prompt string | response string |
|---|---|
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | def checkKey(dict, key):
if key in dict:
return True
return False
# def helper(s):
# l=len(s)
# if (l==1):
# l=[]
# l.append(s)
# return l
# ch=s[0]
# recresult=helper(s[1:])
# myresult=[]
# myresult.append(ch)
# for st in recresult:
# myresult.append(st)
# ts=ch+st
# myresult.append(ts)
# ret... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, second, first;
cin >> n >> m >> second >> first;
int t[100001][3];
for (int i = 0; i < m; i++) {
cin >> t[i][0] >> t[i][1] >> t[i][2];
}
int step = 1, cur = second, i = 0;
string res;
while (cur != first) {
if (step == t[i][0])... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
int a[100006][3];
int main() {
int n, m, s, f, dir, i;
scanf("%d%d%d%d", &n, &m, &s, &f);
if (s < f)
dir = 1;
else
dir = -1;
for (i = 0; i < m; i++) {
scanf("%d%d%d", &a[i][0], &a[i][1], &a[i][2]);
}
i = 0;
int t = 1;
while (f != s) {
if (a[i][0] == t) {
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
map<int, pair<int, int> > ma;
int h = -1;
for (int i = 0; i < m; i++) {
int a, b, c;
cin >> a >> b >> c;
h = max(h, a);
ma[a] = {b, c};
}
for (int i = 1; i <= h; i... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class P342B {
public static void main(String[] args) {
InputStream inputStream = System.in;
Ou... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.sync_with_stdio(false);
int n, m, s, f, ti, li, ri;
cin >> n >> m >> s >> f;
vector<pair<int, pair<int, int>>> v;
for (int i = 0; i < m; ++i) {
cin >> ti >> li >> ri;
v.push_back(make_pair(ti, make_pair(li, ri)));
}
int curt = 1, vptr ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... |
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main{
public static void main(String[]args)throws IOException{
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
String x=bf.readLine();
String[]xa=x.split(" ");
int n=Integer.parseI... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f;
int t, l, r;
int tt, now;
void deal() {
if (f > s) {
now++;
printf("R");
} else {
now--;
printf("L");
}
}
void work() {
tt = 0;
now = s;
for (int i = 0; i < m; i++) {
scanf("%d %d %d", &t, &l, &r);
if (now == f) continue;
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int on_bit(int x, int pos) {
x |= (1 << pos);
return x;
}
int off_bit(int x, int pos) {
x &= ~(1 << pos);
return x;
}
bool is_on_bit(int x, int pos) { return ((x & (1 << pos)) != 0); }
int flip_bit(int x, int pos) {
x ^= (1 << pos);
return x;
}
int lsb(int x) { ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, add = 0, rund = 1;
cin >> n >> m >> s >> f;
if (s < f)
add = 1;
else
add = -1;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
while (rund < t) {
if (s == f) break;
if (add == 1)
c... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int spies, moments, from, to;
scanf("%d %d", &spies, &moments);
scanf("%d %d", &from, &to);
int sec = 0, current = from, time, left, right;
for (int i = 0; i < (int)(moments); i++) {
sec++;
scanf("%d %d %d", &time, &left, &right);
if (curr... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
struct look {
int t, l, r;
};
look a[100000];
int n, m, s, f;
int main() {
cin >> n >> m >> s >> f;
for (int i = 0; i < m; ++i) cin >> a[i].t >> a[i].l >> a[i].r;
int curLook = 0;
for (int t = 1; s != f; ++t) {
if (curLook < m && a[curLook].t == t) {
int... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.*;
import java.io.*;
import java.math.*;
public class Q1 {
static ArrayList<Integer> adj[],adj2[];
static int color[],cc;
static long mod=1000000007;
static TreeSet<Integer> ts;
static boolean b[],visited[],possible,ans1,ans2;
static Stack<Integer> s;
static int totalnodes,colored,minc;
stat... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.lang.Math;
import java.util.Arrays;
import java.util.Comparator;
public class Main
{
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... |
import java.io.*;
import java.util.*;
public class IEEE {
public static void main(String[] args) throws IOException {
//Scanner sc=new Scanner(System.in);
BufferedReader r=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer ss=new StringTokenizer(r.readLine());
int n=Integer.pars... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Spies
{
public static void main(String[] args) throws IOException
