Live online mirror of The 2019 ICPC Asia Jakarta Regional Contest

5 лет назад, # ^ |

+14

Thanks for the kind testimony, and congrats once again for winning it last year! :)

Ah, I just realized that FBHC finals clashes with the regional. Maybe I should be thankful that I missed it by penalties now, otherwise I would have a big dilemma.

Good luck for the finals! :)

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I_love_Hoang_Yen

5 лет назад, # ^ |

Just curious, what do you think about problems in Vietnam regional last year? And Vietnam 2017 problems.

Sorry for off-topic comment :D

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ko_osaga

5 лет назад, # ^ |

About 2018, I didn't like harder problems (except G). But problems in 2016/2017 was good, so I can recommend Vietnam contest :)

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Tanzir5

5 лет назад, # |

+45

The link to timeanddate.com redirects to 27th October of 2018. I guess someone is still living in 2018.

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5 лет назад, # ^ |

← Rev. 2 →

+44

Sometimes I am wondering why I can be a red coder when I still make this kind of careless mistakes. This is why we shouldn't copy-paste things (I copy-pasted the timeanddate URL from last year).

Anyway, thanks for pointing it out. It is fixed now. Take my upvote!

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aryanc403

5 лет назад, # |

← Rev. 2 →

+13

Please change the link to https://codeforces.me/contests/1252 instead of https://codeforces.me/contest/1252 . Right now it shows "contest has not started".
People can directly register by visiting the first link.

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Nanjo_Qi

5 лет назад, # |

-9

Maybe I'm a new of Codeforces ... Could you tell me if I participate as a team member , will my personal rating change after the contest ... ?

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ouuan

5 лет назад, # ^ |

+17

I think it's unrated.

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huzaifa242

5 лет назад, # |

+16

Just curious if it is rated? if rated how would the rating mechanism work for a team against individual?

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Nanjo_Qi

5 лет назад, # ^ |

← Rev. 2 →

I was just considering this question and wanted to know how rating will be calculated. Obviously it should be fairer if unrated. Thank you :-).

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deleted2

5 лет назад, # |

-39

Is it rated?

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Mahdivelayatipour

5 лет назад, # ^ |

+12

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VivaanGupta

5 лет назад, # |

+26

I am getting this message please fix it "Can't read or parse problem descriptor"

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SorahISA

5 лет назад, # |

Can we see the resolver?

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codgician

5 лет назад, # |

← Rev. 3 →

How to solve problem E? We attempted to solve it using system of difference constraints with SPFA and got TLE. Thanks in advance.

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-14

5 лет назад, # ^ |

+13

Try to maintain feasible intervals backward and then get result forward based on the feasible intervals.

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codgician

5 лет назад, # ^ |

Thanks. I think I nailed it. Orz%%%

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kurzgesagt

5 лет назад, # ^ |

← Rev. 2 →

[deleted]

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TheOneYouWant

5 лет назад, # |

← Rev. 2 →

Can anyone explain the solution to J? I implemented a greedy idea only to get WA on testcase 20.

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ffao

5 лет назад, # ^ |

The main complexity in J is how to place the type-3 blocks. I decided to iterate over number of type-3 blocks to place, and then for each possibility place the blocks so that we get the fewest number of odd segments of soil using a dp (because we can fill even segments with type-2s completely, they are strictly better than odd segments).

After the type-3 blocks are placed you can place type-1 and type-2 greedily.

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squarepants

4 года назад, # ^ |

I have another idea. Consider all parts of consecutive cells not containing rocks. There are exactly $$$r + 1$$$ parts. Let $$$dp[i][j][1]$$$ be the maximum number of ghosts that can be repelled using at most $$$j$$$ blocks of type $$$1$$$ and last cell of this part covered by a block of type $$$3$$$. Similarly $$$dp[i][j][0]$$$ be the maximum number of ghosts that can be repelled using at most $$$j$$$ blocks of type $$$1$$$ and last cell of this part not covered by a block of type $$$3$$$. Then the dp transitions are given by

$$$ dp[p][k][0] = \max_{j}\big(dp[p - 1][k - j][0] + jg_1 + \Bigl\lfloor \frac{a_i - j}{2} \Bigr\rfloor g_2, dp[p - 1][k - j][1] + j g_1 + \Bigl\lfloor \frac{a_i - j}{2} \Bigr\rfloor g_2 + g_3 \big)$$$

