anyways… its okay. thank you for your reply…
May be I was the unlucky one to have some microscopic level test case missing
i have faced such cases before.
But I think that I should concentrate more on the editorial solution now because this is how we make proper use coding, to work less ourself and let the machine think of all the cases. That’s what programmers do. 
No! There’s a guy with more than 1000 lines of code and that too is not passing first subtask. 
Can anyone explain why did we need to add 0 to mult set?
Because if 0 was a valid choice of multiplication i.e. if any ratio b[i]/a[i] ==0 was present in the test case, it would have counted. So why are we adding 0 exclusively?
@rajarshi_basu?
My code has passed all test cases and edge cases still I’m stuck at 10 points 
This is my code:- https://www.codechef.com/viewsolution/34420341
Can anybody suggest a possible correction or test case which I couldn’t figure out!?
Well It just fails at one case:
when (a,b,c) —> (p,q,r)
p = ka+d
q = kb+d
r = c+d
It fails in the 3 permutations of this case only I believe
Salute your patience man !!!
1 Like
try:
import random
def randoms():
return random.randint(-10,10)
def goopo(a,b,c,p,q,r):
if a!=0:
if p/a==int(p/a):
if q-b*(p/a)==r-c*(p/a):
return True
if b!=0:
if q/b==int(q/b):
if p-a*(q/b)==r-c*(q/b):
return True
if c!=0:
if r/c==int(r/c):
if q-b*(r/c)==p-a*(r/c):
return True
return False
def checkthreeoftwoproduct(a,b,c,p,q,r):
if a!=0:
x=p/a
else:
x=9999999999
if b!=0:
y=q/b
else:
y=9999999999
if c!=0:
z=r/c
else:
z=9999999999
if x==int(x) and y==int(y) and z==int(z):
if x*y == z or z*y==x or x*z==y:
return True
return False
def checkthreeoftwosum(a,b,c,p,q,r):
x=p-a
y=q-b
z=r-c
if x==y+z or z==x+y or y==x+z:
return True
return False
def checklastsum(a,b,c,p,q,r):
x=p-a
y=q-b
z=r-c
if b+x!=0 and c!=0 :
if ((q/(b+x))==r/c) and r/c == int(r/c):
return True
if c+x!=0 and b!=0 :
if ((r/(c+x))==q/b) and q/b == int(q/b):
return True
if b+z!=0 and a!=0 :
if ((q/(b+z))==p/a) and p/a == int(p/a):
return True
if a+z!=0 and b!=0 :
if ((p/(a+z))==q/b) and q/b == int(q/b):
return True
if c+y!=0 and a!=0:
if ((r/(c+y))==p/a) and p/a == int(p/a):
return True
if a+y!=0 and c!=0 :
if ((p/(a+y))==r/c) and r/c == int(r/c):
return True
return False
def checklastproduct(a,b,c,p,q,r):
if a!=0:
if p/a==int(p/a):
x=p/a
if (q-b*x==r-c) or (r-c*x==q-b):
return True
if b!=0:
if q/b==int(q/b):
x=q/b
if (p-a*x==r-c) or (r-c*x==p-a):
return True
if c!=0:
if r/c==int(r/c) and c!=0:
x=r/c
if (q-b*x==p-a) or (p-a*x==q-b):
return True
return False
def operation(p,q,r,a,b,c,t):
x=p-a*t
y=q-b*t
z=r-c*t
if (x==y==z) and t==int(t):
return True
else:
return False
def hoopo(a,b,c,p,q,r):
if (a-b)!=0:
t1=(p-q)/(a-b)
else:
t1=9999999999.9
if (b-c)!=0:
t2=(q-r)/(b-c)
else:
t2=9999999999.9
if (c-a)!=0:
t3=(r-p)/(c-a)
else:
t3=9999999999.9
if operation(p,q,r,a,b,c,t1) and t1==int(t1):
return True
if operation(p,q,r,a,b,c,t2) and t2==int(t2):
return True
