1207A - There Are Two Types Of Burgers

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (Roms)**

```
#include <bits/stdc++.h>
using namespace std;
int t;
int b, p, f;
int h, c;
int main() {
cin >> t;
for(int tc = 0; tc < t; ++tc){
cin >> b >> p >> f;
cin >> h >> c;
b /= 2;
if(h < c){
swap(h, c);
swap(p, f);
}
int res = 0;
int cnt = min(b, p);
b -= cnt, p -= cnt;
res += h * cnt;
cnt = min(b, f);
b -= cnt, f -= cnt;
res += c * cnt;
cout << res << endl;
}
return 0;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
n, m = map(int, input().split())
a = [[] for i in range(n)]
for i in range(n):
a[i] = list(map(int, input().split()))
ans = []
for i in range(n - 1):
for j in range(m - 1):
if(a[i][j] * a[i][j + 1] * a[i + 1][j] * a[i + 1][j + 1] > 0):
a[i][j] = 2
a[i + 1][j] = 2
a[i][j + 1] = 2
a[i + 1][j + 1] = 2
ans.append([i, j])
cnt1 = 0
for i in range(n):
for j in range(m):
if(a[i][j] == 1):
cnt1 += 1
if(cnt1 != 0):
print(-1)
else:
print(len(ans))
for x in ans:
print(x[0] + 1, x[1] + 1)
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution (adedalic)**

```
const val INF64 = 1e18.toLong()
fun main(args: Array<String>) {
val tc = readLine()!!.toInt()
for (t in 1..tc) {
val (n, a, b) = readLine()!!.split(' ').map { it.toInt() }
val s = readLine()!!
val d = Array(n + 1) { arrayOf(INF64, INF64) }
d[0][0] = b.toLong()
for (pos in 0 until n) {
if (s[pos] == '0') {
d[pos + 1][0] = minOf(d[pos + 1][0], d[pos][0] + a + b)
d[pos + 1][1] = minOf(d[pos + 1][1], d[pos][0] + 2 * (a + b))
d[pos + 1][1] = minOf(d[pos + 1][1], d[pos][1] + a + 2 * b)
d[pos + 1][0] = minOf(d[pos + 1][0], d[pos][1] + 2 * a + b)
} else {
d[pos + 1][1] = minOf(d[pos + 1][1], d[pos][1] + a + 2 * b)
}
}
println(d[n][0])
}
}
```

1207D - Number Of Permutations

Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution (Roms)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = int(3e5) + 99;
const int MOD = 998244353;
int mul(int a, int b){
return (a * 1LL * b) % MOD;
}
int sum(int a, int b){
return (a + b) % MOD;
}
int n;
pair<int, int> a[N];
int f[N];
int main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d%d", &a[i].first, &a[i].second);
f[0] = 1;
for(int i = 1; i < N; ++i)
f[i] = mul(i, f[i - 1]);
int res = f[n];
for(int c = 0; c < 2; ++c){
int d = 1;
sort(a, a + n);
int l = 0;
while(l < n){
int r = l + 1;
while(r < n && a[l].first == a[r].first) ++r;
d = mul(d, f[r - l]);
l = r;
}
res = sum(res, MOD - d);
for(int i = 0; i < n; ++i)
swap(a[i].first, a[i].second);
}
sort(a, a + n);
int l = 0;
int d = 1;
while(l < n){
int r = l + 1;
while(r < n && a[l].first == a[r].first) ++r;
map<int, int> m;
for(int i = l; i < r; ++i) ++m[a[i].second];
for(auto p : m) d = mul(d, f[p.second]);
l = r;
}
for(int i = 1; i < n; ++i)
if(a[i - 1].second > a[i].second) d = 0;
res = sum(res, d);
printf("%d\n", res);
return 0;
}
```

