CodeChef June Long Challenge 2020
Operations on a Tuple
(TTUPLE)
Problem Description
These type of Questions are generally appear in Long challenge due to their various edge cases involve which take more time to find out.
Examples
Example 1:
2 3 5 7 6 5 10 8 6 3 9 7 8
Examples
Example 1:
2 3 5 7 6 5 10 8 6 3 9 7 8
3 5 7 6 5 10
+3
Examples
Example 1:
2 3 5 7 6 5 10 8 6 3 9 7 8
+3
3 5 7 6 5 10
8 6 3
9 7 4
9 7 8
+1
x2
So how to approach this Problem
Observe
So how to approach this Problem
Observe
1. We can only have 4 answers 0, 1, 2, and 3 for any tuple
So how to approach this Problem
Observe
1. We can only have 4 answers 0, 1, 2, and 3 for any tuple
2. Now take a look on these answers how we can get these answers.
So how to approach this Problem
Observe
1. We can only have 4 answers 0, 1, 2, and 3 for any tuple
2. Now take a look on these answers how we can get these answers.
Let's try to understand them one by one.
So how to approach this Problem
Case 1: when answer will be 0
Initial triple = Target Triple
So how to approach this Problem
Case 1: when answer will be 0
Initial triple = Target Triple
p = a
q = b
r = c
(1,2,3)
(1,2,3)
So how to approach this Problem
For all other answers(1,2,3) we will try to understand it by different scenarios.
So how to approach this Problem
Now there can be three possible cases:
CASE 1: where there is only one different number.
a) (1,2,3) (2,2,3)
b) (1,2,3) (1,4,3)
c) (1,2,3) (1,2,4)
p != a
or
q != b
or
r != c
So how to approach this Problem
Now there can be three possible cases:
CASE 1: where there is only one different number.
a) (1,2,3) (2,2,3)
b) (1,2,3) (1,4,3)
c) (1,2,3) (1,2,4)
CASE 2: where there are only two different number.
a) (1,2,3) (2,5,3)
b) (1,2,3) (1,5,4)
p != a
or
q != b
or
r != c
p != a & q != b
or
q != b & r != c
or
p != a & r != c
So how to approach this Problem
Now there can be three possible cases:
CASE 1: where there is only one different number.
a) (1,2,3) (2,2,3)
b) (1,2,3) (1,4,3)
c) (1,2,3) (1,2,4)
CASE 2: where there are only two different number.
a) (1,2,3) (2,5,3)
b) (1,2,3) (1,5,4)
CASE 3: where there are three different number.
a) (1,2,3) (4,5,6)
p != a & q != b
or
q != b & r != c
or
p != a & r != c
p != a
or
q != b
or
r != c
p != a & q != b & r != c
Case Wise solution
CASE 1: where there is only one different number.
a) (1,2,3) (2,2,3)
b) (1,2,3) (1,4,3)
c) (1,2,3) (1,2,4)
No matter what answer will be 1 in this case.
Case Wise solution
CASE 2: where there are only two different number.
a) (1,2,3) (2,5,3)
b) (1,2,3) (1,5,4)
Answers possible for this case.
1
2
Case Wise solution
CASE 2: where there are only two different number.
Answers possible for this case.
1
2
If the two numbers that are different can be converted into corresponding target values in a single move.
if answer is not 1 it will be definitely 2.
Case Wise solution
CASE 2: where there are only two different number.
Answers possible for this case.
1
2
If the two numbers that are different can be converted into corresponding target values in a single move.
if answer is not 1 it will be definitely 2.
For example (2,3,4) -> (5,6,4)
Case Wise solution
CASE 2: where there are only two different number.
Answers possible for this case.
1
2
If the two numbers that are different can be converted into corresponding target values in a single move.
if answer is not 1 it will be definitely 2.
For example (2,3,4) -> (5,6,4)
Add 3 to the two different numbers hence answer would be 1.
+3
(5,6,4)
Case Wise solution
CASE 2: where there are only two different number.
Answers possible for this case.
1
2
If the two numbers that are different can be converted into corresponding target values in a single move.
if answer is not 1 it will be definitely 2.