if operation(p,q,r,a,b,c,t3) and t3==int(t3):
return True
return False
def operations2(p,q,r,a,b,c,k):
if (a+k==p) and b+k!=0 and c+k!=0:
if (q/(b+k)==r/(c+k)) and r/(c+k)==int(r/(c+k)):
return True
if (b+k==q) and a+k!=0 and c+k!=0:
if (p/(a+k)==r/(c+k)) and r/(c+k)==int(r/(c+k)):
return True
if (c+k==r) and b+k!=0 and a+k!=0:
if (q/(b+k)==p/(a+k)) and p/(a+k)==int(p/(a+k)):
return True
return False
def joopo(a,b,c,p,q,r):
k1=p-a
k2=q-b
k3=r-c
if operations2(p,q,r,a,b,c,k1):
return True
if operations2(p,q,r,a,b,c,k2):
return True
if operations2(p,q,r,a,b,c,k3):
return True
return False
for _ in range(int(input())):
a,b,c=map(int,input().split())
p,q,r=map(int,input().split())
#a=randoms()
#b=randoms()
#c=randoms()
#p=randoms()
#q=randoms()
#r=randoms()
#print(a,b,c)
#print(p,q,r)
equals=0
if p==a:
equals+=1
if q==b:
equals+=1
if r==c:
equals+=1
if (equals==3):
steps=0
elif (equals==2):
steps=1
elif (equals==1):
x=p-a
y=q-b
z=r-c
if a!=0:
xx=p/a
else:
xx=9999999999
if b!=0:
yy=q/b
else:
yy=9999999999
if c!=0:
zz=r/c
else:
zz=9999999999
if x==y!=0 or y==z!=0 or z==x!=0:
steps=1
elif ((xx==yy!=9999999999 and xx==int(xx) and zz==1) or (yy==zz!=9999999999 and yy==int(yy) and xx==1) or (zz==xx!=9999999999 and zz==int(zz) and yy==1)):
steps=1
else:
steps=2
elif (equals==0):
x=p-a
y=q-b
z=r-c
if a!=0:
xx=p/a
else:
xx=9999999999
if b!=0:
yy=q/b
else:
yy=9999999999
if c!=0:
zz=r/c
else:
zz=9999999999
if x==y==z!=0 or (xx==yy==zz!=9999999999 and xx==int(xx)):
steps=1
elif x==y!=0 or y==z!=0 or z==x!=0:
steps=2
elif ((xx==yy!=9999999999 and xx==int(xx) and zz==1) or (yy==zz!=9999999999 and yy==int(yy) and xx==1) or (zz==xx!=9999999999 and zz==int(zz) and yy==1)):
steps=2
elif checklastproduct(a,b,c,p,q,r):
#print(10)
steps=2
elif checklastsum(a,b,c,p,q,r):
#print(11)
steps=2
elif checkthreeoftwosum(a,b,c,p,q,r):
#print(12)
steps=2
elif checkthreeoftwoproduct(a,b,c,p,q,r):
#print(13)
steps=2
elif joopo(a,b,c,p,q,r):
#print(14)
steps=2
elif goopo(a,b,c,p,q,r):
#print(15)
steps=2
elif hoopo(a,b,c,p,q,r):
#print(16)
steps=2
elif a==0 and b==0 and c==0:
steps=2
else:
steps=3
print(steps)
except EOFError:
pass
why i am getting WA???
yes they did for sure let x=p/a y=q/b and z=r/c
if xy==z or yz==x or x*z==y:
they can be done in 2 steps
We only need to consider the multiply-add case. Why?
Let's say:
(p + i) * j = a, Eg: (6 + 3) * 4 = 36
p * j + i * j = a 6 * 4 + 3 * 4 = 36
Replace (i * j) by k Replace (3 * 4) by 12
p * j + k = a 6 * 4 + 12 = 36
So we have replaced add-multiply with multiply-add. And this can be done for any number so we need to only check the multiply-add condition.
4 Likes
I got 90 during the contest… and now after much frustration found an edge case that wasn’t working. Putting it here in case some finds it useful.
1
10 20 30
8 13 18
Output:
3
My code was giving 2. Basically it was considering non-integer multiplication.
10 * 0.5 + 3 = 8
20 * 0.5 + 3 = 13
30 * 0.5 + 3 = 18
2 Likes
My reaction just after looking at the code - “BC kya hai yeh?” .