Idea: adedalic, Ne0n25, BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
int main()
{
cout << "?";
for(int i = 1; i <= 100; i++)
cout << " " << i;
cout << endl;
cout.flush();
int res1;
cin >> res1;
cout << "?";
for(int i = 1; i <= 100; i++)
cout << " " << (i << 7);
cout << endl;
cout.flush();
int res2;
cin >> res2;
int x = 0;
x |= (res1 & (((1 << 7) - 1) << 7));
x |= (res2 & ((1 << 7) - 1));
cout << "! " << x << endl;
cout.flush();
return 0;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 500043;
const int K = 750;
int a[N];
int sum[K][K];
int main()
{
int q;
scanf("%d", &q);
for(int i = 0; i < q; i++)
{
int t, x, y;
scanf("%d %d %d", &t, &x, &y);
if(t == 1)
{
a[x] += y;
for(int i = 1; i < K; i++)
sum[i][x % i] += y;
}
else
{
if(x >= K)
{
int ans = 0;
for(int i = y; i <= 500000; i += x)
ans += a[i];
printf("%d\n", ans);
}
else
printf("%d\n", sum[x][y]);
}
}
return 0;
}
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution (PikMike)**

```
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < int(n); i++)
using namespace std;
const int AL = 26;
const int N = 400 * 1000 + 13;
struct node{
int nxt[AL];
node(){
memset(nxt, -1, sizeof(nxt));
}
int& operator [](int x){
return nxt[x];
}
};
struct node_at{
int nxt[AL];
int p;
char pch;
int link;
int go[AL];
node_at(){
memset(nxt, -1, sizeof(nxt));
memset(go, -1, sizeof(go));
link = p = -1;
}
int& operator [](int x){
return nxt[x];
}
};
int cntnm;
node trienm[N];
int cntqr;
node_at trieqr[N];
int add_string(string s){
int v = 0;
for (auto it : s){
int c = it - 'a';
if (trieqr[v][c] == -1){
trieqr[cntqr] = node_at();
trieqr[cntqr].p = v;
trieqr[cntqr].pch = c;
trieqr[v][c] = cntqr;
++cntqr;
}
v = trieqr[v][c];
}
return v;
}
int go(int v, int c);
int get_link(int v){
if (trieqr[v].link == -1){
if (v == 0 || trieqr[v].p == 0)
trieqr[v].link = 0;
else
trieqr[v].link = go(get_link(trieqr[v].p), trieqr[v].pch);
}
return trieqr[v].link;
}
int go(int v, int c) {
if (trieqr[v].go[c] == -1){
if (trieqr[v][c] != -1)
trieqr[v].go[c] = trieqr[v][c];
else
trieqr[v].go[c] = (v == 0 ? 0 : go(get_link(v), c));
}
return trieqr[v].go[c];
}
int add_letter(int v, int c){
if (trienm[v][c] == -1){
trienm[cntnm] = node();
trienm[v][c] = cntnm;
++cntnm;
}
return trienm[v][c];
}
vector<int> g[N];
int tin[N], tout[N], T;
void dfs_init(int v){
tin[v] = T++;
for (auto u : g[v])
dfs_init(u);
tout[v] = T;
}
int f[N];
void upd(int v, int val){
for (int i = tin[v]; i < N; i |= i + 1)
f[i] += val;
}
int get(int x){
int sum = 0;
for (int i = x; i >= 0; i = (i & (i + 1)) - 1)
sum += f[i];
return sum;
}
int sum(int v){
return get(tout[v] - 1) - get(tin[v] - 1);
}
int n, m;
int nm[N], qr[N];
vector<int> nms[N];
vector<int> reqs[N];
int ans[N];
void dfs(int v, int cur){
upd(cur, 1);
for (auto it : nms[v])
for (auto q : reqs[it])
ans[q] = sum(qr[q]);
forn(i, AL) if (trienm[v][i] != -1)
dfs(trienm[v][i], go(cur, i));
upd(cur, -1);
}
int main(){
cntqr = 0;
trieqr[cntqr++] = node_at();
cntnm = 0;
trienm[cntnm++] = node();
char buf[N];
scanf("%d", &n);
forn(i, n){
int t;
scanf("%d", &t);
if (t == 1){
scanf("%s", buf);
nm[i] = add_letter(0, buf[0] - 'a');
}
else{
int j;
scanf("%d%s", &j, buf);
--j;
nm[i] = add_letter(nm[j], buf[0] - 'a');
}
nms[nm[i]].push_back(i);
}
scanf("%d", &m);
forn(i, m){
int j;
scanf("%d%s", &j, buf);
--j;
reqs[j].push_back(i);
qr[i] = add_string(buf);
}
for (int v = 1; v < cntqr; ++v)
g[get_link(v)].push_back(v);
T = 0;
dfs_init(0);
dfs(0, 0);
forn(i, m)
printf("%d\n", ans[i]);
}
```

