Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct dancer {
int g;
int p;
int t;
int id;
};
int n, xs, ys;
map<int, vector<dancer>> groups;
map<int, pair<int, int>> finalPos;
bool s1(dancer a, dancer b) {
if (a.g != b.g) return a.g > b.g;
if (a.g == 1)
return a.p < b.p;
else
return a.p > b.p;
}
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 10;
void file(string s) {
freopen((s + ".in").c_str(), "r", stdin),
freopen((s + ".out").c_str(), "w", stdout);
}
long long read() {
long long f = 1, a = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f =... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
/**
* @author Don Li
*/
public class RooterSong {
int N = (int) 1e5;
int n, w, h;
int[... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.*;
import java.util.*;
public class B {
/*
3 5 5
2 1 2
2 2 3
2 3 4
*/
static InputStream is;
static int X,Y;
public static void main(String[] args) throws IOException {
is = System.in;
int n = ni();
X = ni();
Y = ni();
TreeMap<Integer,ArrayList<d>> tm = new TreeMap<>();
d[] ds ... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | var l1 = readline().split(' ');
var n = +l1[0];
var w = +l1[1];
var h = +l1[2];
var src = new Array(n);
var dest = new Array(n);
for(var i = 0; i < n; i++) {
var line = readline().split(' ');
var p = +line[1], t = +line[2];
var x1, y1, x2, y2;
if(line[0] == '1') { // vertical
x1 = p; y1 = -t;
... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const long long MAX = 1000005;
long long n;
long long g[MAX], p[MAX], t[MAX], b[MAX];
basic_string<long long> a[2][2 * MAX];
long long w, h;
bool cmp(long long i, long long j) { return p[i] < p[j]; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int n, h, w, g[N], p[N], t[N], ansx[N], ansy[N];
vector<int> v[3 * N];
int main() {
scanf("%d%d%d", &n, &w, &h);
for (int i = 0; i < n; i++) {
scanf("%d%d%d", g + i, p + i, t + i);
v[p[i] - t[i] + N].push_back(i);
}
fill(ansx, ansx + N... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, w, h;
struct ed1 {
int x, y, clas, id, ANSx, ANSy;
} p[100099];
struct ed2 {
int x, y, clas;
} ans[100099];
bool cmp1(const ed1 &A, const ed1 &B) {
if (A.clas != B.clas) return A.clas < B.clas;
return A.x < B.x;
}
bool cmp2(const ed2 &A, const ed2 &B) {
if ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
public class Codeforces849D {
static class DancerComparator implements Comparator<int[]> {
public int compare(int[] o1, int[] o2) {
int l1 = o1[1] - o1[2];
int l2 = o2[1] - o2... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int sum[maxn], g[maxn], p[maxn], t[maxn], ans[maxn][2];
vector<int> type[maxn << 1];
struct node {
int x, y, id;
};
struct node tmpans[maxn];
struct node tmpst[maxn];
bool cmp(node t1, node t2) { return t1.y == t2.y ? t1.x > t2.x : t1.y < t2.y; ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
vector<int> pt[2][maxn + maxn];
int g[maxn], p[maxn], t[maxn], anid[maxn];
pair<int, int> ans[maxn];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, w, h;
cin >> n >> w >> h;
for (int i = 0; i < n; ++i) {
cin >> g[i... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > arr1[200100][2];
int arr[100009][3];
int ans[100009][2];
int main() {
int n, w, h;
cin >> n >> w >> h;
int a, b, c;
for (int f = 0; f < n; f++) {
cin >> a >> b >> c;
arr1[c - b + 100000][a - 1].push_back({b, f});
arr[f][0] = a;
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
using namespace std;
const int maxn = 1e5 + ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <class A, class B>
A cvt(B x) {
stringstream ss;
ss << x;
A y;
ss >> y;
return y;
}
int n, w, h;
int main() {
scanf("%d %d %d", &n, &w, &h);
map<int, vector<pair<int, int> > > diag, exits;
for (int i = 0; i < (int)(n); i++) {