But HATTTS OFFF dude for so much work. How many hours each day it took you to come up with this?
I don’t think this is an Easy-Medium problem. It is harder than that because of too many cases and no neat logic to cover them all.
The cases for 2 elements are simple, for 3 elements I solved for any pair all the equations a * d1 + d2 = x, (a + d1) * d2 = x, a + d1 + d2 = x, a * d1 * d2 = x, based on a two-bit mask that represents whether operation 1 and 2 was done or not, and see if any satisfied the equation of the remaining element, based on two cases, one in which everyone did the two operations (++ and ** operations not allowed), or in which at least one element was doing exactly one operation.
Somebidy please tell my error the code above …
is there any specific data structure which can be used to solve this problem regardless of making all these cases and all
i saw a post regarding this on geek for geeks where he has used this concept of taking all cases into consideration at a time and move forward using BFS … i would really like to know about a more specific solution to this
please help me with my code, i used BFS
#include<bits/stdc++.h>
using namespace std;
typedef long long int inti;
struct node{
inti a,b,c,p,q,r,step;
};
int main(){
inti t;
cin>>t;
while(t–){
queue que;
node n;
cin>>n.a>>n.b>>n.c;
cin>>n.p>>n.q>>n.r;
n.step=0;
que.push(n);
inti ans=3;
if(n.p==n.q&&n.q==n.r){
if(n.p==0)
ans=1;
else
ans=2;
}
while(!que.empty()){
node temp=que.front();
que.pop();
if(temp.step<ans){
if(temp.a==temp.p && temp.b==temp.q && temp.c==temp.r)
ans=min(ans,temp.step);
else{
if(temp.a!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.p/temp.a;
one.step=temp.step+1;
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.b!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.q/temp.b;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.c!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.r/temp.c;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.p-temp.a!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.p-temp.a;
one.step=temp.step+1;
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
if(temp.q-temp.r!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.q-temp.b;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
if(temp.r-temp.c!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.r-temp.c;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
}
}
}
cout<<ans<<"\n";
}
return 0;
}
please help me with my code, i used BFS
i ran all known cases which were running fine, but answer coming is WA
#include<bits/stdc++.h>
using namespace std;
typedef long long int inti;
struct node{
inti a,b,c,p,q,r,step;
};
int main(){
inti t;
cin>>t;
while(t–){
queue que;
node n;
cin>>n.a>>n.b>>n.c;
cin>>n.p>>n.q>>n.r;
n.step=0;
que.push(n);
inti ans=3;
if(n.p==n.q&&n.q==n.r){
if(n.p==0)
ans=1;
else
ans=2;
}
while(!que.empty()){
node temp=que.front();
que.pop();
if(temp.step<ans){
if(temp.a==temp.p && temp.b==temp.q && temp.c==temp.r)
ans=min(ans,temp.step);
else{
if(temp.a!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.p/temp.a;
one.step=temp.step+1;
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.b!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.q/temp.b;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c;
que.push(one);
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.c!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.r/temp.c;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b;
one.c=temp.c*q;
que.push(one);
one.a=temp.a*q;
one.b=temp.b*q;
one.c=temp.c*q;
que.push(one);
}
if(temp.p-temp.a!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.p-temp.a;
one.step=temp.step+1;
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
if(temp.q-temp.r!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.q-temp.b;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c;
que.push(one);
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
if(temp.r-temp.c!=0){
node one;
one.p=temp.p;
one.q=temp.q;
one.r=temp.r;
int q=temp.r-temp.c;
one.step=temp.step+1;
one.a=temp.a;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b;
one.c=temp.c+q;
que.push(one);
one.a=temp.a+q;
one.b=temp.b+q;
one.c=temp.c+q;
que.push(one);
}
}
}
}
cout<<ans<<"\n";
}
return 0;
}
Still scored 90… 
@rajarshi_basu
@i_64
The editorial mentions you are restricted to add some thing which completes at least 1 pair , i.e. (p-a) || (q-b) || (r-c) but we can have a case
3 5 7
15 21 27
Add 2 to all and multiply all by 3
Can any one proove that what editorial says covers this case also?
1 5 7
7 27 37
should be 2 and *5 then +2