Similarly, we can come up with dp transitions for $$$dp[i][j][1]$$$. Computing these dp values would give time complexity $$$\mathcal{O}(rk^2)$$$. But key observation is that, if we separate $$$j$$$ term and $$$k$$$ term and convert it into a range max query. So, the run time becomes $$$\mathcal{O}(rk)$$$.

My code

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chemthan

5 лет назад, # ^ |

← Rev. 4 →

We need at most O(number of rocks) type-1.

It turns out the test cases are weak, lol.

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BenW

5 лет назад, # ^ |

← Rev. 6 →

may i look at your code? i couldn't see the other's code. it's for learning purpose since i keep stucking on test case 23 after debugging on test case 11,19 & 22

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kurzgesagt

5 лет назад, # ^ |

← Rev. 2 →

[deleted]

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FLEA

5 лет назад, # |

How to solve problem K?

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5 лет назад, # ^ |

For a subsequence $$$[l,r]$$$, you can count how many $$$A$$$ and $$$B$$$ in first element and second element of $$$f(l,r,A,B)$$$. Let them be $$$x_1,y_1$$$ and $$$x_2,y_2$$$. For example, with substring $$$ABA$$$, $$$x_1=2,y_1=3,x_2=1,y_2=2$$$.

For query type $$$2$$$, use segment tree to calculate $$$x_i$$$ and $$$y_i$$$ of subsequence $$$[l,r]$$$. To build the tree, we need to combine $$$x_i$$$ and $$$y_i$$$ of two segment, which could be figured out by draft.

For query type $$$1$$$, use lazy update on segment tree. When we flip a segment odd number of times, simply $$$swap(x_1, y_2)$$$ and $$$swap(x_2, y_1)$$$.

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5 лет назад, # ^ |

Can I see your code ? Thank you very much!

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5 лет назад, # ^ |

Yes

#include <bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 8;
 
typedef long long ll;
 
ll MOD = 1e9 + 7;
 
struct node {
    ll x1, y1, x2, y2;
    bool lazy;
    node(ll x1 = 1, ll y1 = 0, ll x2 = 0, ll y2 = 1) {
        this->x1 = x1;
        this->y1 = y1;
        this->x2 = x2;
        this->y2 = y2;
        lazy = false;
    }
};
 
node t[N << 2];
char s[N];
int n, q;
 
void apply(node &a, node b, node c) {
    a.x1 = (b.x1 * c.x1 + b.y1 * c.x2) % MOD;
    a.y1 = (b.x1 * c.y1 + b.y1 * c.y2) % MOD;
    a.x2 = (b.x2 * c.x1 + b.y2 * c.x2) % MOD;
    a.y2 = (b.x2 * c.y1 + b.y2 * c.y2) % MOD;
}
 
void build(int k, int l, int r){
    if (l == r) {
        if (s[l] == 'A') {
            t[k].x1 = t[k].y1 = t[k].y2 = 1;
            t[k].x2 = 0;
        } else {
            t[k].x1 = t[k].x2 = t[k].y2 = 1;
            t[k].y1 = 0;
        }
        return;
    }
    int m = (l + r) >> 1;
    build(k << 1, l, m);
    build(k << 1 ^ 1, m + 1, r);
    apply(t[k], t[k << 1 ^ 1], t[k << 1]);
}
 
void lazy(int k, int l, int r) {
    if (t[k].lazy) {
        swap(t[k].x1, t[k].y2);
        swap(t[k].x2, t[k].y1);
        if (l < r) {
            t[k << 1].lazy ^= 1;
            t[k << 1 ^ 1].lazy ^= 1;
        }
        t[k].lazy = false;
    }
}
 