For example (2,3,4) -> (6,9,4)
Multiply by 3 to the two different numbers hence answer would be 1.
x3
(6,9,4)
typedef long long ll
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
// When there are two different numbers
ll case1()
{
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 1;
return 2;
}
Returns true if we can convert the two different numbers in a single move
for this Example: (1,2,3) -> (3,4,3)
Here N = 1, M = 2
n = 3, m = 4
N
n
M
m
Let's move to our next Case
When all three numbers are different.
p != a
q != b
r != c
Let's move to our next Case
When all three numbers are different.
p != a
q != b
r != c
There are three cases that could be possible:
when answer is 1
when answer is 2
when answer is 3
Let's move to our next Case
When all three numbers are different.
There are three cases that could be possible:
when answer is 1
when answer is 2
when answer is 3
We will discuss all three cases one by one
Case 1: when the answer is 1
We need to convert all three numbers to target numbers
Case 1: when the answer is 1
We need to convert all three numbers to target numbers
We need a single operation either add a number or multiply a number so that three of them converts to the target numbers
Implies
Case 1: when the answer is 1
We need to convert all three numbers to target numbers
We need a single operation either add a number or multiply a number so that three of them converts to the target numbers
Implies
Implies
p -> a
q -> b
r -> c
(p,q,r) + d => a, b, c
or
(p,q,r) X d => a, b, c
Code for this case
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;p != a
q != b
r != c
Case 1: when the answer is 1 (Code)
Code for this case
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;p != a
q != b
r != c
Case 1: when the answer is 1 (Code)
p q r
a b c
Code for this case
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;p != a
q != b
r != c
Case 1: when the answer is 1 (Code)
p q r
a b c
Code for this case
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;p != a
q != b
r != c
Case 1: when the answer is 1 (Code)
p q r
a b c
Code for this case
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;p != a
q != b
r != c
Case 1: when the answer is 1 (Code)
p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
This is rather a long case
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)
(p + x) * d = a \newline
(q + x) * d = b \newline
(r + x) * d = c \newline
OR
(p * d) + x = a \newline
(q * d) + x = b \newline
(r * d) + x = c \newline
Example
4 2 3
5 3 4
10 6 8
+1
x2
Example
4 2 3
8 4 6
10 6 8
x2
+2
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)
(p + x) * d = a \newline
(q + x) * d = b \newline
(r + x) * d = c \newline
OR
(p * d) + x = a \newline
(q * d) + x = b \newline
(r * d) + x = c \newline
p * d - q * d = a - b => d = (a - b) / (p - q)
-
p * d - q * d = a - b => d = (a - b) / (p - q)
-
p * d - q * d = a - b => d = (a - b) / (p - q)
-
p * d - q * d = a - b => d = (a - b) / (p - q)
-
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)
(p + x) * d = a \newline
(q + x) * d = b \newline
(r + x) * d = c \newline
OR
(p * d) + x = a \newline
(q * d) + x = b \newline
(r * d) + x = c \newline
p * d - q * d = a - b => d = (a - b) / (p - q)
p * d - q * d = a - b => d = (a - b) / (p - q)
p * d - q * d = a - b => d = (a - b) / (p - q)
p * d - q * d = a - b => d = (a - b) / (p - q)
q * d - r * d = b - c => d = (b - c) / (q - r)
q * d - r * d = b - c => d = (b - c) / (q - r)
p * d - r * d = a - c => d = (a - c) / (p - r)
p * d - r * d = a - c => d = (a - c) / (p - r)
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)(Code)
if (
p != q && q != r && r != p && (a-b)%(p-q) == 0 && (b-c)%(q-r) == 0 && (a-c)%(p-r) == 0
&&
((a-b)/(p-q) == (b-c)/(q-r)) && ((b-c)/(q-r) == (a-c)/(p-r))
)
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)(Code)
if (
p != q && q != r && r != p && (a-b)%(p-q) == 0 && (b-c)%(q-r) == 0 && (a-c)%(p-r) == 0
&&
((a-b)/(p-q) == (b-c)/(q-r)) && ((b-c)/(q-r) == (a-c)/(p-r))
)
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.2)(Code)
if (
p != q && q != r && r != p && (a-b)%(p-q) == 0 && (b-c)%(q-r) == 0 && (a-c)%(p-r) == 0
&&
((a-b)/(p-q) == (b-c)/(q-r)) && ((b-c)/(q-r) == (a-c)/(p-r))
)
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
First check if any two numbers can be converted to target numbers in one move and third number with extra move.
p q
a b
r
c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
First check if any two numbers can be converted to target numbers in one move and third number with extra move.