int g, p, t;
sc... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
if (x < y) x = y;
}
inline int readChar();
template <class T = int>
inline T readInt();
template <class T>
inl... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
typedef pair<set<pii>, set<pii> > line;
map<int, line> lines;
int n, w, h;
void insert_dancer(int type, int pos, int t, int i) {
int line_number = pos - t;
if (type == 1)
lines[line_number].first.insert(pii(pos, i));
else if (type == 2)... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 2e5 + 10;
map<int, vector<tuple<int, int, int>>> mc[3];
pair<int, int> ans[N];
vector<tuple<int, int, int>> check(const vector<tuple<int, int, int>>& vp1,
const vector<tuple<int, int, int>>& vp2... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.FilterInputStream;
import java.io.BufferedInputStream;
import java.io.InputStream;
/**
* @author khokharnikunj8
*/
public class Main {
public static void main(... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
vector<int> arr[N][2];
vector<pair<int, int>> mp[2 * N];
vector<array<int, 3>> v;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n, w, h;
cin >> n >> w >> h;
for (int i = 0; i < n; i++) {
int g, p, t;... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const long long INF = 1e18;
const int irand(int lo, int hi) {
return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long lrand(long long lo, long long hi) {
return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
te... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.*;
import java.util.*;
import java.math.BigInteger;
import java.util.Map.Entry;
import static java.lang.Math.*;
public class D extends PrintWriter {
void test(int[] a, int[] b) {
int n = a.length;
int m = b.length;
for (int t = 0; t <= n + m + 4; t++) {
for (in... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
static const int MAXN = 100004;
static const int MAXW = 100003;
static const int MAXT = 100002;
int n, w, h;
int g[MAXN], p[MAXN], t[MAXN];
std::vector<int> s[MAXW + MAXT];
int ans_x[MAXN], ans_y[MAXN];
int main() {
scanf("%d%d%d", &n, &w, &h);
for (int i = 0; i < n; ++i) {
scanf("%d%d%... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int read() {
int s = 0;
char c = getchar(), lc = '+';
while (c < '0' || '9' < c) lc = c, c = getchar();
while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
return lc == '-' ? -s : s;
}
void write(int x) {
if (x < 0) putchar('... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct person {
int id, pos;
person(int a, int b) : id(a), pos(b) {}
bool operator<(const person &o) const { return pos < o.pos; }
};
int main() {
unordered_map<int, vector<person>> a;
int n, w, h;
cin >> n >> w >> h;
for (int i = 0; i < n; i++) {
int g, p... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100210, A = 100100;
int n, w, h;
vector<int> ver[maxn], hor[maxn];
vector<int> vid[maxn], hid[maxn];
deque<int> mp[maxn << 1];
vector<int> mpy[maxn << 1];
pair<int, int> ans[maxn];
int main() {
cin >> n >> w >> h;
for (register int i = 1; i <= n; ++i) {... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > x(400000);
int main(void) {
int k;
int N;
int M;
scanf("%i %i %i", &k, &M, &N);
int a[k][3];
int i;
for (i = 0; i < k; i++) {
scanf("%i %i %i", &a[i][0], &a[i][1], &a[i][2]);
a[i][2] += 150000;
x[a[i][2] - a[i][1]].push_back(i)... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
bool comp(const pair<pair<long long, long long>, long long>& a,
const pair<pair<long long, long long>, long long>& b) {
if (a.first.first == b.first.first)
return a.first.second < b.first.second;
else
return a.first.first < b.first.first;
}
int main() ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
ifstream in("file.in");
