I was trying to solve a question and I got into a few obstacles that I failed to solve, starting off here is the question: Codeforces - 817D
Now I tried to brute force it, using a basic get min and max for each segment of the array I could generate and then keeping track of them I subtract them and add them together to get the final imbalance, this looked good but it gave me a time limit exceeded cause brute forcing n*(n+1)/2 subsegments of the array given n is 10^6 , so I just failed to go around it and after like a couple of hours of not getting any new ideas I decided to see a solution that I could not understand anything in to be honest :/ , here is the solution:
#include <bits/stdc++.h>
#define ll long long
int a[1000000], l[1000000], r[1000000];
int main(void) {
int i, j, n;
scanf("%d",&n);
for(i = 0; i < n; i++) scanf("%d",&a[i]);
ll ans = 0;
for(j = 0; j < 2; j++) {
vector<pair<int,int>> v;
v.push_back({-1,INF});
for(i = 0; i < n; i++) {
while (v.back().second <= a[i]) v.pop_back();
l[i] = v.back().first;
v.push_back({i,a[i]});
}
v.clear();
v.push_back({n,INF});
for(i = n-1; i >= 0; i--) {
while (v.back().second < a[i]) v.pop_back();
r[i] = v.back().first;
v.push_back({i,a[i]});
}
for(i = 0; i < n; i++) ans += (ll) a[i] * (i-l[i]) * (r[i]-i);
for(i = 0; i < n; i++) a[i] *= -1;
}
cout << ans;
}
I tried tracing it but I keep wondering why was the vector used , the only idea I got is he wanted to use the vector as a stack since they both act the same(Almost) but then the fact that I don't even know why we needed a stack here and this equation ans += (ll) a[i] * (i-l[i]) * (r[i]-i); is really confusing me because I don't get where did it come from.
Well thats a beast of a calculation. I must confess, that i don't understand it completely either. The problem with the brute force solution is, that you have to calculate values or all over again.
In a slightly modified example, you calculate the following values for an input of 2, 4, 1 (i reordered it by "distance")
[2, *, *] (from index 0 to index 0), imbalance value is 0; i_min = 0, i_max = 0
[*, 4, *] (from index 1 to index 1), imbalance value is 0; i_min = 1, i_max = 1
[*, *, 1] (from index 2 to index 2), imbalance value is 0; i_min = 2, i_max = 2
[2, 4, *] (from index 0 to index 1), imbalance value is 2; i_min = 0, i_max = 1
[*, 4, 1] (from index 1 to index 2), imbalance value is 3; i_min = 2, i_max = 1
[2, 4, 1] (from index 0 to index 2), imbalance value is 3; i_min = 2, i_max = 1
where i_min and i_max are the indices of the element with the minimum and maximum value.
For a better visual understanding, i wrote the complete array, but hid the unused values with *
So in the last case [2, 4, 1], brute-force looks for the minimum value over all values, which is not necessary, because you already calculated the values for a sub-space of the problem, by calculating [2,4] and [4,1]. But comparing only the values is not enough, you also need to keep track of the indices of the minimum and maximum element, because those can be reused in the next step, when calculating [2, 4, 1].
The idead behind this is a concept called dynamic programming, where results from a calculation are stored to be used again. As often, you have to choose between speed and memory consumption.
So to come back to your question, here is what i understood :
the arrays l and r are used to store the indices of the greatest number left or right of the current one
vector v is used to find the last number (and it's index) that is greater than the current one (a[i]). It keeps track of rising number series, e.g. for the input 5,3,4 at first the 5 is stored, then the 3 and when the 4 comes, the 3 is popped but the index of 5 is needed (to be stored in l[2])
then there is this fancy calculation (ans += (ll) a[i] * (i-l[i]) * (r[i]-i)). The stored indices of the maximum (and in the second run the minimum) elements are calculated together with the value a[i] which does not make much sense for me by now, but seems to work (sorry).
at last, all values in the array a are multiplied by -1, which means, the old maximums are now the minimums, and the calculation is done again (2nd run of the outer for-loop over j)
This last step (multiply a by -1) and the outer for-loop over j are not necessary but it's an elegant way to reuse the code.
