didn't get it here
//no of combos should range between 1 to 2000 so max no of combos
//no of burgers order should be min of 2 and combos should be greater than sodas
//above line means sodas min value is 3 and max greater than 2000
//profit can be calculated by defining fxns for each items and profit fxn has addition of all.


}