ofstream out("file.out");
struct om {
int p, t, id;
} a[100005], b[100005];
int n, w, h, cnt1[200005], cnt2[200005], ans[100005], v[200005], k, m, first,
nibab[100005];
pair<int, int> fk[100005];
queue<int> q[200005];
bool comp(om a, om b) { re... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > f[3][200100];
pair<int, int> ans[100100], res[100100];
int g[100100], p[100100], t[100100];
int n, h, w;
void work(vector<pair<int, int> > &f, vector<pair<int, int> > &g) {
int n = f.size(), m = g.size();
for (int i = 0; i < n; ++i) {
if (m >... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 100;
const int maxm = 100000 + 10;
const int mod = (int)1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LINF = (1LL << 60);
const double PI = acos(-1.0);
struct node {
int x, y, id;
node() {}
bool operator<(const node &r) const { return ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int geta = 100000;
int p[100010], x[100010], y[100010];
vector<int> hoge[2][200010];
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
int n, w, h;
cin >> n >> w >> h;
for (int i = 0; i < n; i++) {
int g, t;
cin >> g >> p[i] >> t;
g--;
ho... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > vec1(312345), vec2(312346);
int g[123456], p[123456], t[123456];
int cons;
int h, w;
int computeup(int i) {
int val1, val;
vector<int>::iterator it;
val1 = t[i] - p[i] + cons;
it = upper_bound(vec1[val1].begin(), vec1[val1].end(), p[i]);
val =... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const double pi = 3.141592653589793;
vector<pair<int, int> > vec[111111][2];
map<int, int> mm;
int c;
int ara[111111][2];
int ans[111111];
pair<int, int> point[111111];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, w, h;
cin >> n >> w >> h;
int a, x... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | //package codeforces.constructive_algorithms.cf_848_B_rooster_song;
/**
* Created by nafee on 2/14/18.
*/
import java.io.*;
import java.util.*;
public class Main {
static void redirectIO() throws IOException
{
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream("... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100100;
struct kk {
int g, x, t, id;
int z;
kk() {}
kk(int gg, int xx, int tt, int idd) {
g = gg;
x = xx;
t = tt;
id = idd;
z = x - t;
}
};
bool menor(const kk &a, const kk &b) {
if (a.z == b.z) {
return a.x < b.x;
}
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int d = 150000;
const int N = 400000;
const int M = 102;
unsigned long long cs = 1;
struct T {
int g, p, t, i;
T() {}
bool operator<(const T &other) const {
return make_pair(p, make_pair(g, t)) <
make_pair(other.p, make_pair(other.g, other.t));
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, w, h;
struct rec {
int g, p, t, id;
} x[111116], y[111116];
bool cmp_x(rec x, rec y) {
if (x.p - x.t != y.p - y.t) return x.p - x.t < y.p - y.t;
return (2 * x.g - 3) * x.p < (2 * y.g - 3) * y.p;
}
bool cmp_y(rec x, rec y) { return x.id < y.id; }
int read() {
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7, M = 2e6;
const long long mod = 1e9 + 7;
inline int read() {
int ret = 0;
char ch = getchar();
bool f = 1;
for (; !isdigit(ch); ch = getchar()) f ^= !(ch ^ '-');
for (; isdigit(ch); ch = getchar()) ret = (ret << 1) + (ret << 3) + ch - 48;
r... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct point {
int x, y, x_end, y_end, id;
point() {}
point(int a, int b, int c, int d, int e) {
x = a;
y = b;
x_end = c;
y_end = d;
id = e;
}
} number[100005], temp[100005];
void cp(point &a, point &b) { a = point(b.x, b.y, b.x_end, b.y_end, b.i... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | /* Author: Ronak Agarwal */
import java.io.* ; import java.util.* ; import java.math.* ;
