one tip: Please hope realistic

I think you forgot that you are 1634 instead of 1834...

prolly you should pray not falling back to cyan...

Solve A-F

quite a similar graph with me but keep practicing

If someone is honest with himself, he doesn't give a shit to them.

shayan did this countless times :)))

umm, the stream starts 5 minutes after the contest, so even if you submit using the answer, it won't count to the contest submissions (or something like that idk)

i need to solve 4-5 problems to keep my rating :o. help.

we all need it for edu rounds

u should stop trolling on me and get life

I had the same issue bye bye CM_

Input and output are one of the slowest parts of any program. It is good if you add fast io as a template.

In problem D, we have n, m <= 10 ^ 6, which are huge and require fast io. Another thing, is to use "\n" for the next line, don't use endl for it. Sine endl also flushes your output and is slower.

$$$10^7$$$ is supposed to be $$$0.1$$$ seconds as I estimate. Just think about $$$10^8$$$ ~ $$$1$$$ second.

if(y <= -2) pno else pyes

even i thought the same untill i thought some thing like

if a ="qwerty" b = "qwft"

then ur ans will be 7 but actual answer 8

i tried lcs but failed

Use binary search on the max minimum -- the choices are predetermined unless the pairs are {1, 1} or {-1, -1}, during binary search, decide which movie to assign it to

typedef array<int, 2> ai;
void solve(){
    int n;
    cin >> n;
    vector<ai> v(n);
    for (int i = 0; i < n; i++) cin >> v[i][0];
    for (int i = 0; i < n; i++) cin >> v[i][1];
    int sum1 = 0, sum2 = 0;
    int l = n*-1, r = n;
    int ans = l;
    while (l<=r) {
        int mid = (l+r)/2;
        int temp1 = 0, temp2 = 0;
        for (auto [a, b] : v) {
            if (a==b && a != 0) {
                continue;
            }
            else if (a>b) temp1 += a;
            else temp2 += b;
        }
        for (auto [a, b] : v) {
            if (a==b && a != 0) {
                if (a==1) {
                    if (temp1 < mid) temp1 += a;
                    else temp2 += a;
                }
                else if(a == -1) {
                    if (temp1 > mid) temp1 += a;
                    else temp2 += a;
                }
            }
        }
        if (temp1 >= mid && temp2 >= mid) {
            l = mid+1;
            ans = max(ans, mid);
        }
        else r = mid-1;
    }
    cout << ans << endl;
}

Observation 1: If $$$a[i] \ne b[i]$$$, it's always optimal to take the review of the bigger of the two.

Observation 2: If $$$a[i] = b[i]$$$, you can't greedily assign yet. However, since they are equal, you can apply this decision to either of them.

Keep track of the 1 == 1, -1 == -1, and then distribute those plus and minuses after you've totaled what you were forced to take from each in Observation 1.

The main objective is making the smaller rating movie's rating as bigger as possible.At first all the 1 0 and 0 1 pairs can be counted(If 1 0 then first movie gets +1 if 0 1 then second movie gets +1) because increasing rating of a particular movie will be always good as here both smaller and greater count rating movie get positive rating.We can avoid -1 0 and 0 -1 pairs as decreasing a movie rating is not efficient at all.Then we can come to the calculation of 1 1 pairs.If already counted first movie rating is smaller then we can consider this movie otherwise the second movie will be considered.For -1 -1 pair we can always consider the movie the rating of which is greater so that it doesn't reduce the value of smaller rating movie.

Consider each pair of $$$(a_i, b_i)$$$. If $$$a_i \ne b_i$$$, then it's always advantageous to take the larger one, because the smaller one is either $$$0$$$ or $$$-1$$$, which does not improve the final score if chosen. The pairs $$$(a_i, b_i) = (0, 0)$$$ are meaningless, so we're left with $$$x$$$ pairs of $$$(-1, -1)$$$ and $$$y$$$ pairs of $$$(1, 1)$$$, and we want to assign those pairs to the first group and the second group, which can be done greedily.

For example, this test:

1
2
-1 0 
1 1 

Same for D.

if B<-1 then NO simple observation

Its always optimal to melt a tool after forging (gives x experience and bi additional units of that metal).

Now, if we forge ith tool k times using metal j, we would use — ai+(ai-bi)+(ai-bi)...-bi = k(ai-bi) units of metal j

(the last minus comes when we melt after last forging) __

So for each metal, we can start forging in the increasing order of (ai-bi). Still figuring out how to bring this down from O(n*m) :(

I explained my solution in this comment https://mirror.codeforces.com/blog/entry/130882?#comment-1164494

You can make a dp vector of size 1e6+100 which represents the number of experience if you have i ingots. For c <=1e6 you can print the answer, otherwise you can make it smaller than 1e6 by forging and melting as much as possible with the pair that has the smallest a-b and smallest a among them. My solution: https://mirror.codeforces.com/contest/1989/submission/267753254

yeah i was confused too

Happened with me... I spent 40 minutes on B with 35 thinking about lcs due to this..Realised it very late that insertions in the resulting string could only be made at ends or the beginning so we had to check the max length of substring of b present in a as a subsequence..