Hope this helps a bit.
Pictures of the code and errors:
https://puu.sh/sPat1/da5b011ac6.png
https://puu.sh/sPavZ/ad81dc6386.png
Questions : This is a program that generates 600 numbers from 1 to 49 , 3 times. 1800 numbers total. How can I check through all the numbers I mean how many times 1 is included, 2 ,3 ,4 ,5 ,6 . . . 48, 49.. <- IN ONLY ONE OF THE ARRAYS a[100][6] for example. Please help.
You can create another array/vector with indexes from [0..49] and increment the desired index every time you meet a value in an original array:
int arr[100][6];
//fill it
int check[50];
for(int i = 0; i < 50; i++)
check[i] = 0;
for(int i = 0; i < 100; i++)
for(int j = 0; j < 6; j++)
check[arr[i][j]]++;
//now check has a list of occurences for 0, 1, 2, ..., 49
As for code style, I would suggest to go away from magic constants like 100, 50, and use something like
const int max_random = 50;
const int mat_width = 100;
const int mat_height = 6;
Also, for C++ the usage of std::vector is encouraged instead of usual arrays - STL is created just for the purpose of leaving most of the pointers/low-level operations under the hood.
I want to generate a random number from a specific range in each iteration
for (int i = 0; i < 10; i++)
{
v1[i] = rand() % n;
}
This code will generate a number between 0 and 9. However, I do not want the selected number to be the same as the index i. For example, if I am in the first iteration (ie: i == 1), I want the random number to be either 0, 2, 3, 4, 5, 6, 7, 8, 9 and not 1.
Can someone help me in this?
If you want to exclude one number, just remove the greatest number from the set of possible choices, and if you happen to pick the one you didn't want, choose the greatest number instead.
For example, if you want to pick a number between 0 and 9, but it should not be the number 1, then pick a number between 0 and 8. If you pick 1, choose 9 instead.
Try something like:
for (int i = 0; i < 10; i++)
{
do
{
v1[i] = rand() % n;
} while (v1[i] == i);
}
Given the numbers you are using the do-loop will only cycle more than once about every ten calls.
I have an assignment where I'm given a string S containing the letters 'R' and 'K', for example "RRRRKKKKRK".
I need to obtain the maximum number of 'R's that string could possibly hold by flipping characters i through j to their opposite. So:
for(int x = i; x < j; x++)
{
if S[x] = 'R'
{
S[X] = 'S';
}
else
{
S[X] = 'R';
}
}
However, I can only make the above call once.
So for the above example: "RRRRKKKKRK".
You would have i = 4 and j = 8 which would result in: "RRRRRRRRKR" and you would then output the number of R's in the resulting string: 9.
My code partially works, but there are some cases that it doesn't. Can anyone figure out what is missing?
Sample Input
2
RKKRK
RKKR
Sample Output
4
4
My Solution
My solution which works only for the first case, I don't know what I'm missing to complete the algorithm:
int max_R = INT_MIN;
for (int i = 0; i < s.size(); i++)
{
for (int j = i + 1; j < s.size(); j++)
{
int cnt = 0;
string t = s;
if (t[j] == 'R')
{
t[j] = 'K';
}
else
{
t[j] = 'R';
}
for (int b = 0; b < s.size(); b++)
{
if (t[b] == 'R')
{
cnt++;
if (cnt > max_R)
{
max_R = cnt;
}
}
}
}
}
cout << max_R << endl;
How about turning this into the Maximum subarray problem which has O(n) solution?
Run through the string once, giving 'K' a value of 1, and 'R' a value of -1.