import static java.lang.Math.min ;
import static java.lang.Math.max ;
import static java.lang.Math.abs ;
/* Thread is created here to increase the stack size of the java code so that recursive dfs can be performed */
public class C... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double g = 10.0, eps = 1e-7;
const int N = 200000 + 10, maxn = 60000 + 10, inf = 0x3f3f3f;
int gg[N], p[N], t[N], x[N];
vector<int> vx[N], vy[N];
pair<int, int> ans[N];
bool comp(int x, int y) { return p[x] < p[y... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | n,w,h = map(int,input().split())
D = []
original = []
for i in range(n):
g,p,t = map(int,input().split())
a = p-t
p = p if g == 1 else -p
original.append(())
D.append((a,p,i))
D.sort()
from bisect import bisect
res = [None]*n
i = 0
while i < len(D):
a = D[i][0]
j = bisec... | PYTHON3 |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, w, h;
vector<int> ansx(100000), ansy(100000), g(100000), p(100000), t(100000), x, y;
int main() {
map<int, vector<int>> s;
x.clear();
y.clear();
s.clear();
cin >> n >> w >> h;
for (int i = 0; i < n; i++) {
cin >> g[i] >> p[i] >> t[i];
s[p[i] - t[i... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > vec[2][200005];
pair<int, int> ans[200005];
int main() {
int i, j, n, w, h, opt, p, t, sz1, sz2;
cin >> n >> w >> h;
for (i = 1; i <= n; i++) {
cin >> opt >> p >> t;
vec[opt - 1][p - t + 100000].push_back(make_pair(p, i));
}
for (i ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.tie(0)->sync_with_stdio(0);
cin.exceptions(ios::badbit | ios::failbit);
int n, w, h;
cin >> n >> w >> h;
map<int, pair<vector<pair<int, int>>, vector<pair<int, int>>>> a;
for (auto i = 0; i < n; ++i) {
int g, p, t;
cin >> g >> p >> t;
... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.util.Map.Entry;
import java.io.BufferedReader... | JAVA |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, w, h;
int zsuv = 100001;
int g[100000], p[100000], t[100000];
vector<pair<pair<int, int>, int> > groups[250000];
vector<pair<int, int> > finish[250000];
vector<int> ishky[250000];
pair<int, int> res[100000];
int main() {
scanf("%d %d %d", &n, &w, &h);
for (int i ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int N, W, H, g[100009], p[100009], t[100009], ansx[100009], ansy[100009];
vector<int> v[200009];
int main() {
cin >> N >> W >> H;
for (int i = 0; i < N; i++) {
cin >> g[i] >> p[i] >> t[i];
v[p[i] - t[i] + 100000].push_back(i);
}
for (int i = 0; i <= 200000; ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
void in(int& a);
void in(long long& a);
struct no {
int x, id;
no(int a, int b) {
x = a;
id = b;
}
bool operator<(const no& b) const { return x < b.x; }
};
int n, m, w, h, t, tt, x, y, an[100010], as[100010];
vector<no> p[100010 << 2];
vector<int> q[100010 <... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, w, h;
struct node {
int ty, pos, t, id, to;
} po[100005] = {{0}};
bool cmp(node x, node y) {
if (x.pos - x.t < y.pos - y.t) return true;
if (x.pos - x.t > y.pos - y.t) return false;
if (x.ty > y.ty) return true;
if (x.ty < y.ty) return false;
if (x.pos > ... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
vector<pair<int, int> > v[2][2 * MAXN];
vector<int> tmp;
pair<int, int> res[MAXN];
int main() {
int a, b, c, d, e, f, k, q, m, n, w, h;
scanf("%d%d%d", &n, &w, &h);
for (a = 0; a < n; a++) {
scanf("%d%d%d", &b, &c, &d);
v[b - 1][MAXN +... | CPP |
848_B. Rooter's Song | Wherever the destination is, whoever we meet, let's render this song together.
On a Cartesian coordinate plane lies a rectangular stage of size w × h, represented by a rectangle with corners (0, 0), (w, 0), (w, h) and (0, h). It can be seen that no collisions will happen before one enters the stage.
