### lucifer1004's blog

By lucifer1004, history, 4 months ago,

## Problem A — Retype

We only have two options, thus

$ans=K-1+\min(N + 1, K - S + N - S + 1)$

Time complexity is $O(1)$.

Code (C++)
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

class Solution {
public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
long long N, K, S;
cin >> N >> K >> S;
cout << K - 1 + min(N + 1, K - S + N - S + 1) << endl;
}
};

int main() {
int t;
cin >> t;
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


## Problem B — Boring Numbers

All problems such that require counting of numbers within range $[L,R]$ can be transformed into solving for $[0,R]$ and $[0,L]$ separately, and taking their difference as the final answer.

Now suppose $X$ has $D$ digits and we want to count boring numbers within $[0,X]$.

First, let's consider all numbers with $d<D$ digits. For $d$ digits, we can generate $5^d$ boring nubmers since we have $5$ options for each position (the most significant nubmer must be add so it cannot be $0$). So all numbers with $d<D$ digits make a contribution of $\sum_{i<D}5^i$.

Then we consider numbers with $D$ digits and are no larger than $X$.

Start from the most significant digit, and suppose that we are at the $i$-th digit now.

• If $X[i]$ does not satisfy the requirement of parity, we just need to count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $b[X[i]]$), then add to the total number $b[X[i]]\cdot5^{D-i}$. Since for these $b[X[i]]$ numbers, the following $D-i$ digits can be chose arbitrarily. In this case, we can stop right here.
• Otherwise, we first count the digits that are smaller than $X[i]$ and can satisfy the parity (we can precalculate such numbers in $a[X[i]]$) and add to the total number $a[X[i]]\cdot5^{D-i}$. Then we are going to count boring numbers that have exactly same $i$ digits as $X$ and continue our processing. Note that if $i=D$, we need to add $1$ to the total number, since this means $X$ itself is a boring number.

Time complexity is $O(\log R)$ if we exclude the precalculations.

Code (C++)
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;
ll five[20], pre[20];
int a[10] = {0, 0, 1, 1, 2, 2, 3, 3, 4, 4};
int b[10] = {0, 1, 1, 2, 2, 3, 3, 4, 4, 5};

class Solution {
ll count(ll x) {
string s = to_string(x);
int n = s.size();
ll ans = pre[n - 1];
for (int i = 1; i <= n; ++i) {
int c = s[i - 1] - '0';
if (c % 2 != i % 2) {
ans += five[n - i] * b[c];
break;
} else {
ans += five[n - i] * a[c];
if (i == n)
ans++;
}
}
return ans;
}

public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
ll L, R;
cin >> L >> R;
cout << count(R) - count(L - 1) << endl;
}
};

int main() {
int t;
cin >> t;
five[0] = 1;
for (int i = 1; i < 20; ++i)
five[i] = five[i - 1] * 5;
pre[0] = 0;
for (int i = 1; i < 20; ++i)
pre[i] = pre[i - 1] + five[i];
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


## Problem C — Rugby

Apparently, we can solve for $x$ and $y$ independently.

First consider $y$. Since all the people will be in the same row, this becomes a classical problem in which we just need to take the median of $Y_i$ as the meeting place.

Then we consider $x$. It is obvious that once we determine the starting point $x$, the optimal movement is determined. The leftmost person will go to the leftmost cell, and the rest follow.

Thus we can solve this problem via ternary search. In order to prove the correctness, we need to prove that $dist(x)$ has only one extreme point, which is also its minimum point. (If we consider integer points, there might be two, but the two must be $x$ and $x+1$).

Obviously, when $x+N-1\leq\min(X_i)$, $dist(x)$ decreases with $x$. While when $x\geq\max(X_i)$, $dist(x)$ increases with $x$.

We then observe that, when we move the starting point from $x$ to $x+1$, there will be $k(x)$ people who will move $1$ less, and $N-k(x)$ people who will move $1$ more. So $dist(x+1)-dist(x)=N-2\cdot k(x)$. During the process where $x$ moves from $-\infty$ to $\infty$, $k(x)$ goes to $0$ from $N$, and will never increase. So $dist(x+1)-dist(x)$ will increase from $-N$ to $N$ and will never increase. So $dist(x)$ will take its extreme value (also its minimum) at the minimum $x$ that makes $dist(x+1)-dist(x)\geq0$.

The final time complexity is $O(N\log N+N\log MAX)$, in which $MAX$ is our search range.

Code (C++), Ternary search
#include <algorithm>
#include <climits>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

class Solution {
public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
int N;
cin >> N;
vector<int> X(N), Y(N);
for (int i = 0; i < N; ++i)
cin >> X[i] >> Y[i];
sort(Y.begin(), Y.end());
ll ylo = 0;
for (int yi : Y)
ylo += abs(yi - Y[N / 2]);
sort(X.begin(), X.end());
ll l = -2e9, r = 2e9;
ll xlo = LLONG_MAX;
auto dist = [&](ll start) {
ll ret = 0;
int idx = 0;
for (int xi : X) {
ret += abs(start + idx - xi);
idx++;
}
xlo = min(xlo, ret);
return ret;
};
while (l <= r) {
ll ml = l + (r - l) / 3, mr = r - (r - l) / 3;
ll dl = dist(ml), dr = dist(mr);
if (dl <= dr)
r = mr - 1;
if (dl >= dr)
l = ml + 1;
}
cout << ylo + xlo << endl;
}
};

int main() {
int t;
cin >> t;
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


We can also do binary search on $dist(x+1)-dist(x)$, or $k(x)$, and the solution is very similar.