E.g For 'RKRRKKKKRKK' you produce an array -> [-1, 1, -1, -1, 1, 1, 1, 1, -1, 1, 1] -> [-1, 1, -2, 4, -1, 2] (I grouped consecutive -1s and 1s to be more clear)
Apply Kadane's algorithm on the generated array. What you get from doing this is the maximum number of 'R's you can obtain from flipping 'K's.
Continuing with the example, you find that the maximum subarray is [4, -1, 2] with a sum of 5.
Now add the absolute value of the negative values outside this subarray with the sum of your maximum subarray to obtain your answer.
In our case, only -1 and -2 are negative and outside the subarray. We get |-1| + |-2| + 5 = 8
Try to carefully think about your solution. Do you understand, what it does?
First, let’s forget that the input file may contain multiple tests, so let’s get rid of the while loop. Now, we have just two for loops. The second one obviously just counts R’s in the processed string. But what does the first one do?
The answer is that the first loop flips all the letters from the second one (i.e. which has index 1) till the end of the string. We can see that in the first testcase:
RKKRK
it is indeed the optimal solution. The string turns into RRRKR and we get four R’s. But in the second case:
RKKR
the string turns into RRRK and we get three R’s. While if we flipped just the letters from 2 to 3 (i.e. indices 1 to 2) we could get RRRR which has four R’s.
So your algorithm always flips letters from index 1 to the end, but this is not always optimal. What can we do? How do we know which letters to flip? Well, there are some smart solutions, but the easiest is to just try all possible combinations!
You can flip all the letters from 0 to 1, count the number of R’s, remember it. Get back to the original string, flip letters from 0 to 2, count R’s, remember it and so on till you flip from 0 to n-1. Then you flip letters from 1 to 2, from 1 to 3, etc. And the answer is the largest value you remembered.
This is horribly inefficient, but this works. After you get more practice in solving algorithmic problems, get back to this task and try to figure out more efficient solutions. (Hint: if you consider building the optimal answer incrementally, that is by going through the string char by char and transforming the optimal solution for the substring s[0..i] into the optimal solution for s[0..i+1] you can arrive to a pretty straightforward O(n^2) algorithm. This can be enhanced to O(n), but this step is slightly more involved.)
Here is the sketch of this solution:
def solve(s):
answer = 0
for i in 0..(n-1)
for j in i..(n-1)
t = copy(s) # we will need the original string later
flip(t, i, j) # flip letters from i to j in t
c = count_R(t) # count R's in t
answer = max(answer, c)
return answer
Ok, I really don't know how to frame the question properly because I barely have any idea how to describe what I want in one sentence and I apologize.
Let me get straight to the point and you can just skip the rest cause I just want to show that I've tried something and not coming here to ask a question on a whim.
I need an algorithm that produces 6 random numbers where it may not produce more than 2 consecutive numbers in that sequence.
example: 3 3 4 4 2 1
^FINE.
example: 3 3 3 4 4 2
^NO! NO! WRONG!
Obviously, I have no idea how to do this without tripping over myself constantly.
Is there a STL or Boost feature that can do this? Or maybe someone here knows how to concoct an algorithm for it. That would be awesome.
What I'm trying to do and what I've tried.(the part you can skip)
This is in C++. I'm trying to make a Panel de Pon/Tetris Attack/Puzzle League whatever clone for practice. The game has a 6 block row and 3 or more matching blocks will destroy the blocks. Here's a video in case you're not familiar.
When a new row comes from the bottom it must not come out with 3 horizontal matching blocks or else it will automatically disappear. Something I do not want for horizontal. Vertical is fine though.