On the sides of t... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct node {
int g, p, t, idx;
node(){};
node(int g_, int p_, int t_, int idx_) : g(g_), p(p_), t(t_), idx(idx_) {}
};
int n, W, H;
node A[100007];
vector<node> G[100007 << 1];
pair<int, int> ans[100007];
void work(const vector<node> &E) {
vector<int> Pv, Ph;
map... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int ans[55] = {0, 0, 3, 3, 3, 15, 3, 3,
29, 17, 255, 5, 83, 4095, 43, 3,
63, 15, 63, 63, 1023, 63, 3, 63,
27, 15, 4095, 39, 4095, ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
#pragma comment(linker, "/stack:64000000")
using namespace std;
template <typename first>
inline first abs(const first& a) {
return a < 0 ? -a : a;
}
template <typename first>
inline first sqr(const first& a) {
return a * a;
}
const long double PI = 3.1415926535897932384626433832795;
const ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int main() {
int k;
scanf("%d", &k);
if (k == 2)
printf("1 1\n");
else if (k == 3)
printf("1 0 1\n");
else if (k == 4)
printf("0 0 1 1\n");
else if (k == 5)
printf("0 0 1 0 1\n");
else if (k == 6)
printf("0 0 0 0 1 1\n");
else if (k == 7)
printf("0 1 0 1 ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const char *ans[51] = {
"", "", "1", "1", "1", "1 2 3",
"1 4 5", "2 3 4", "1 2 7", "3 5 6", "2 3 8", "1 8 10",
"1 2 10", "3 5 8", "1 11 12", "3 4 12", "10 12 15", "4 12 16",
"4 11 16", "3 9 10", ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long P[55] = {0,
0,
3,
5,
12,
20,
48,
106,
177,
318,
931,
1757,
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
const long long TAB[51] = {0LL,
0LL,
6LL,
10LL,
24LL,
40LL,
96LL,
212LL,
354LL,... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int a[] = {0, 0, 1, 1, 1, 2, 1, 1, 14, 8, 4, 2, 41, 13, 21, 1, 22,
4, 19, 19, 4, 2, 1, 16, 13, 4, 35, 19, 4, 2, 41, 4, 87, 41,
115, 2, 59, 31, 49, 8, 28, 4, 31, 44, 50, 13, 151, 16, 91, 56, 14},
n;
int main() {
scanf("%d", &n);
for (int i = ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
struct Matrix {
bool X[50][50];
};
int N, M;
long long P;
long long A[50];
bool C[50];
Matrix G[50];
Matrix operator*(Matrix A, Matrix B) {
Matrix C;
memset(C.X, 0, sizeof(C.X));
for (int k = 0; k < N; k++)
for (int i = 0; i < N; i++)
if (A.X[i][k])
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int MAX_K = 50;
vector<int> primes;
vector<long long> prime_divisors;
int PRIMES_COUNT, k;
int A[MAX_K][MAX_K], E[MAX_K][MAX_K], T[MAX_K][MAX_K], W[MAX_K][MAX_K];
inline void precalc() {
for (int i = 0; i < MAX_K; ++i) {
E[i][i] = 1;
}
const int SIEVE_SIZE =... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n;
const double eps = 1e-8;
int A[55][55], t[55][55];
void mul(int (&a)[55][55], int (&b)[55][55], int (&c)[55][55]) {
int t[55][55];
memset(t, 0, sizeof(A));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (a[i][j])
for (int k = 1; ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
using namespace rel_ops;
const double PI = acos((double)-1);
int ts, ts2, ts3, ts4;
int n, m;
long long gcd(long long x, long long y) {
long long t;
for (; y != 0;) {
t = x % y;
x = y;
y = t;
}
return x;
}
double sqr(double x) { return x * x; }
long long... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int a[] = {0, 0, 1, 1, 1, 2, 1, 1, 14, 8, 4, 2, 41, 13, 21, 1, 22,
4, 19, 19, 4, 2, 1, 16, 13, 4, 35, 19, 4, 2, 41, 4, 87, 41,
115, 2, 59, 31, 49, 8, 28, 4, 31, 44, 50, 13, 151, 16, 91, 56, 14},
n;