void update(int k, int l, int r, int x, int y) {
    lazy(k, l, r);
    if (l > y || r < x) return;
    if (l >= x && r <= y) {
        t[k].lazy ^= 1;
        lazy(k, l, r);
        return;
    }
    int m = (l + r) >> 1;
    update(k << 1, l, m, x, y);
    update(k << 1 ^ 1, m + 1, r, x, y);
    apply(t[k], t[k << 1 ^ 1], t[k << 1]);
}
 
node get(int k, int l, int r, int x, int y) {
    lazy(k, l, r);
    if (l > y || r < x) return node();
    if (l >= x && r <= y) return t[k];
    int m = (l + r) >> 1;
    node t1 = get(k << 1, l, m, x, y);
    node t2 = get(k << 1 ^ 1, m + 1, r, x, y);
    node t0;
    apply(t0, t2, t1);
    return t0;
}
 
int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
    freopen(".inp", "r", stdin);
    freopen(".out", "w", stdout);
    #endif
    cin >> n >> q;
    cin >> s + 1;
    build(1, 1, n);
    while (q--) {
        int id, l, r;
        cin >> id >> l >> r;
        if (id == 1) {
            update(1, 1, n, l, r);
        } else {
            ll a, b;
            cin >> a >> b;
            node t0 = get(1, 1, n, l, r);
            ll A = (t0.x1 * a + t0.y1 * b) % MOD;
            ll B = (t0.x2 * a + t0.y2 * b) % MOD;
            cout << A << " " << B << endl;
        }
    }
}

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5 лет назад, # ^ |

Can you explain the x1 y1 x2 y2 in detail? Thanks in advance!

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5 лет назад, # ^ |

Look at the function in the statement. The return value is $$$(A,B)$$$. $$$x_1,y_1$$$ are number of original $$$A$$$ and $$$B$$$ in $$$A$$$ returned, $$$x_2,y_2$$$ are number of original $$$A$$$ and $$$B$$$ in $$$B$$$ returned.

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5 лет назад, # ^ |

← Rev. 2 →

Thank you ！The x1 y1 x2 y2 is like a 2 x 2 matrix . I try to use the segment tree to maintain the multiplication of the matrix,but I don't know how to change the matrix when the we flip a segment odd number of times. Now , I know it. If we flip a segment odd number of times , actually it's the transpose of this matrix. Thank you.

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KhozmoS

5 лет назад, # ^ |

ok but i can't understand the transition or combinations of nodes in Segment Tree.ej why ( a.x1 = (b.x1 * c.x1 + b.y1 * c.x2) % MOD; )

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5 лет назад, # ^ |

For a segment $$$[l,r]$$$, $$$f(l,r,A,B)=(x_1\times A+y1\times b, x_2\times A + y_2\times B)=(A',B')$$$

When combine segment $$$[l_2,r_2]$$$ next to $$$[l_1,r_1]$$$ and , $$$A'$$$ and $$$B'$$$ of the first segment become $$$A$$$ and $$$B$$$ of input of $$$f(l_2,r_2,A,B)$$$. You can make some drafts to figure out the equations.

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dtc03012

5 лет назад, # ^ |

use sqrt decomposition!

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bkdn23.ILoveUhiuhiu

14 месяцев назад, # ^ |

Can I see your code ? Thank you very much!

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kurzgesagt

5 лет назад, # ^ |

← Rev. 2 →

-8

[deleted]

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katana_handler

5 лет назад, # |

how and where to submit the solutions if we want to upsolve???

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hieplpvip

5 лет назад, # |

← Rev. 2 →

Could someone give me advice on how to solve problem G ?

EDIT: Solved using segment tree

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BenW

5 лет назад, # ^ |

← Rev. 7 →

i keep getting WA on test case 2. could you please help me to find my mistake? i couldn't find my bug.

EDIT: Solved, i did a silly mistake, i thought that Ai would be in decreasing order, i didn't read the problem carefully

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docardoso

5 лет назад, # |

Could someone provide insights on problem H? I got stuck on test case 5 with my solution: https://ideone.com/at2GS0 .