p q
a b
r
c
Example
3 5 6
15 25 7
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
First check if any two numbers can be converted to target numbers in one move and third number with extra move.
p q
a b
r
c
Example
3 5 6
15 25 7
x5
+1
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
typedef long long ll;
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
typedef long long ll;
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
typedef long long ll;
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
a - p = d
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
a - p = d
p q r
d
+
a, q+d, r+d
e
+
or
e
x
a, b,c
Example
3 5 6
5 7 8
5 14 16
+2
x2
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
b - q = d
p q r
d
+
a, q+d, r+d
e
+
or
e
x
a, b,c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
c - r = d
p q r
d
+
a, q+d, r+d
e
+
or
e
x
a, b,c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
a - p = d
p q r
d
+
p, s, t
a, b,c
e
+
or
e
x
s = b - d
t = c - d
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)
p q r
a b c
d
a - p = d
p q r
d
+
p, s, t
a, b,c
e
+
or
e
x
s = b - d
t = c - d
Example
3 5 7
3 10 14
5 12 16
x2
+2
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.3)(code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(q+a-p, r+a-p, b,c) || check1(p+c-r, q+c-r, a,b) || check1(p+b-q, r+b-q, a,c))
return 2;
if (check1(q, r, b-(a-p),c-(a-p)) || check1(p, q, a-(c-r),b-(c-r)) || check1(p, r, a-(b-q),c-(b-q)))
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)
p q r
a b c
d
a - p = d
p q r
d
+
a, q+d, r
e
+
or
e
x
a, b,c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)
p q r
a b c
d
a - p = d
p q r
d
+
a, q+d, r
e
+
or
e
x
a, b,c
Example
1 2 3
4 15 9
1 2 3
3
+
4, 5, 3
3
x
4, 15, 9
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)
p q r
a b c
d
a - p = d
p q r
d
+
a, q+d, r
e
+
or
e
x
a, b,c
p
q
r
q
p
r
r
p
q
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)
p q r
a b c
d
a - p = d
p q r
p, s, c
a, b,c
Example
1 2 3
4 9 9
1 2 3
3
x
1, 6, 9
3
+
4, 9, 9
d
+
e
+
or
e
x
s = b - d
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)
p q r
a b c
d
a - p = d
p q r
p, s, c
a, b,c
d
+
e
+
or
e
x
s = b - d
p
q
r
q
p
r
r
p
q
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)(code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(q+a-p, r, b,c) || check1(p+c-r, q, a,b) || check1(p+b-q, r, a,c) || check1(q, r+a-p, b,c) || check1(p, q+c-r, a,b) || check1(p, r+b-q, a,c))
return 2;
if (check1(q, r, b-(a-p),c) || check1(p, q, a-(c-r),b) || check1(p, r, a-(b-q),c) || check1(q, r, b,c-(a-p)) || check1(p, q, a,b-(c-r)) || check1(p, r, a,c-(b-q)))
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.4)(code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (check1(q+a-p, r, b,c) || check1(p+c-r, q, a,b) || check1(p+b-q, r, a,c) || check1(q, r+a-p, b,c) || check1(p, q+c-r, a,b) || check1(p, r+b-q, a,c))
return 2;
if (check1(q, r, b-(a-p),c) || check1(p, q, a-(c-r),b) || check1(p, r, a-(b-q),c) || check1(q, r, b,c-(a-p)) || check1(p, q, a,b-(c-r)) || check1(p, r, a,c-(b-q)))
return 2;Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.5)
p q r
a b c
d
a/p = d
p q r
d
x
a, qxd, rxd
e
+
or
e
x
a, b,c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.5)
p q r
a b c
d
a/p = d
p q r
d
x
a, qxd, rxd
e
+
or
e
x
a, b,c
Example
3 5 7
6 10 14
6 12 16
x2
+2