[DELETED]

In solution for C) editorial says that: If s[pos]=1 then we have to stay on the height 2, but in the sample test 1

`8 2 5 00110010`

It is optimal to go to height 1, after index 3, where s[3] = 1, and s[4] = 0 (as shown in the problem figure). What am I missing here?The "s" array in the editorial corresponds to intervals (so s[3] corresponds to the height of the interval (3, 4)), but the dp array "d" corresponds to the height at specific x-values.

In other words, because s[3] = 1, the road needs to be at height 2 at x=4, meaning that d[4][0] doesn't exist. s[4] = 0 instead means that d[5][0] and d[5][1] are both OK.

It depends on what's the value of a and b. whether to get down at the gap of two crossroads or not, depends on how many piller you need to add. case 1: If you get down, then it takes 2 more pipes. which adds 2*a more values case 2: if you don't, then it takes k more pillers. which adds k*b more values

take the min of case1 and case 2. That's all.

I didn't get the 2nd query of F ? What sum is supposed to calculate ?

sum of all those elements whose indices give remainder y when divided by x.

During contest, after solving first 5 problems, spent 1 hour working on problem F looking for a solution which was faster than O(n^(3/2)) because I assumed, incorrectly, that O(n^(3/2)) cannot pass... I even designed an O(n^(4/3)) algorithm to solve a simpler version of the problem where all the numbers are fixed from the beginning and never modified. (The solution is similar to the solution in the editorial except that the value of K is n^(2/3) instead of sqrt(n)). I then spent all the time trying to improve my algorithm to support modification to the sequence...

Honestly, I also felt that problem E was much easier than problem C and D.

I think 500000 is too large that my friend sxd666 got FST only because he used long long instead of int.

Can anyone explain more elaborately :For queries with small x (x≤K), we may notice two things:

k=4 1 2 3 4 5 6 7 8 then 1 2 3 4 is only possible for <=k and each can have atmost ai-1 remainder, hence total approximately k^2 queries.

IN ques.D for counting of cnt1,given is cnt1=c1!*c2*!..cn! my doubt is while calculating c1! ,it is not necessary that the corrsponding elements in the another column will be all different and give different permutations..so we must divide it by those factorials..

Even if you consider the second column, even two sets of identical pairs will need to be considered twice so there is no need to divide.

I was thinking the same thing. The third sample case is as follows

The given answer is 4. However, there are only 3 ways to arrange the above pairs (3!/2!). What am I misunderstanding?

Can someone explain the question C please?

An online solution to problem G :

We build the suffix automaton for the trie which is given, and then we build a segment tree for every node (endpos set) . For a query, we do the heavy-chain decomposition for the trie, and then we query the segment tree sum.

Total time complexity is $$$\mathrm O(n\log^2n)$$$.

My code : https://codeforces.com/contest/1207/submission/59299826

This method use so much memory that it will get MLE on test 21 (https://codeforces.com/contest/1207/submission/59296517), but you can get AC by replacing the array with

`std::map`

.Sorry for my poor English.

Why your code links to submission by different user?

The editorial solution for C,

should it be

why the solution ignores the 0 to 1 transition when the s[i] == '1'?

When s[i] = '0', there are 4 possibilities, you can reach 'pos+1', that is two transitions each from height 1 and 2.

When s[i] = '1', there is only one 1 possibility, you can reach 'pos+1', that is using one gas pipeline and 2 pillars.

Pictorial illustration

but When s[i] = '1', why it can not be height 1 to height 2 transition? which uses 2 pipes and 2 pillar

Because you can only have zig-zagged line in the interval (i, i+1) when s[i] = '0', when s[i] = '1', you cannot break the line, it must be linear coming from the left side. Hence you can only modify dp[i+1][1].

ok, I got it. Thank you.

