To Increment or Not to Increment

-

To Increment or Not to Increment

So, I re-did POJ 2352 (Stars) today, and got a time limit exceeded error.
See the post earlier for a recap.
I spent a hour debugging while loops, no luck. It's the only source of infinite loops, after all.
ARRGGHH

So, here's my code:

#include <cstdio>
using namespace std;
int BIT[33001];
int ans[15001];
void incr(int ind)
{
while(ind<33000)
{ 
BIT[ind]++;
ind += (ind&(-ind));
}
}
int get(int ind)
{
int res=0;
while(ind)
{
res += BIT[ind];
ind -= (ind&(-ind));
}
return res;
}
int main()
{
int n; scanf("%d",&n);
for(int i=0;i<n;i++)
{
int x, y; scanf("%d %d",&x, &y);
ans[get(x)]++;
incr(x);
}
for(int i=0;i<n;i++)
printf("%d\n",ans[i]);
}
Ahem. See that disturbing red line over there?
"No?"
Maybe you're color blind. Okay.
But anyways, this little monster here:
ans[get(x)]++;
Read that line ten times, then read this line:
ans[get(++x)]++;
What's the difference?
"There's no difference!"
Okay, so you are plainly blind then.
But, the x, passed as an argument to get(), has a preincrement operator at its front.
Read the code again, particularly the get() part and the output part.
What difference does it make?
"I don't know!"
Well it solved the TLE problem. But why? It shouldn't make a problem.


Answer

Hmm! So you think you can cheat yourself out by looking at the "Answer" section, eh? Well too bad for you, it's not here today yet. So think about it more.

:P

Comments

Post a Comment

Popular posts from this blog

USACO Training "subset": Subset Sums

-
USACO Training "subset": Subset Sums Problem Statement Subset Sums JRM For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical. For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical: {3} and {1,2} This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions). If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum: {1,6,7} and {2,3,4,5} {2,5,7} and {1,3,4,6} {3,4,7} and {1,2,5,6} {1,2,4,7} and {3,5,6} Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways. Your program must calculate the answer, not look it up from a table. PROGRAM NAME: s
Read more

USACO 2018 Open: Talent Show

-
Image
USACO 2018 Open Talent Show In Short Given N objects with value Vi and weight Wi, a collection of elements is said to have total value-to-weight ratio as (the sum of their value) divided by (the sum of their weight). Find the best possible value-to-weight ratio of these X objects, given the restraint that the collection must have total weight at least W. Return that ratio multiplied by 1000 and floored. Analysis It would be very difficult to calculate this optimal ratio - we'd have to either try all possible collections - O(2^n), where n is at most 250, about 10 to the power of 75 calculations, OR use a greedy heuristic - get the objects with the best ratio until it satisfies the weight limit. Neither of those are too to our liking, so how about searching for it? The search space is linear and not infinite - it asks for the ratio to the closest thousandths. The search space is sorted as well - if ratio X is achievable, then ratio Y < X is also achievable, if
Read more

OpenJudge 1768: Kadane's Algorithm on 2D Arrays

-
OpenJudge 1768: Kadane's Algorithm on 2-D Arrays Simple Version Given a 2-D matrix of integers, find the sub-matrix of this whose sum is maximized, and find that sum. The size of the matrix is no larger than 100 by 100. Analysis O(N^4) - Works, but naive     That's true, an naive O(N^4) version would work. We simply need to calculate (two dimensional!) prefix sums first. A quick review to what 2-D prefix sums are:     The 2-D prefix sum of two coordinates, X and Y, is the sum of all elements that have X-coordinate less than or equal to X, and Y coordinate less than or equal to Y.     It can be calculated through tabulation by removing overlapping areas.     Let P be the prefix sum array and V be the array of values, so the formula is as follows:     P[x, y] = V[x, y] - P[x-1, y] - P[x, y-1] + P[x-1, y-1].     We need to add the P[x-1, y-1] at the end since subtracting P[x-1, y] and P[x, y-1] we had overlapped the P[x-1, y-1]. Then, all we need to
Read more