I tried Solving ICPC India Online prelims Question D but I am not to find the correct logic
https://www.codechef.com/ICPCOL2025/problems/ICPC25P4TMP
I Would like to hear what was the logic from the people who solved this
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I tried Solving ICPC India Online prelims Question D but I am not to find the correct logic
https://www.codechef.com/ICPCOL2025/problems/ICPC25P4TMP
I Would like to hear what was the logic from the people who solved this
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Auto comment: topic has been updated by Erikasa (previous revision, new revision, compare).
Auto comment: topic has been updated by Erikasa (previous revision, new revision, compare).
Auto comment: topic has been updated by Erikasa (previous revision, new revision, compare).
cry burger
ok! Next order?
Hint: try to solve it in exactly 2 operations.
Lets call $$$k$$$ the number of indices $$$i$$$ that satisfy $$$i \leq \frac{n}{2}$$$ and $$$p[i] \leq \frac{n}{2}$$$. Then the number of indices $$$i\leq \frac{n}{2}, p[i] \gt \frac{n}{2}$$$ is $$$\frac{n}{2}-k$$$, the number of indices $$$i \gt \frac{n}{2}, p[i] \leq \frac{n}{2}$$$ is $$$\frac{n}{2}-k$$$, and the number of indices $$$i \gt \frac{n}{2}, p[i] \gt \frac{n}{2}$$$ is $$$k$$$. This gives us a way to do it in exactly 2 operations.
Since we want the minimum number of operations, we only need to figure out if it is possible in 1 operation. Hopefully you are able to solve this case :)
Think about it as a 2d plot, where you draw a cross in the middle of the plot. Then you choose (left top + right bottom) and (left bottom + right top)
Thanks i think i got how to do it
ans <= 2 is always possible