I am getting wrong Answer on Test 13 in

Problem 1207C Gas Pipeline. My logic is that i am using dp[j][0] which is the cost to bring pipeline to height 1 at position j and dp[j][1] for cost to bring pipeline to height 2 at j. if Character at j is 1 dp[j][0] is -1 which means that it is impossible to bring pipeline to height 1. I am using MAX_VALUE in place of -1 in dp[j-1][0] if dp[j-1][0] is -1 which means it has a very high cost to build the pipeline from there.Link to my Solution..

https://codeforces.com/contest/1207/submission/59357734

Can someone tell me why my submission is giving WA for question D. Thanks in advance

https://codeforces.com/contest/1207/submission/59385418

Here's the mistake. you are subtracting something from modded value. In some case value of tot will be less than (cnt1 + cnt2 — cnt12) or (cnt1 + cnt2) that's why your final answer is landing to negative value.

replace your lines with below lines.

Oh right Thanks a lot . After one hour of struggling it was the MOD :p

Hey, I understood the approach to calculate c1,c2. Can you elaborate on how to calculate c12.

It would be much helpful. Thank you :)

c12 is the case when both the first and second elements should be sorted in order. So just sort the pair and check if the new sequence is sorted according to both first and second element. Then just check for the repeating pairs and subtract it from c1 and c2.

Could You please elaborate a bit..Thankyou!!

F is not worth 2100.

What do you mean? For each problem you get 1 point for solving it.

UPDATE: OK, I see. Tags with numbers of points have been added.

Could anyone give me a greedy solution for problem C. I came up with this approach in the contest but I can't implement it correctly. Thanks in advance!

https://codeforces.com/contest/1207/submission/59389852

thank u

https://codeforces.com/contest/1207/submission/59315647

In the editorial of Problem F it states that: "Note that, as in most problems related to sqrt-heuristics, it may be optimal to choose the constant that is not exactly sqrt(N), but something similar to it (but most solutions should pass without tuning the constant)"

Could anyone please explain why it may not be optimal to choose exactly sqrt(N)?

The two parts of the algorithm have O(K) and O(N/K) complexities but the constant factor will not necessarily be the same (e.g. we do a modulo and use more memory for the small x part and always update K elements, while the naive algorithm for large x just sums N/x values which is only at most N/K... many things to consider).

Pushing K on one side or the other of sqrt(N) will increase the time spent on one part and reduce the other, so because of the constant factors the optimal sum of complexities might be for 0.7sqrt(N) or 3.5sqrt(N) or whatever. Here sqrt(N) is 750 but experimentaly i got the lowest execution time with K around 400.

Do people really test this in competition though ? Or just guess and round a bit up or down ?

Never mind. Solved it.

[deleted]

I've used greedy approach to solve problem C. It fails on test case 13. Please tell me where it goes wrong. Code

can someone help me in problem D,i got the concept how to solve it but i am not getting how to calculate cnt12

Please someone explain the problem E !!

I couldn't understand from the editorial.

1) In each query generate numbers in the way to check some bits in any $$$i$$$ case (if we are sure that $$$k_{th}$$$ bit is '1' in choosed number and '0' in answer, then $$$x$$$ has $$$k_{th}$$$ bit also '1', otherwise it is '0' — it follows from the properties of XOR).

2) Let's fix first $$$a$$$ bits as '1' and choose any combinations in another $$$14 - a$$$ bits. We have to generate $$$100$$$ different combinations, so $$$2^{14 - a}$$$ must be better than $$$100$$$ => $$$14 - a = 7, a = 7$$$ ($$$2^7 = 128 > 100$$$).

So we fix first $$$7$$$ bits as $$$1111111$$$ and increase another, getting the following sequence: $$$11111110000000_2$$$, $$$11111110000001_2$$$, $$$11111110000010_2$$$ and etc.

3) Using the same algorithm generate another $$$100$$$ numbers, but visa versa: $$$00000001111111_2$$$, $$$00000011111111_2$$$, $$$00000010111111_2$$$ and etc.