I've tried to accomplish just that and it appears I can't get it right. When I start the game chunks of blocks are missing because it detects a match when it shouldn't. My method is more than likely heavy handed and too convoluted as you'll see.
enum BlockType {EMPTY, STAR, UP_TRIANGLE, DOWN_TRIANGLE, CIRCLE, HEART, DIAMOND};
vector<Block> BlockField::ConstructRow()
{
vector<Block> row;
int type = (rand() % 6)+1;
for (int i=0;i<6;i++)
{
row.push_back(Block(type));
type = (rand() % 6) +1;
}
// must be in order from last to first of the enumeration
RowCheck(row, diamond_match);
RowCheck(row, heart_match);
RowCheck(row, circle_match);
RowCheck(row, downtriangle_match);
RowCheck(row, uptriangle_match);
RowCheck(row, star_match);
return row;
}
void BlockField::RowCheck(vector<Block> &row, Block blockCheckArray[3])
{
vector<Block>::iterator block1 = row.begin();
vector<Block>::iterator block2 = row.begin()+1;
vector<Block>::iterator block3 = row.begin()+2;
vector<Block>::iterator block4 = row.begin()+3;
vector<Block>::iterator block5 = row.begin()+4;
vector<Block>::iterator block6 = row.begin()+5;
int bt1 = (*block1).BlockType();
int bt2 = (*block2).BlockType();
int bt3 = (*block3).BlockType();
int bt4 = (*block4).BlockType();
int type = 0;
if (equal(block1, block4, blockCheckArray))
{
type = bt1 - 1;
if (type <= 0) type = 6;
(*block1).AssignBlockType(type);
}
else if (equal(block2, block5, blockCheckArray))
{
type = bt2 - 1;
if (type <= 0) type = 6;
(*block2).AssignBlockType(type);
}
else if (equal(block3, block6, blockCheckArray))
{
type = bt3 - 1;
if (type == bt3) type--;
if (type <= 0) type = 6;
(*block3).AssignBlockType(type);
}
else if (equal(block4, row.end(), blockCheckArray))
{
type = bt4 - 1;
if (type == bt3) type--;
if (type <= 0) type = 6;
(*block4).AssignBlockType(type);
}
}
Sigh, I'm not sure if it helps to show this...At least it shows that I've tried something.
Basically, I construct the row by assigning random block types, described by the BlockType enum, to a Block object's constructor(a Block object has blockType and a position).
Then I use a RowCheck function to see if there's 3 consecutive blockTypes in one row and I have do this for all block types. The *_match variables are arrays of 3 Block objects with the same block type. If I do find that there are 3 consecutive block types then, I just simply subtract the first value by one. However if I do that I might end up inadvertently producing another 3 match so I just make sure the block types are going in order from greatest to least.
Ok, it's crappy, it's convoluted and it doesn't work! That's why I need your help.
It should suffice to keep record of the previous two values, and loop when the newly generated one matches both of the previous values.
For an arbitrary run length, it would make sense to size a history buffer on the fly and do the comparisons in a loop as well. But this should be close to matching your requirements.
int type, type_old, type_older;
type_older = (rand() % 6)+1;
row.push_back(Block(type_older));
type_old = (rand() % 6)+1;
row.push_back(Block(type_old));
for (int i=2; i<6; i++)
{
type = (rand() % 6) +1;
while ((type == type_old) && (type == type_older)) {
type = (rand() % 6) +1;
}
row.push_back(Block(type));
type_older = type_old;
type_old = type;
}
Idea no 1.
while(sequence doesn't satisfy you)
generate a new sequence
Idea no 2.
Precalculate all allowable sequences (there are about ~250K of them)
randomly choose an index and take that element.
The second idea requires much memory, but is fast. The first one isn't slow either because there is a veeery little probability that your while loop will iterate more than once or twice. HTH
Most solutions seen so far involve a potentially infinite loop. May I suggest a different approch?
// generates a random number between 1 and 6
// but never the same number three times in a row
int dice()
{
static int a = -2;
static int b = -1;
int c;
if (a != b)
{
// last two were different, pick any of the 6 numbers
c = rand() % 6 + 1;
}
else
{
// last two were equal, so we need to choose from 5 numbers only
c = rand() % 5;
// prevent the same number from being generated again
if (c == b) c = 6;
}
a = b;
b = c;
return c;
}
The interesting part is the else block. If the last two numbers were equal, there is only 5 different numbers to choose from, so I use rand() % 5 instead of rand() % 6. This call could still produce the same number, and it also cannot produce the 6, so I simply map that number to 6.