Code (C++), Binary search
#include <algorithm>
#include <climits>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

class Solution {
public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
int N;
cin >> N;
vector<int> X(N), Y(N);
for (int i = 0; i < N; ++i)
cin >> X[i] >> Y[i];
sort(Y.begin(), Y.end());
ll ylo = 0;
for (int yi : Y)
ylo += abs(yi - Y[N / 2]);
sort(X.begin(), X.end());
ll l = -2e9, r = 2e9;
ll xlo = LLONG_MAX;
auto dist = [&](ll start) {
ll ret = 0;
int idx = 0;
for (int xi : X) {
ret += abs(start + idx - xi);
idx++;
}
xlo = min(xlo, ret);
return ret;
};
while (l <= r) {
ll mid = (l + r) / 2;
ll delta = dist(mid + 1) - dist(mid);
if (delta >= 0)
r = mid - 1;
else
l = mid + 1;
}
cout << ylo + xlo << endl;
}
};

int main() {
int t;
cin >> t;
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


Actually, we can also apply the median method on $x$. But we need to substitute $X_i$ with $X_i-i$ after the first sort, and then do a second sort. Detailed explanation can be seen in the official analysis.

Code (C++), Two-pass sorting for x
#include <algorithm>
#include <climits>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;

class Solution {
public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
int N;
cin >> N;
vector<int> X(N), Y(N);
for (int i = 0; i < N; ++i)
cin >> X[i] >> Y[i];
sort(Y.begin(), Y.end());
ll ylo = 0;
for (int yi : Y)
ylo += abs(yi - Y[N / 2]);
sort(X.begin(), X.end());
for (int i = 0; i < N; ++i)
X[i] -= i;
sort(X.begin(), X.end());
ll xlo = 0;
for (int xi : X)
xlo += abs(xi - X[N / 2]);
cout << ylo + xlo << endl;
}
};

int main() {
int t;
cin >> t;
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


## Problem D — Friends

If we build a graph with the strings, we will have too many edges.

So instead we can build a graph with different letters (in this problem we have $C=26$ letters). We will save this graph in an adjacent matrix.

The initial setting is $d[i][j]=\infty$ and $d[i][i]=0$. Then we enumerate on all $N$ strings. If two different letters $a$ and $b$ both occur in the same string $S$, we set $d[a][b]=d[b][a]=1$. The meaning is that, if we have a string $S'$ with $a$ and another string $S"$ with $b$, we can build a chain $S'\to S\to S"$ which has $1$ middle point.

Then we do Floyd-Warshall on this adjacent matrix. Now $d[i][j]$ means the minimum middle points that are needed to build a chain from $i$ to $j$.

For each query, we enumerate on letters in $S[X_i]$ and $S[Y_i]$, and the final answer will be

$\min_{p\in S[X_i],q\in S[Y_i]}d[p][q] + 2$

If the answer is $\infty$, we just output $-1$.

The total time complexity is $O((N+Q)L^2+C^3)$, in which $C=26$ is the size of the alphabet.

Code (C++)
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
const int INF = 0x3f3f3f3f;

class Solution {
public:
void solve(int case_num) {
cout << "Case #" << case_num << ": ";
int N, Q;
cin >> N >> Q;
vector<string> S(N + 1);
for (int i = 1; i <= N; ++i)
cin >> S[i];
vector<vector<int>> d(26, vector<int>(26, INF));
for (string s : S)
for (char c1 : s)
for (char c2 : s)
if (c1 != c2)
d[c1 - 'A'][c2 - 'A'] = 1;
for (int k = 0; k < 26; ++k)
for (int i = 0; i < 26; ++i) {
if (i == k)
continue;
for (int j = 0; j < 26; ++j) {
if (j == i || j == k)
continue;
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}
}

for (int i = 1; i <= Q; ++i) {
int X, Y;
cin >> X >> Y;
int ans = INF;
bool found = false;
for (char c1 : S[X]) {
for (char c2 : S[Y]) {
if (c1 == c2) {
cout << 2 << " ";
found = true;
break;
}
ans = min(ans, d[c1 - 'A'][c2 - 'A'] + 2);
}
if (found)
break;
}
if (!found)
cout << (ans == INF ? -1 : ans) << " ";
}
cout << endl;
}
};

int main() {
int t;
cin >> t;
for (int i = 1; i <= t; ++i) {
Solution solution = Solution();
solution.solve(i);
}
}