{
InputStreamReader a = new InputStreamReader(System.in);
BufferedReader buf = new BufferedReader(a);
String[] temp = buf.readLine().split... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
void FAST() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int Rselect(vector<int>&, int, int, int);
int partition(vector<int>&, int, int);
void scanc(vector<char>& v, long long n) {
for (int i = 0; i < n; i++) {
char num;
cin >> num;
v.push_back(... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int n, m, st, ed, dir, ok;
int te[100005], lf[100005], rt[100005];
int main() {
scanf("%d%d%d%d", &n, &m, &st, &ed);
for (int i = (0); i < (m); i++) scanf("%d%d%d", &te[i], &lf[i], &rt[i]);
for (int i = 1, j = 0; st != ed; i++) {
while (j < m && te[j] < i) ++j;
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<bool> v;
vector<int> ans;
int n, t, s, f, a, b, c, curs = 1, curr;
scanf("%d%d%d%d", &n, &t, &s, &f);
curr = s;
if (s < f) {
while (t--) {
scanf("%d%d%d", &a, &b, &c);
if (curr == f) continue;
if (curs < a) {
if ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int M = 1000000000 + 7;
const double eps = 1e-10;
struct node {
int t, l, r;
} num[100005];
int cmp(node a, node b) { return a.t < b.t; }
char ans[1000005];
int main() {
int n, m, s, f;
while (~scanf("%d %d %d %d", &n, &m, &s, &f)) {
for (int i = 1; i <= m; ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
int n, m, s, f;
int t, x, y;
int step;
int main() {
scanf("%d %d %d %d", &n, &m, &s, &f);
step = 1;
int temp;
for (int i = 0; i < m; i++) {
temp = s;
scanf("%d %d %d", &t, &x, &y);
while (step != t) {
step++;
if (temp == f) {
printf("\n");
return ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
public class XeniaSpecies implements Runnable {
static class State {
int t, a, b;
public State(int t, int a, int b) {
this.t = t;
this.a... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
long long co;
int arp[100004];
int non;
int main(int argc, char const* argv[]) {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m, s, f;
cin >> n >> m >> s >> f;
long long a[m][3];
for (int i = 0; i < m; i++) {
cin >> a[i][0];
cin >> a[i][... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int n, m, s, f;
int t[MAXN], l[MAXN], r[MAXN];
bool check(long long p, long long l, long long r) {
return (p - l) * (p - r) <= 0;
}
int main(int argc, char const *argv[]) {
scanf("%d%d%d%d", &n, &m, &s, &f);
for (int i = 0; i < m; ++i) scanf("... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | n,m,s,f = map(int,raw_input().split())
flag = 1
direct = {
1 : "R",
-1 : "L"
}
if s > f :
flag = -1
res = ""
pre_t = 0
for i in range(m) :
t,l,r = map(int,raw_input().split())
if s == f:
break
if t != (pre_t + 1) :
num = 0
if ( t - pre_t - 1) > abs(f-s):
num ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) {
FastReader in = new FastReader();
int n = in.nextInt();
int m = in.nextInt();
int s = in.nextInt();
int f = in.nextInt();
int cs = 1;
for(int i = 0; i < m; i++) {
int t = in.nextInt();
int l = in.ne... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.util.*;
import java.math.*;
public class B implements Runnable {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer st;
static Random rnd;
class Segment {
int time, l, r;
public Segment(int time, int l, int r) {
this.time = time;
this.l = l;
this.... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
struct steps {
int l, r, step;
};
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
vector<steps> v(m);
char left = 'L', right = 'R', none = 'X';
for (int i = 0; i < m; i++) cin >> v[i].step >> v[i].l >> v[i].r;
int i, j, st = 1;
if (s < f) {
for (i ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
long long int inf = 1e18;
long long int p = 998244353;
long long int phi(long long int n) {
long long int result = n;
for (long long int i = 2; i * i <= n; i++) {
if (n % i == 0) {
while (n % i == 0) n /= i;
result -= result / i;
}
}
if (n > 1) r... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.ArrayBlockingQueue;
public class B
{
String line;
StringTokenizer inputParser;
BufferedReader is;
FileInputStream fstream;
DataInputStream in;
String FInput="";
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
* Created with IntelliJ IDEA.
* User: shiwangi
* Date: 9/7/13
* Time: 2:11 PM
* To change this template use File | Settings | File Templates.