int main() {
scanf("%d", &n);
for (int i = ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int k, cnt, c[52];
long long n, divisor[52];
struct matrix {
int n, m, mat[52][52];
} result, shift, cur;
matrix operator*(const matrix &a, const matrix &b) {
matrix ret;
memset(ret.mat, 0, sizeof(ret.mat));
for (int i = 0; i < a.n; i++)
for (int j = 0; j < b.m; j++)
for (int ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 51;
int a[maxn], c[maxn];
int N;
struct Mat {
int mat[maxn][maxn];
Mat operator*(Mat M) {
Mat ret;
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j) {
ret.mat[i][j] = 0;
int now = 0;
for (int k = 0; k < N; ++k... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int P[5] = {2, 3, 7, 61, 24251};
long long N, K, X[100000];
int n, i, j, c[51], a[51], Tot;
struct Matrix {
int m[51][51];
Matrix operator*(const Matrix &A) const {
Matrix C;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
C.m[i][... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
int v = 0, f = 1;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = (c & 15);
while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
x = v * f;
}
inline void read(lo... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int ans[] = {0, 0, 1, 1, 1, 2, 1, 1, 14, 8, 4, 2, 41,
13, 21, 1, 22, 4, 19, 19, 4, 2, 1, 16, 13, 4,
35, 19, 4, 2, 41, 4, 87, 41, 115, 2, 59, 31, 49,
8, 28, 4, 31, 44, 50, 13, 151, 16, 91, 56, 14};
int main() {
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int i, K, Cnt;
long long N, p, d;
long long P[100005];
int A[55][55], B[55][55], C[55], E[55][55], D[55][55][55];
void Mul1() {
int i, j, k;
for (i = 1; i <= K; ++i)
for (j = 1; j <= K; ++j) E[i][j] = 0;
for (i = 1; i <= K; ++i)
for (j = 1; j <= K; ++j)
if (A[i][j])
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
long long int siguiente(long long int len, long long int n, long long int k) {
k &= n;
long long int x = 0;
while (k) {
x ^= k & 1;
k >>= 1;
}
n >>= 1;
n |= x << (len - 1);
return n;
}
const int tope = 10000;
bool testea(long long int len, long long in... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
typedef int arr[52][52];
arr a;
long long step;
int b[52], p[1000];
int n, cnt;
void matrixmul1(arr a, arr b, arr c) {
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++) c[i][j] = (c[i][j] + a[i][k] * b[k][j]) & 1;
}
void matr... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
const long long C[51] = {0,
0,
3ll,
5ll,
9ll,
23ll,
54ll,
83ll,
150ll,
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const long long DIVISORS[][100] = {
{},
{},
{},
{},
{
3,
5,
},
{},
{
3,
7,
},
{},
{
3,
5,
17,
},
{
7,
73,
},
{
3,
11,
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt, a[55][55], b[55][55], e[55][55], c[55], res[55];
typedef int arr[55][55], arr1[55];
arr tmp[55];
unsigned int tmpx[55], tmpy[55], tmpx1[55], tmpy1[55];
void mul1(arr1 &res, arr &a) {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
typedef int mat[55][55];
int c[55];
vector<long long> vlist;
mat g;
int n;
void mult(mat &res, mat &a, mat &b) {
static mat c;
memset(c, 0, sizeof c);
for (int i = (1); i <= (n); i++)
for (int j = (1); j <= (n); j++)
for (int k = (1); k <= (n); k++)
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt, a[55][55], b[55][55], e[55][55], c[55], res[55];
typedef int arr[55][55], arr1[55];
arr tmp[55];
void mul1(arr1 &res, arr &a) {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) c[i] = (a[i][j] * res[j] + c[i]) & 1;
m... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual soluti... | JAVA |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n;
const int MAX_N = 50;
long long answer[MAX_N + 1];
int main() {
answer[2] = 3LL;
answer[3] = 5LL;
answer[4] = 25LL;
answer[5] = 59LL;
answer[6] = 97LL;
answer[7] = 239LL;
answer[8] = 95LL;
answer[9] = 119LL;
answer[10] = 507LL;
answer[11] = 3487LL... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
void Get(int &T) {
char C;
bool F = 0;
for (; C = getchar(), C < '0' || C > '9';)
if (C == '-') F = 1;
for (T = C - '0'; C = getchar(), C >= '0' && C <= '9'; T = T * 10 + C - '0')
;
F && (T = -T);
}
int N;
void Init() { Get(N); }
int T[55][55];
void Mul(in... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | m={2: '11', 3: '101', 4: '1001', 5: '01001', 6: '100001', 7: '1000001', 8: '01110001', 9: '000100001', 10: '0010000001', 11: '01000000001', 12: '100101000001', 13: '1011000000001', 14: '10101000000001', 15: '100000000000001', 16: '0110100000000001', 17: '00100000000000001', 18: '000000100000000001', 19: '11001000000000... | PYTHON |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxk = 51;
struct matrix {
bool e[maxk][maxk];
int sz;
void print() {
for (int i = 0, _n = (sz); i < _n; i++) {
for (int j = 0, _n = (sz); j < _n; j++) {
printf("%d", e[i][j]);
}
printf("\n");
}
printf("\n");
}
};
matr... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, len = 0;
long long m = 1, a[2000];
bool A[60][60], B[60][60], f[60], g[60];
bool c[60];
void find_factor() {
for (int i = 1; i <= n; ++i) m <<= 1;
--m;
long long k = m;
for (long long i = 2; i * i <= k; ++i)
if (k % i == 0) {
a[++len] = m / i;
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
const int Ans[] = {0, 0, 1, 1, 1, 2, 1, 1, 14, 8, 4, 2, 41,
13, 21, 1, 22, 4, 19, 19, 4, 2, 1, 16, 13, 4,
35, 19, 4, 2, 41, 4, 87, 41, 115, 2, 59, 31, 49,
8, 28, 4, 31, 44, 50, 13, 151, 16, 91, 56, 14};
int n;
int ... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int N = 55;
struct matrix_t {
bitset<N> row[N], col[N];
int n;
void resize(int _n) {
n = _n;
for (int i = 1; i <= n; ++i) {
row[i].reset();
col[i].reset();
}
}
void output() {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt;
typedef bool arr[55][55], arr1[55];
arr a, b, e, tmp[55];
arr1 c, res;
void mul1(arr1 &res, arr &a) {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) c[i] = ((a[i][j] * res[j]) + c[i]) & 1;
memcpy(res, c, sizeof(res... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class SeqGen implements Runnable {
private void solve() throws IOException {
long[] answer = new long[51];
answer[2] = 3L;
answer... | JAVA |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt, a[55][55], b[55][55], e[55][55], c[55], res[55];
typedef int arr[55][55], arr1[55];
arr tmp[55];
void mul1(arr1 &res, arr &a) {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) c[i] = (a[i][j] * res[j] + c[i]) & 1;
m... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int a[51][6] = {
{0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {2, 1, 0, 0, 0, 0},
{3, 2, 0, 0, 0, 0}, {4, 3, 0, 0, 0, 0}, {5, 3, 0, 0, 0, 0},
{6, 5, 0, 0, 0, 0}, {7, 6, 0, 0, 0, 0}, {8, 6, 5, 4, 0, 0},
{9, 5, 0, 0, 0, 0}, {10, 7, 0, 0,... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 55;
struct Tmat {
int v[maxn][maxn];
};
Tmat ans, e, ee;
int c[maxn], res[maxn];
long long fac[maxn];
int n, tot;
long long p;
void mul(Tmat a, Tmat b, Tmat &c) {
