first input will be number of test cases t
, then given two numbers a
and b
you have to perform i
operations such that,
a
if i
is odda
if i
is even now print YES if a
can become equal to b
and NO if it can't
when I tried to submit my solution i
got error that time limit is exceeded.
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t, a, b;
cin >> t;
while (t)
{
cin >> a >> b;
int flag = 1;
while (a != b && a < b)
{
if (flag % 2 == 0)
{
a += 2;
}
else
{
a += 1;
}
flag++;
}
if (a == b)
cout << "YES" << endl;
else
cout << "NO" << endl;
t--;
}
return 0;
}
You don't need to actually iterate from a
to b
, just use the following observations to solve:
a
increases by 3
. So after an even number of operations (let number be 2k
), a
increases by 3k
.2k+1
), a
increases by 3k+1
. As you can see, a
can either be increased by 3k
or 3k+1
, which implies that b
will be reachable from a
if (ba) mod 3 = (0 or 1)
(and obviously, if b>a
). You can check this in O(1)
complexity.
The inner loop can be simplified into a math expression. We can come up with the model using some examples.
For example, if a = 10, b = 20.
Loop 1 a += 1 (11)
Loop 2 a += 2 (13)
Loop 3 a += 1 (14)
Loop 4 a += 2 (16)
...
We can see that after two operations, a increases by 3. Which can be generalized, that after the 2n operation (which represents an even number), that a increases by 3n. On the other hand after the 2n + 1 operation (odd numbers), a is equal to an increase of 3n + 1. Thus, a can be increased either by 3n or 3n+1. Thus, 3n or 3n+1 has to be able to be equal to the difference of a and b, which in this case is 10.
Thus either
1. 3 has to divide b - a, (b-a)%3 = 0
2. 3 has to divide b - a with a remainder of 1, (b-a)%3=1
Thus, instead of a loop, using a mathematical expression can simply the runtime to O(1) per input.
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