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5 лет назад, # ^ |

I think it is the accuracy problem of floating point numbers.

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sabertooth

5 лет назад, # ^ |

can you review this.. Stuck on test case 5 Code

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docardoso

5 лет назад, # ^ |

I think you made the same mistake as I: relying on a floating point value instead of sticking with integers, which led to some rounding error. This is a diff of my AC code and a previous version of it which also got WA on test case 5: https://www.diffchecker.com/sckBvjUO .

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5 лет назад, # ^ |

you need long double

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deermen001

5 лет назад, # ^ |

Can you tell why we needed to use long double instead of double?

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5 лет назад, # ^ |

this test data: 1 999999999 999999999 if you use double,your answer will lose 0.5

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Priyansh31dec

4 года назад, # ^ |

Thanks a lot for this example test case. Can you give some insights about knowing when to use normal double and when to use long double and is it good if we start using only long double from now on? I am really curious about this precision thing here. The double value can have a precision of up to 10^(-11) but still, it is going wrong for a single decimal, any reason why?

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steinum

5 лет назад, # |

can anyone give me one small test case for k? I use square root decomposition .

here my solution

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5 лет назад, # ^ |

This test case ,your output is different with mine.

Click me

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soloSquad1999

5 лет назад, # |

← Rev. 3 →

Sorry if I misunderstood something. But this contest was not in gym, right? So why I could not see another team 's code?

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https://github.com/jonathanirvings/icpc-jakarta-2019/blob/master/problem_analysis.pdf

5 лет назад, # |

where can I find the solution

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21August

5 лет назад, # |

+23

Will the submissions be visible?

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25shubham

5 лет назад, # |

pls someone write editorial blog, it will help in upsolve the problem..

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nirvana

5 лет назад, # ^ |

The judge has shared problem analysis of the contest in their github. The link is already added below the Contest Material tab when you open a problem from the contest.

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lamnhh

5 лет назад, # |

How to solve problem I?

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chemthan

5 лет назад, # ^ |

← Rev. 3 →

The annoying problem, no innovative idea. The obvious thing we notice is just find the way to reach the border. Other thing I can see is that to go through two touching circles we need to follow their tangent. So we can draw lines between each pairs of circle then get some intersections,...

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tfg

5 лет назад, # ^ |

You can draw something like voronoi diagram (but not really) by defining a half-plane using a separator between two circles (for example, a line that's exactly between two circles). By using this you can find some point from this graph you can get to and then traverse it until you can get to the end point.

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NeiH

5 лет назад, # |

explain me solve problem H?? please

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NeiH

5 лет назад, # ^ |

← Rev. 2 →

sorry. I'm not use long double type so WA test 5

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5 лет назад, # ^ |

← Rev. 2 →

in fact,before i found the problem，i had WA 5 times

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5 лет назад, # |

Auto comment: topic has been updated by jonathanirvings (previous revision, new revision, compare).

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its_aks_ulure

5 лет назад, # |

Please make the test cases visible.

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anonymous_Bunny

5 лет назад, # |

Please make testcases and other's solutions available

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21August

5 лет назад, # |

Can you please make all sources visible?

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YU1

5 лет назад, # |

I try a different DP way to solve problem B but it cant pass the test 5. Can anyone make the test cases visible QAQ.

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crazeann

5 лет назад, # |

I need help on problem H, got WA on test 40. Here is my code. I have tried on my own test case with 8000 numbers of N, but did not spot any mistake. what do i miss?!!

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flashmt

5 лет назад, # ^ |

Line 64, why do you compare with only the last one?

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toxic_hack

5 лет назад, # |

Why are the test cases and solutions still not visible!?

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IsmailToukhy

5 лет назад, # |

← Rev. 2 →

why TLE? problem K

is that because of using vectors in representing the matrices or what?