register int i, j, k;
for (i = 0; i < n; ++i)
for (j = 0; j < n; ++j)
for (k =... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 55;
long long p[maxn];
int c[maxn];
int n, ps;
long long len;
void init() { scanf("%d", &n); }
void mkdivisor() {
len = (1LL << n) - 1;
ps = 0;
long long x = len;
for (long long i = 2; i * i <= x; i++)
if (x % i == 0) {
p[ps++] = len / i;
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | import static java.lang.Math.*;
import static java.util.Arrays.*;
import java.util.*;
public class B {
Scanner sc = new Scanner(System.in);
int[][] pp = {
{1},
{1},
{1},
{2},
{1},
{1},
{1,2,7},
{4},
... | JAVA |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
int K, n;
long long S, q[1005];
struct matrix {
bool a[55][55];
matrix(bool x = 0) {
memset(a, 0, sizeof(a));
for (int i = 0; i < K; i++) a[i][i] = x;
}
friend matrix operator*(const matrix& A, const matrix& B) {
matrix C;
for (int i = 0; i < K; i++)
for (int k = 0... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, cnt;
long long z[60];
bool c[60];
struct matrix {
bool z[52][52];
matrix() { memset(z, false, sizeof(z)); }
matrix operator*(const matrix &a) const {
matrix ans;
for (int b = 1; b <= n; b++)
for (int c = 1; c <= n; c++)
if (z[c][b])
... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt, a[55][55], b[55][55], e[55][55], c[55], res[55];
typedef int arr[55][55], arr1[55];
void mul1(arr1 &res, arr &a) {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) c[i] = (a[i][j] * res[j] + c[i]) & 1;
memcpy(res, c,... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > mul(vector<vector<long long> > a,
vector<vector<long long> > b) {
vector<vector<long long> > res(
((long long)a.size()), vector<long long>(((long long)b[0].size()), 0));
for (long long i = 0; i < ((long lon... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
vector<long long> a;
bool check(int n, const vector<int>& p, long long S) {
bool a[n][n], b[n][n], c[n], d[n];
memset(a, 0, sizeof a);
memset(b, 0, sizeof b);
memset(c, 0, sizeof c);
memset(d, 0, sizeof d);
for (int i = 0; i < n; i++) c[i] = 1;
for (int i = 0;... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int N = 60;
using ll = long long;
inline ll count(ll x) {
for (int i = 32; i; i >>= 1) x ^= x >> i;
return x & 1;
}
int n;
struct Arr {
ll a[N];
Arr operator+(const Arr &b) const {
Arr c;
for (int i = 0; i < n; i++) {
c.a[i] = 0;
for (int j... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int kMaxK = 50;
int k, c[kMaxK];
long long p;
vector<long long> fact;
struct Matrix {
int v[kMaxK][kMaxK];
void clear() { memset(v, 0, sizeof(v)); }
Matrix() { clear(); }
} e, f, a, b;
Matrix operator*(const Matrix &a, const Matrix &b) {
Matrix c;
for (int i... | CPP |
86_E. Long sequence | A sequence a0, a1, ... is called a recurrent binary sequence, if each term ai (i = 0, 1, ...) is equal to 0 or 1 and there exist coefficients <image> such that
an = c1·an - 1 + c2·an - 2 + ... + ck·an - k (mod 2), for all n ≥ k. Assume that not all of ci are zeros.
Note that such a sequence can be uniquely recovere... | 2 | 11 | #include <bits/stdc++.h>
long long S, d[55 * 55];
int n, cnt, a[55][55], b[55][55], e[55][55], c[55], res[55];
void mul1() {
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) c[i] = (a[i][j] * res[j] + c[i]) & 1;
memcpy(res, c, sizeof(res));
}
void mul() {
memset(b, 0, sizeof(... | CPP |
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