Solution with simple do-while loop (good enough for most cases):
vector<Block> row;
int type = (rand() % 6) + 1, new_type;
int repetition = 0;
for (int i = 0; i < 6; i++)
{
row.push_back(Block(type));
do {
new_type = (rand() % 6) + 1;
} while (repetition == MAX_REPETITION && new_type == type);
repetition = new_type == type ? repetition + 1 : 0;
type = new_type;
}
Solution without loop (for those who dislike non-deterministic nature of previous solution):
vector<Block> row;
int type = (rand() % 6) + 1, new_type;
int repetition = 0;
for (int i = 0; i < 6; i++)
{
row.push_back(Block(type));
if (repetition != MAX_REPETITION)
new_type = (rand() % 6) + 1;
else
{
new_type = (rand() % 5) + 1;
if (new_type >= type)
new_type++;
}
repetition = new_type == type ? repetition + 1 : 0;
type = new_type;
}
In both solutions MAX_REPETITION is equal to 1 for your case.
How about initializing a six element array to [1, 2, 3, 4, 5, 6] and randomly interchanging them for awhile? That is guaranteed to have no duplicates.
Lots of answers say "once you detect Xs in a row, recalculate the last one until you don't get an X".... In practice for a game like this, that approach is millions of times faster than you need for "real-time" human interaction, so just do it!
But, you're obviously uncomfortable with it and looking for something more inherently "bounded" and elegant. So, given you're generating numbers from 1..6, when you detect 2 Xs you already know the next one could be a duplicate, so there are only 5 valid values: generate a random number from 1 to 5, and if it's >= X, increment it by one more.
That works a bit like this:
1..6 -> 3
1..6 -> 3
"oh no, we've got two 3s in a row"
1..5 -> ?
< "X"/3 i.e. 1, 2 use as is
>= "X" 3, 4, 5, add 1 to produce 4, 5 or 6.
Then you know the last two elements differ... the latter would take up the first spot when you resume checking for 2 elements in a row....
vector<BlockType> constructRow()
{
vector<BlockType> row;
row.push_back(STAR); row.push_back(STAR);
row.push_back(UP_TRIANGLE); row.push_back(UP_TRIANGLE);
row.push_back(DOWN_TRIANGLE); row.push_back(DOWN_TRIANGLE);
row.push_back(CIRCLE); row.push_back(CIRCLE);
row.push_back(HEART); row.push_back(HEART);
row.push_back(DIAMOND); row.push_back(DIAMOND);
do
{
random_shuffle(row.begin(), row.end());
}while(rowCheckFails(row));
return row;
}
The idea is to use random_shuffle() here. You need to implement rowCheckFails() that satisfies the requirement.
EDIT
I may not understand your requirement properly. That's why I've put 2 of each block type in the row. You may need to put more.
I think you would be better served to hide your random number generation behind a method or function. It could be a method or function that returns three random numbers at once, making sure that there are at least two distinct numbers in your output. It could also be a stream generator that makes sure that it never outputs three identical numbers in a row.
int[] get_random() {
int[] ret;
ret[0] = rand() % 6 + 1;
ret[1] = rand() % 6 + 1;
ret[2] = rand() % 6 + 1;
if (ret[0] == ret[1] && ret[1] == ret[2]) {
int replacement;
do {
replacement = rand() % 6 + 1;
} while (replacement == ret[0]);
ret[rand() % 3] = replacement;
}
return ret;
}
If you wanted six random numbers (it's a little difficult for me to tell, and the video was just baffling :) then it'll be a little more effort to generate the if condition:
for (int i=0; i<4; i++) {
if (ret[i] == ret[i+1] && ret[i+1] == ret[i+2])
/* three in a row */
If you always change ret[1] (the middle of the three) you'll never have three-in-a-row as a result of the change, but the output won't be random either: X Y X will happen more often than X X Y because it can happen by random chance and by being forced in the event of X X X.