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && check1(q*(a/p),r*(a/p),b,c))
||
(q != 0 && b%q == 0 && check1(p*(b/q),r*(b/q),a,c))
||
(r != 0 && c%r == 0 && check1(p*(c/r),q*(c/r),a,b))
)
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && check1(q*(a/p),r*(a/p),b,c))
||
(q != 0 && b%q == 0 && check1(p*(b/q),r*(b/q),a,c))
||
(r != 0 && c%r == 0 && check1(p*(c/r),q*(c/r),a,b))
)
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && check1(q*(a/p),r*(a/p),b,c))
||
(q != 0 && b%q == 0 && check1(p*(b/q),r*(b/q),a,c))
||
(r != 0 && c%r == 0 && check1(p*(c/r),q*(c/r),a,b))
)
return 2;p q r
a b c
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
p q r
a b c
d
a/p = d
p q r
d
x
a, qxd, r
e
+
or
e
x
a, b,c
Example
2 3 4
6 9 4
6 11 6
x3
+2
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && (check1(q*(a/p),r,b,c) || check1(q,r*(a/p),b,c)))
||
(q != 0 && b%q == 0 && (check1(p*(b/q),r,a,c) || check1(p,r*(b/q),a,c)))
||
(r != 0 && c%r == 0 && (check1(p*(c/r),q,a,b) || check1(p,q*(c/r),a,b)))
)
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && (check1(q*(a/p),r,b,c) || check1(q,r*(a/p),b,c)))
||
(q != 0 && b%q == 0 && (check1(p*(b/q),r,a,c) || check1(p,r*(b/q),a,c)))
||
(r != 0 && c%r == 0 && (check1(p*(c/r),q,a,b) || check1(p,q*(c/r),a,b)))
)
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if ((p != 0 && a%p == 0 && (check1(q*(a/p),r,b,c) || check1(q,r*(a/p),b,c)))
||
(q != 0 && b%q == 0 && (check1(p*(b/q),r,a,c) || check1(p,r*(b/q),a,c)))
||
(r != 0 && c%r == 0 && (check1(p*(c/r),q,a,b) || check1(p,q*(c/r),a,b)))
)
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (1.6)
p q r
a b c
d
a/p = d
p q r
d
x
p, s, c
e
+
or
e
x
a, b,c
Example
2 3 4
2 5 6
4 10 6
+2
x2
b = s*d
=> s = b/d
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (p != 0 && a%p == 0 && a != 0 && ((b%(a/p) == 0 && check1(q,r,b/(a/p),c)) || (c%(a/p) == 0 && check1(q,r,b,c/(a/p)))))
return 2;
if (q != 0 && b%q == 0 && b != 0 && ((a%(b/q) == 0 && check1(p,r,a/(b/q),c)) || (c%(b/q) == 0 && check1(p,r,a,c/(b/q)))))
return 2;
if (r != 0 && c%r == 0 && c != 0 && ((a%(c/r) == 0 && check1(p,q,a/(c/r),b)) || (b%(c/r) == 0 && check1(p,q,a,b/(c/r)))))
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (p != 0 && a%p == 0 && a != 0 && ((b%(a/p) == 0 && check1(q,r,b/(a/p),c)) || (c%(a/p) == 0 && check1(q,r,b,c/(a/p)))))
return 2;
if (q != 0 && b%q == 0 && b != 0 && ((a%(b/q) == 0 && check1(p,r,a/(b/q),c)) || (c%(b/q) == 0 && check1(p,r,a,c/(b/q)))))
return 2;
if (r != 0 && c%r == 0 && c != 0 && ((a%(c/r) == 0 && check1(p,q,a/(c/r),b)) || (b%(c/r) == 0 && check1(p,q,a,b/(c/r)))))
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 2 (Code)
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
if (p != 0 && a%p == 0 && a != 0 && ((b%(a/p) == 0 && check1(q,r,b/(a/p),c)) || (c%(a/p) == 0 && check1(q,r,b,c/(a/p)))))
return 2;
if (q != 0 && b%q == 0 && b != 0 && ((a%(b/q) == 0 && check1(p,r,a/(b/q),c)) || (c%(b/q) == 0 && check1(p,r,a,c/(b/q)))))
return 2;
if (r != 0 && c%r == 0 && c != 0 && ((a%(c/r) == 0 && check1(p,q,a/(c/r),b)) || (b%(c/r) == 0 && check1(p,q,a,b/(c/r)))))
return 2;
p q r
a b c
p q r
a b c
OR
Code for this case
p != a
q != b
r != c
Case 1: when the answer is 3
Thats'it we have covered all the cases whee answer could be 2 now if our input does not satisfy any of these cases that means we will need minimum 3 operations.