• +13

 » 4 months ago, # |   0 Auto comment: topic has been updated by lucifer1004 (previous revision, new revision, compare).
 » 4 months ago, # |   0 Your editorial for D is nice, idk why I didnt think like this! Can you tell where I was getting error in C, i used basic terneray search like in this editorial for X: Code#include using namespace std; void fastio() { cin.tie(nullptr); cin.sync_with_stdio(false); } using LL = long long; using LD = long double; const LL MOD = 998244353; const LL INF = 1e9; const LL N = 3e5+1; #define S second #define F first #define vt vector LL f (LL x, LL n, vt> &pt) { // put on x, x+1, ... x+n-1 LL tot = 0; for (LL i = 0; i < n; ++i) { tot += abs(x+i - pt[i].F); } return tot; } int main() { fastio(); LL T; cin >> T; for (LL _t = 1; _t <= T; ++_t) { LL n; cin >> n; vt> pt(n); for (LL i = 0; i < n; ++i) { cin >> pt[i].F >> pt[i].S; } for (auto &e : pt) { swap(e.F, e.S); } sort(pt.begin(), pt.end()); LL ymed, ycost = 0; if (n&1) { ymed = pt[n/2].F; } else { ymed = (pt[n/2].F+pt[n/2-1].F)/2; } for (auto &e : pt) { swap(e.F, e.S); ycost += abs(e.S - ymed); } sort(pt.begin(), pt.end()); LL p0 = -2e9, p3 = 2e9, p1 = p0 + (p3-p0)/3, p2 = p3 - (p3-p0)/3; LL xmed = INF; while (p0+3 < p3) { LL xstep1 = f(p1, n, pt), xstep2 = f(p2, n, pt); if (xstep1 >= xstep2) { p0 = p1; } else { p3 = p2; } p1 = p0 + (p3-p0)/3; p2 = p3 - (p3-p0)/3; } LL xstep = INF; for (LL i = p0; i <= p3; ++i) { xstep = min(xstep, f(i, n, pt)); } // print print: cout << "Case #" << _t << ": " << xstep+ycost << "\n"; } } 
•  » » 4 months ago, # ^ |   0 Your INF is too small for the large test set.
•  » » » 4 months ago, # ^ |   0 oh no! thanks!
 » 4 months ago, # | ← Rev. 2 →   0 I guess I'm doing the same thing in problem C as everyone and as told in Analysis (The Median Thing). But this doesn't work. Please help! Thanks Code public static void main(String[] args) throws IOException { // TODO Auto-generated method stub int t = scn.nextInt(); int x = 1; while (t > 0) { int n = scn.nextInt(); Point[] arr = new Point[n]; for (int i = 0; i < n; ++i) { arr[i] = new Point(scn.nextInt(), scn.nextInt(), i); } System.out.println("Case #" + x + ": " + solve(n, arr)); t--; x++; } } public static long solve(int n, Point[] arr) { Arrays.sort(arr, new yCom()); int yMed = arr[n / 2].y; long ans = 0; Arrays.sort(arr, new xCom()); int xMed = arr[n / 2].x; int i=n/2; while(i>=0) { int x=arr[i].x; int y=arr[i].y; ans+=Math.abs(yMed-y); ans+=Math.abs(xMed-x); xMed--; i--; } xMed=arr[n/2].x+1; i=n/2+1; while(i { @Override public int compare(Point arg0, Point arg1) { // TODO Auto-generated method stub return arg0.x - arg1.x; } } public static class yCom implements Comparator { @Override public int compare(Point arg0, Point arg1) { // TODO Auto-generated method stub return arg0.y - arg1.y; } } public static class Point { int x; int y; int id; Point(int x, int y, int id) { this.x = x; this.y = y; this.id = id; } } static Reader scn = new Reader(); static OutputStream out = new BufferedOutputStream(System.out); 
•  » » 4 months ago, # ^ |   0 For $X$, you should use $X_i-i$ to substitute $X_i$ after sorting, sort again, and then do the median trick.
•  » » » 4 months ago, # ^ |   0 I saw in the official editorial that for problem C, x-coordinates, we first transform them into x[i] — i, and then instead of now having to find the line, we just need to converge all the x — coordinates to the median of this transformed array. Can you please help me with this thing that what is the intuition and how does this manage to solve the thing that they would be in a single line after this?
 » 4 months ago, # |   0 Auto comment: topic has been updated by lucifer1004 (previous revision, new revision, compare).
 » 6 weeks ago, # |   0 Thank you so much for Question B...It is really really helpful |:D|
 » 5 weeks ago, # | ← Rev. 7 →   0 Problem B, Why is a array different from b array? I don't understand. In both cases we have to count the valid digits that are smaller than X[i]. Could you explain why they are different please?EDITED:I think I got it:so b array is for when the current digit isn't valid. So let's say 3 is our digit and it should be on even position. That means there are 2 and 0 digits valid smaller than 3. Which is 2 now for a array is when the current digit is valid.Again 3 should be odd. It is. So smaller valid digits than 3 is only 1. which counts to 1. That's why they are different right ?I still don't understand how someone can come up with such an algorithm so fast in a competition. I think it takes months, years of practice to be so bright to come up with this.