First some comments on the above solutions.
There is nothing wrong with the techniques that involve rejecting a random value if it isn't satisfactory. This is an example of rejection sampling, a widely used technique. For example, several algorithms for generating a random gaussian involve rejection sampling. One, the polar rejection method, involves repeatedly drawing a pair of numbers from U(-1,1) until both are non-zero and do not lie outside the unit circle. This throws out over 21% of the pairs. After finding a satisfactory pair, a simple transformation yields a pair of gaussian deviates. (The polar rejection method is now falling out of favor, being replaced by the ziggurat algorithm. That too uses a rejection sampling.)
There is something very much wrong with rand() % 6. Don't do this. Ever. The low order bits from a random number generator, even a good random number generator, are not quite as "random" as are the high order bits.
There is something very much wrong with rand(), period. Most compiler writers apparently don't know beans about producing random numbers. Don't use rand().
Now a solution that uses the Boost random number library:
vector<Block> BlockField::ConstructRow(
unsigned int max_run) // Maximum number of consecutive duplicates allowed
{
// The Mersenne Twister produces high quality random numbers ...
boost::mt19937 rng;
// ... but we want numbers between 1 and 6 ...
boost::uniform_int<> six(1,6);
// ... so we need to glue the rng to our desired output.
boost::variate_generator<boost::mt19937&, boost::uniform_int<> >
roll_die(rng, six);
vector<Block> row;
int prev = 0;
int run_length = 0;
for (int ii=0; ii<6; ++ii) {
int next;
do {
next = roll_die();
run_length = (next == prev) ? run_length+1 : 0;
} while (run_length > max_run);
row.push_back(Block(next));
prev = next;
}
return row;
}
I know that this already has many answers, but a thought just occurred to me. You could have 7 arrays, one with all 6 digits, and one for each missing a given digit. Like this:
int v[7][6] = {
{1, 2, 3, 4, 5, 6 },
{2, 3, 4, 5, 6, 0 }, // zeros in here to make the code simpler,
{1, 3, 4, 5, 6, 0 }, // they are never used
{1, 2, 4, 5, 6, 0 },
{1, 2, 3, 5, 6, 0 },
{1, 2, 3, 4, 6, 0 },
{1, 2, 3, 4, 5, 0 }
};
Then you can have a 2 level history. Finally to generate a number, if your match history is less than the max, shuffle v[0] and take v[0][0]. Otherwise, shuffle the first 5 values from v[n] and take v[n][0]. Something like this:
#include <algorithm>
int generate() {
static int prev = -1;
static int repeat_count = 1;
static int v[7][6] = {
{1, 2, 3, 4, 5, 6 },
{2, 3, 4, 5, 6, 0 }, // zeros in here to make the code simpler,
{1, 3, 4, 5, 6, 0 }, // they are never used
{1, 2, 4, 5, 6, 0 },
{1, 2, 3, 5, 6, 0 },
{1, 2, 3, 4, 6, 0 },
{1, 2, 3, 4, 5, 0 }
};
int r;
if(repeat_count < 2) {
std::random_shuffle(v[0], v[0] + 6);
r = v[0][0];
} else {
std::random_shuffle(v[prev], v[prev] + 5);
r = v[prev][0];
}
if(r == prev) {
++repeat_count;
} else {
repeat_count = 1;
}
prev = r;
return r;
}
This should result in good randomness (not reliant of rand() % N), no infinite loops, and should be fairly efficient given the small amount of numbers that we are shuffling each time.
Note, due to the use of statics, this is not thread safe, that may be fine for your usages, if it is not, then you probably want to wrap this up in an object, each with its own state.