Complete Code
ll p,q,r;
ll a,b,c;
const int maxN = 1e5;
ll arr[maxN];
ll arr1[maxN];
bool check1(ll N, ll M, ll n, ll m)
{
if (N - n == M - m)
return true;
else if (N != 0 && M != 0 && n % N == 0 && m % M == 0 && n/N == m/M)
return true;
return false;
}
ll case2()
{
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 1;
return 2;
}
ll case3()
{
ll g,h;
for (int i = 1; i < min(a,b); ++i)
{
if (i >= 1000)
break;
if (p % i == 0)
{
g = p;
break;
}
else
h = i;
}
if ((p - a == q - b) && (q - b == r - c))
return 1;
if (p != 0 && q != 0 && r != 0 && a%p == 0 && b%q == 0 && c%r == 0 && a/p == b/q && b/q == c/r)
return 1;
if (p != q && q != r && r != p && (a-b)%(p-q) == 0 && (b-c)%(q-r) == 0 && (a-c)%(p-r) == 0 && ((a-b)/(p-q) == (b-c)/(q-r)) && ((b-c)/(q-r) == (a-c)/(p-r)))
return 2;
if (check1(p,q,a,b) || check1(p,r,a,c) || check1(q,r, b,c))
return 2;
if (check1(q+a-p, r+a-p, b,c) || check1(p+c-r, q+c-r, a,b) || check1(p+b-q, r+b-q, a,c))
return 2;
if (check1(q, r, b-(a-p),c-(a-p)) || check1(p, q, a-(c-r),b-(c-r)) || check1(p, r, a-(b-q),c-(b-q)))
return 2;
if ((p != 0 && a%p == 0 && check1(q*(a/p),r*(a/p),b,c)) || (q != 0 && b%q == 0 && check1(p*(b/q),r*(b/q),a,c)) || (r != 0 && c%r == 0 && check1(p*(c/r),q*(c/r),a,b)))
return 2;
if (check1(q+a-p, r, b,c) || check1(p+c-r, q, a,b) || check1(p+b-q, r, a,c) || check1(q, r+a-p, b,c) || check1(p, q+c-r, a,b) || check1(p, r+b-q, a,c))
return 2;
if (check1(q, r, b-(a-p),c) || check1(p, q, a-(c-r),b) || check1(p, r, a-(b-q),c) || check1(q, r, b,c-(a-p)) || check1(p, q, a,b-(c-r)) || check1(p, r, a,c-(b-q)))
return 2;
if ((p != 0 && a%p == 0 && (check1(q*(a/p),r,b,c) || check1(q,r*(a/p),b,c))) || (q != 0 && b%q == 0 && (check1(p*(b/q),r,a,c) || check1(p,r*(b/q),a,c))) || (r != 0 && c%r == 0 && (check1(p*(c/r),q,a,b) || check1(p,q*(c/r),a,b))))
return 2;
if (p != 0 && a%p == 0 && a != 0 && ((b%(a/p) == 0 && check1(q,r,b/(a/p),c)) || (c%(a/p) == 0 && check1(q,r,b,c/(a/p)))))
return 2;
if (q != 0 && b%q == 0 && b != 0 && ((a%(b/q) == 0 && check1(p,r,a/(b/q),c)) || (c%(b/q) == 0 && check1(p,r,a,c/(b/q)))))
return 2;
if (r != 0 && c%r == 0 && c != 0 && ((a%(c/r) == 0 && check1(p,q,a/(c/r),b)) || (b%(c/r) == 0 && check1(p,q,a,b/(c/r)))))
return 2;
return 3;
}
ll solve()
{
cin >> p >> q >> r;
cin >> a >> b >> c;
std::vector<int> diff;
if (p != a)
diff.push_back(0);
if (q != b)
diff.push_back(1);
if (r != c)
diff.push_back(2);
if (diff.size() == 0)
return 0;
if (diff.size() == 1)
return 1;
if (diff.size() == 2)
return case2();
return case3();
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ll tc;
cin >> tc;
for (ll i = 0;i < tc;i++)
{
cout << solve() << "\n";
}
return 0;
}A Thought About the Question
I know this question is all about finding out the different cases and then generalization of these cases.
It definitely takes time to finding out all the cases.
That's why it appears in Long Challenge.
This question teaches not to give up and try to figure out the different edge cases because that's what software Engineering Demands from you, that's what good companies look in to you.
Any Doubts
If you still have any doubts you can comment your doubts in the comment section.
If you have a better approach than this you can also comment it out.
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QWERT
By gauravsahu