Code

#pragma GCC optimize ("unroll-loops")

#include <bits/stdc++.h>

using namespace std;

#define ff      first
#define ss      second
#define all(v)  (v.begin(), v.end())

#define row         vector<long long>
#define matrix      vector<row>
#define ROWS(v)     (v).size()
#define COLS(v)     (v)[0].size()

#define DBG1(x)     cerr<<#x<<":"<<(x)<<'\n';
#define DBG2(x, y)  cerr<<#x<<":"<<(x)<<"  "<<#y<<":"<<(y)<<'\n';
#define DBG3(x,y,z) cerr<<#x<<":"<<(x)<<"  "<<#y<<":"<<(y)<<"  "<<#z<<":"<<(z)<<'\n';

typedef long long lld;

const int N = 1e5 + 9, OO = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, q;
char s[N];

matrix TR[N<<2], arr[2];
bool lazy[N<<2];
matrix ans(2, row(2, 0)), c(2, row(2, 0));
matrix identity = {{1, 0},
                   {0, 1}};

matrix mul(const matrix& a, const matrix& b){
    c[0][0] = (a[0][0] * b[0][0] + a[0][1] * b[1][0]) % MOD;
    c[0][1] = (a[0][0] * b[0][1] + a[0][1] * b[1][1]) % MOD;
    c[1][0] = (a[1][0] * b[0][0] + a[1][1] * b[1][0]) % MOD;
    c[1][1] = (a[1][0] * b[0][1] + a[1][1] * b[1][1]) % MOD;
    return c;
}

inline void edit(int p){
    swap(TR[p][0][0], TR[p][1][1]);
    swap(TR[p][0][1], TR[p][1][0]);
}

inline void push(int p){
    lazy[p] = 0;
    edit(p*2);
    lazy[p*2] ^= 1;
    edit(p*2+1);
    lazy[p*2+1] ^= 1;
}

matrix build(int l = 0, int r = n-1, int p = 1){
    if(l == r) return TR[p] = arr[s[l]-'A'];
    int m = (l + r) >> 1;
    return TR[p] = mul(build(l, m, p*2), build(m+1, r, p*2+1));
}

matrix update(int l, int r, int ql, int qr, int p = 1){
    if(r < ql || l > qr) return TR[p];
    if(l >= ql && r <= qr){
        edit(p);
        lazy[p] ^= 1;
        return TR[p];
    }
    if(lazy[p]) push(p);
    int m = (l + r) >> 1;
    return TR[p] = mul(update(l, m, ql, qr, p*2), update(m+1, r, ql, qr, p*2+1));
}

matrix get(int l, int r, int ql, int qr, int p = 1){
    if(r < ql || l > qr) return identity;
    if(l >= ql && r <= qr) return TR[p];
    if(lazy[p]) push(p);
    int m = (l + r) >> 1;
    return mul(get(l, m, ql, qr, p*2), get(m+1, r, ql, qr, p*2+1));
}

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //fastin;
    scanf("%d %d", &n, &q);
    scanf("%s", s);

    arr[0] = {{1, 0},
              {1, 1}};

    arr[1] = {{1, 1},
              {0, 1}};

    build();

    while(q--){
        int t, l, r;
        scanf("%d %d %d", &t, &l, &r);
        --l, --r;
        if(t == 1) update(0, n-1, l, r);
        else{
            lld a, b;
            scanf("%lld %lld", &a, &b);
            ans[0][0] = a;
            ans[0][1] = b;
            ans = mul(ans, get(0, n-1, l, r));
            printf("%lld %lld\n", ans[0][0], ans[0][1]);
        }
    }

    return 0;
}

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andrewtheodore

5 лет назад, # ^ |

← Rev. 2 →

+21

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mrtempo

5 лет назад, # |

Test cases for problem F cannot be more broken. You don't need to consider subtrees isomorphism in any way. Basically, if you're a tree centroid, then you're okay. Just maximize with its number of children, and you'll pass.

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AmanGupta_

5 лет назад, # |

What's wrong with this code. It fails on test-5 of Even Path. https://codeforces.me/contest/1252/submission/64762718

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vivace_jr

5 лет назад, # |

jonathanirvings Test cases or other's solutions are not visible, please fix

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vivace_jr

5 лет назад, # ^ |

MikeMirzayanov Please fix

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