10891 - Game of Sum

[Larry]

My AC code doesn't handle overflow as a special case..

[Towhid]

Matched all the I/O given by jagadish
At last got ac just changing my INF from 2e50 to 1000000000000000.
Don't know what happened

[Towhid]

Matched all the I/O given by jagadish
At last got ac just changing my INF from 2e50 to 1000000000000000.
Don't know what happened
Well, thanks a lot........

[polone]

10891 is a simple DP

You can cut the digits down from left side or right side
and count the diffence between two players by turns

I got ac by this way

the dp should be O(n^3)

[rasel4all]

I have already read that one. But I did not understand. I need some brief descriptio. to solve this one. So please anyone can give me any good hints to solve this one. Thanks.

[Abednego]

Did you notice that if you try to find a winning strategy for a given game state, all of the previous moves that have been made up to that point don't matter?

Knowing this, note that at any point, there are only 2n-1 possible moves. And how many possible game configurations are there? If there aren't too many, can you compute them all?

[rasel4all]

Can u explain by using some sample data. Please

[Abednego]

Suppose your input is

Code: Select all

4 -20 8 -1 -2

Then n=5 is the length of the sequence.
You have 2n-1 = 9 possible moves:

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take 4
take 4 -20
take 4 -20 8
take 4 -20 8 -1
take 4 -20 8 -1 -2
take -20 8 -1 -2
take 8 -1 -2
take -1 -2
take -2

In each of these cases, your opponent will be left with a sub-array (a piece of the original array). Then your opponent will make a move, and you will get a smaller sub-array. Then you make a move again, etc. until nothing is left.

How many different sub-arrays of the original 5-element array are there? If you think about it for 10 seconds, you will see that there are 16 (including the sub-array of length 0). In general, an array of length n has n(n+1)/2+1 sub-arrays. Each of them corresponds to a game situation. So if n is at most 100, then there are fewer than 100000 possible game situations.

Now note that if you start with a sub-array of length X and make a move, then you will get a sub-array of length smaller than X. Suppose that you already know how to solve this probem for sub-arrays of length smaller than X. To solve it for a sub-array of length X, you should just try to make all of the possible moves and pick the one that gives the best value, assuming that your opponent knows how to play optimally after that.

Finally, use induction to get from X=0 up to X=n. Does this make any sense?

[rasel4all]

Thank u very much. I got it now. Thanks again for your help.

[DP]

Code: Select all

accepted

[Martin Macko]

can someone tell me why i am getting TLE?
i did memorization for this problem but still TLE.
Here is my code:

I don't know why you are getting TLE, but anyway you code is wrong. Try the following input:

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6
15 -16 17 18 -19 13
0

The correct output is:

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28

But your answer is 35.

[DP]

thnx a lot. Martin.

[beno]

Hi, it's the first time I submit the solution to the online-judge on Problem "10891 - Game of Sum". But the "Runtime Error" result really confused me since I can get the correct solution of the sample input. Here is my code, hope someone can give me a tip. Thank you.

Code: Select all

#include <iostream>
#include <fstream>
#include <cassert>

using namespace std;

#define MAXLEN 1000

int bet[MAXLEN];

/* return maximum sum of subsequence in sequence arr 
* @param	arr		sequence to search 
* @param	n		size of sequence
*/
int maxSubseqSum(int seq[], int n);

int main(void)
{
	int n, total;
	ifstream fin("e.in");

	assert(fin);

	while (fin >> n) {
		if (n == 0) break;

		for (int i = 0; i < n; i++)
			fin >> bet[i];

		total = maxSubseqSum(bet, n);
		cout << total << endl;
	}

	return 0;
}

int maxSubseqSum(int seq[], int n)
{
	int sum, maxsum;

	assert(n >= 0);

	maxsum = sum = seq[0];
	for (int i = 1; i < n; i++) {
		if (sum < 0)
			sum = 0;
		sum += seq[i];

		if (maxsum < sum)
			maxsum = sum;
	}

	return maxsum;
}

[smithdwsn]

I think you should recheck this part of your code. In that, there is some mistakes.
int maxSubseqSum(int seq[], int n);

int main(void)
{
int n, total;
ifstream fin("e.in");

assert(fin);

while (fin >> n) {
if (n == 0) break;

for (int i = 0; i < n; i++)
fin >> bet;

total = maxSubseqSum(bet, n);
cout << total << endl;
}

return 0;
}

[kbr_iut]

polone wrote,

the dp should be O(n^3)

mine is O(N^2)