4) From first query answer we check first $$$7$$$ bits using XOR, from another query answer — last $$$7$$$ bits and finally get complete answer.

Oh so, now I get it. Thanks a lot !!

Task F is quite easy to solve and code, but statement is quite hard to understand :D

At first I thought that it is necessary to calculate the sum of all elements $$$a_i = k \cdot x + y$$$, not indexes)

In G instead of adding on path to the root and extracting value from vertex we can do: add to a vertex and calculate sum in a subtree (it can be easily verified to be equivalent)

Another solution for problem G:

There are no more than $$$S$$$ = $$$2 \times sqrt(4 \times 10^5)$$$ different lengths in the set of all patterns, say $$$L_1, L_2, ..., L_S$$$.

When DFS to some trie node $$$v$$$, among all different patterns with length $$$L_i$$$, there is at most one occurrence ending at node $$$v$$$.

Therefore, maintaining number of occurrences of all different patterns can be done in $$$O(S)$$$ whenever DFS moves to next vertex.

So you do an ordinary Aho-Corasick but with the text inside which we search being a trie rather than just a plain text?

I think we can even bound

Sbysqrt(2*4*10^5)but is it enough? Your total compexity isO(S*T). It can be as much as350e6operations. Have you checked if this solution passes the tests?Can someone explain how to calculate c12 more elaborately in Div2D?

Read The Comments

Can someone please explain problem statement of "1207C — Gas Pipeline".

For E

I have an approach but don't know whether it's feasible or not.

Let us denote numbers of $$$1^{st}$$$ query by $$$a_i$$$ and numbers of $$$2^{nd}$$$ query by $$$b_j$$$.So if we select numbers in such a way that xor of any pair ($$$a_i$$$,$$$b_j$$$) is unique. Now let us denote the output to two queries by $$$N$$$ and $$$M$$$ respectively then we can say that $$$N$$$ = $$$a_i \oplus x$$$ and $$$M$$$ = $$$b_j \oplus x$$$.If we now do $$$N \oplus M$$$ then we get $$$a_i \oplus b_j$$$ .Since we selected such numbers that every pair has unique xor we know value of $$$a_i$$$ and $$$b_j$$$ and once we get this we can get $$$x$$$ by $$$N \oplus a_i$$$ or $$$M \oplus b_j$$$.

So my question is there any way to select such numbers that their xor is unique.BledDest

why (x % i)?

R(x,y) is the set of all integers from 1 to 500000 which have remainder y modulo x. Should it be (integers from 1 to 500000) % x = y?

sorry for my poor English

Why am I getting runtime error in D? https://codeforces.com/contest/1207/submission/60137631

Can someone suggest another Dynamic Approach for problem C. Gas Pipeline?

1207E - XOR-угадайка

I solved this problem in a different way than editorial, so i'm sharing it. It's pretty simple too.

For the first query, We simply give numbers from 1 to 100.

now we the judge will return me ($$$Ai$$$

xor$$$X$$$)we

xorall number from 1 to 100 with this number.for the number $$$Ai$$$ it will become ($$$Ai$$$)

xor($$$Ai$$$xor$$$X$$$) = $$$X$$$so the required number is in the list of numbers, let's call this list suspects.

Now we give query numbers from $$$1 * 163$$$ to $$$100 * 163 $$$

The judge will return some value lets call it $$$Y$$$. For each value in the list suspect we check its

xorvalue from ($$$1$$$ x $$$163 $$$) to ($$$100$$$ x $$$163 $$$), if this equals $$$y$$$, then that value of suspect is our answer. Solved without any kind of bitmasks.My Solution : https://codeforces.com/contest/1207/submission/66218060

P.S i choose 163 because its the smalles number who on multiplication with 100 will be less than $$$2^14 -1$$$

We can also solve E just by brute-force. lets say we gave two sequence (Ai,Bi) to the judge now it will return Ai^x and Bi^x ,take xor of these values we get Ai^Bi. If we can generate two lists of size 100 such Ai^Bi is distinct for every Ai,Bi. This can be done by brute force (have a look at my submission https://codeforces.com/contest/1207/submission/6769403) Now we know the Ai,Bi hence x will be just Ai^(Ai^x).