Codeforces Round 1035 (Div. 2) - Editorial

Welcome to the editorial for Codeforces Round 1035 (Div. 2) A,B,C,D .
Below you'll find explanations, code snippets, and approaches for the problems in this contest. link to the contest -- https://codeforces.com/contest/2119

Problem A - Add or XOR

Statement:
You're given two integers a and b. You can apply two operations on a to transform it into b:

Addition: a ← a + 1 (Cost: x)
XOR: a ← a ^ 1 (Cost: y)

You can apply these operations any number of times in any order. The task is to find the minimum cost to convert a into b, or print -1 if it’s not possible.

Constraints-> t(1≤t≤1e4); a,b,x,y (1≤a,b≤100,1≤x,y≤1e7)

Approach:

if a > b, it's only possible when a-b == 1 and a is odd (use one XOR); otherwise, it's impossible.
for rest case compute the number of steps to reach b and minimize cost using a mix of +1 (cost x) and XOR (cost min(x, y)) based on parity.

Code:

if(a>b){
    if((a&1) && a-b==1){
        cout<<y<<endl;
        return ;
    }else{
        cout<<-1<<endl;
        return ;
    }
}
int odd= (b-a + (a%2))/2;
int even = (b-a)-odd;

int s= odd* x+ even*min(y,x);
cout<<s<<endl;

Problem B - Line Segments

Statement:
You are given two points (px,py) and (qx,qy) starting point - (px,py) and we need to perform n operation

In the i-th operation, you must choose any point such that the Euclidean distance between your current position and the point is exactly (a)i, and then move to that point

Determine can we reach (qx,qy) -- answer in Yes/ No.

Constraints-> (1≤n≤1e3); (1≤px,py,qx,qy≤1e7); (1≤ai≤1e4) (1≤t≤1e4)

Approach:
we are given n steps of fixed lengths, and need to check if they can be arranged (in any direction) to exactly reach the target point.

This is possible only if the straight-line distance lies between the maximum possible (S) and minimum possible (2·max − S) total displacements.

Code:

// s = sum(v)
// maxv = max(v);
// then the possible squared distance between start and end lies in [(2×maxv−s)², s²].

int maxx= *max_element(all(v));
int s= accumulate(all(v),(int)0);

int val=0;
val= max(2*maxx-s,val);
s*=s;
val*=val;

int dx= px-qx;
int dy = py-qy;

int sq= dx*dx + dy*dy;

if(sq>=val && sq<= s){
    yes();
    return ;
}
no();

Problem C - A Good Problem

Statement:
You are given n,l,r,k You need to find the lexicographically smallest array a of length n, consisting of integers, such that:

For every 1≤i≤n , l≤ai≤r
a1&a2&…&an=a1⊕a2⊕…⊕an ,where & denotes the bitwise AND operation and ⊕ denotes the bitwise XOR operation

Constraints-> (1≤k≤n≤1e18,1≤l≤r≤1e18); (1≤t≤1e4)

Approach:
Intuition

You need an array where all numbers are between l and r, and doing AND and XOR over the whole array gives the same result.

This only happens when the extra 1s in XOR (from differing bits) don’t mess up the AND — so mostly, the array needs to be kind of uniform or cleverly bit-aligned.

Steps

If n==1: return l;
If n==2: not possible return -1;
If n is odd, we can pick all values as l — the XOR and AND will match — return l;
If n is even:

-- we need to find two values like [l, l, ..., x, x] where the bit alignment allows AND == XOR.

-- So you find the smallest number x in [l, r] such that it's a power of two or nicely aligned for the condition.

-- If such x exists and k falls into the position of x, return it; otherwise, return l.

Code:

// s = sum(v)
// maxv = max(v);
// then the possible squared distance between start and end lies in [(2×maxv−s)², s²].

int maxx= *max_element(all(v));
int s= accumulate(all(v),(int)0);

int val=0;
val= max(2*maxx-s,val);
s*=s;
val*=val;

int dx= px-qx;
int dy = py-qy;

int sq= dx*dx + dy*dy;

if(sq>=val && sq<= s){
    yes();
    return ;
}
no();

Problem D - Token Removing

Statement:
You're given sequences a of length n where each a[i] is between 0 and i.

For each sequence, you perform n steps where, if a[i] ≠ 0, you remove a token from any position in [a[i], i] (if it hasn't been removed yet).

The "weight" of the sequence is the number of valid ways to remove tokens without conflicts, following those rules.

You're asked to find the sum of weights of all valid sequences, modulo m.

Constraints-> (1≤n≤5000,108≤m≤1.01⋅109); (1≤t≤1000)

Approach:
Intuition

We want to count how many ways tokens can be removed for all valid sequences without generating them explicitly.

since removal depends on available positions and past choices, we use dp to track how many tokens are used at each step.

Approach

Use dp[i][j] = number of ways to handle steps i..n with j tokens already removed.

At each step i, carry forward dp[i+1][j] (do nothing), or

try removing a token using available positions (n - i + 1 - j), multiplied by i (possible a[i] choices).

Sum up all dp[1][*] to get the total number of ways, mod m.

Code:

vvi dp(n+2,vi(n+2,0));
dp[n+1][0]=1;

for(int i=n;i>=1;i--){
    for(int j=0;j<=n;j++){
        dp[i][j]= (dp[i][j]+dp[i+1][j])%m;

        if(j==n){
            continue;
        }

        int temp = n-i+1-j;

        if(temp>0){
            int val= (dp[i+1][j]* temp) % m;
            val = (val*i)%m;

            dp[i][j+1]= (dp[i][j+1]+val)%m;
        }

    }
}

// sum of all the element in dp[1] is the answer;

int res = accumulate(all(dp[1]),(int)0) % m;
cout<< res<<endl;

//the code can be improved by using a 1d dp instead of 2d

Closing Notes

If you have any questions or suggestions about the solutions, feel free to leave a comment below!