*/
public class srm1992 {
public static void main(String[] args) thro... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
long long n, m, s, f, t;
map<long long, long long> r, l;
int main() {
cin >> n >> m >> s >> f;
while (m--) {
long long x, y;
cin >> t >> x >> y;
l[t] = x, r[t] = y;
}
t = 1;
while (s != f) {
if (s < f) {
if (!r[t] ||
!((s >= l[t] &&... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, i, x, l, r, t;
string st;
map<int, vector<int> > mp;
cin >> n >> m >> s >> f;
for (i = 1; i <= m; i++) {
cin >> x >> l >> r;
mp[x].push_back(l);
mp[x].push_back(r);
}
int pos;
if (s < f) {
for (t = 1, pos = s; pos... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int t[100002], l[100002], r[100002];
int now = 1;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
for (int i = 0; i < m; i++) cin >> t[i] >> l[i] >> r[i];
int i = 0;
int h = -1;
if (s < f) h = 1;
while (s != f) {
if (t[i] != now || (t[i] == now && !(... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.*;
import java.io.*;
public class a {
static long mod = 1000000007;
public static void main(String[] args) throws IOException {
// Scanner input = new Scanner(new File("input.txt"));
// PrintWriter out = new PrintWriter(new File("output.txt"));
input.init(System.in);
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | def main():
n, m, s, f = map(int, raw_input().split())
d = dict()
for _ in xrange(m):
t, l, r = map(int, raw_input().split())
d[t] = [l, r]
ans = []
t = 1
mv = 'L' if s > f else 'R'
dx = -1 if s > f else 1
while s != f:
if t in d and (d[t][0] <= s <= d[t][1] or d[... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, st, en;
cin >> n >> m >> st >> en;
string ans = "";
int tt = 0;
while (m--) {
int t, l, r;
cin >> t >> l >> r;
if (t != tt + 1) {
while (t != tt + 1) {
tt++;
if (en > st) {
ans += 'R';
st... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | def mi():
return map(int, input().split())
'''
3 5 1 3
1 1 2
2 2 3
3 3 3
4 1 1
10 1 3
'''
n,m,s,f = mi()
t = [0]*m
l = [0]*m
r = [0]*m
for i in range(m):
t[i],l[i],r[i] = mi()
curp = s
curt = 1
i = 0
a = ''
if f>s:
a = (f-s)*'R'
else:
a = (s-f)*'L'
while i<m and curp!=f:
if t[i]=... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, t, l, r, dir, curStep = 1, ctr = 0;
char arr[30];
arr[5 + 1] = 'R', arr[5 - 1] = 'L';
cin >> n >> m >> s >> f;
if (f > s)
dir = 1;
else
dir = -1;
while (s != f) {
if (ctr != m)
cin >> t >> l >> r, ++ctr;
else
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100000;
const int MAX_M = 100000;
const int MAX_L = MAX_N + MAX_M;
char v[MAX_L + 4];
int main() {
int n, m, s, f;
scanf("%d%d%d%d", &n, &m, &s, &f);
int d;
char c;
if (s < f)
d = 1, c = 'R';
else
d = -1, c = 'L';
int t = 0;
while (... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class CF {
public static void main(String[] args) throws Exception {
FastScanner sc = new FastScanner(System.in);
int spies = sc.nextInt();
int checks = sc.nextInt();
int sta... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, s, f, t, l, r;
cin >> n >> m >> s >> f;
map<long long, long long> L, R;
for (int i = 1; i <= m; ++i) {
cin >> t >> l >> r;
L[t] = l;
R[t] = r;
}
string res = "";
for (int i = 1;; ++i) {
l = L[i];
r = R[i];
i... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
struct node {
int t, l, r;
};
node a[100005];
char ans[300005];
bool cmp(node a, node b) { return a.t < b.t; }
int main() {
int n, m, s, f;
int i, j;
scanf("%d%d%d%d", &n, &m, &s, &f);
for (i = 1; i <= m; i++) {
scanf("%d%d%d", &a[i].t, &a[i].l, &a[i].r);
}
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | n,m,s,f=map(int,input().split())
d={}
final=0
for i in range(m):
a=list(map(int,input().split()))
d[a[0]]=[a[1],a[2]]
final=a[0]
z=""
if s<f:
i=0
while s!=f:
i+=1
if i not in d:
s+=1
z+="R"
elif (s not in range(d[i][0],d[i][1]+1)) and (s+1 not in range... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
struct reverse {
bool operator()(const int &left, const int &right) { return (right < left); }
};
struct custom {
bool operator()(const pair<int, int> &left, const pair<int, int> &right) {
return (left.first > right.first);
}
};
void solve() {
int n, m, s, f;
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int nm = 100002;
int n, m, s, f;
int t[nm], x[nm], y[nm];
void nhap() {
scanf("%d%d%d%d", &n, &m, &s, &f);
int i;
for (i = 1; i <= m; ++i) scanf("%d%d%d", &t[i], &x[i], &y[i]);
}
bool kt(int i, int j) { return (i >= x[j] && i <= y[j]); }
void xuli() {
int i = ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
const int MAX_SIZE = 100100;
int f[MAX_SIZE], l[MAX_SIZE], r[MAX_SIZE];
int n, m;
int from, to;
int main() {
scanf("%d%d%d%d", &n, &m, &from, &to);
for (int i = 1; i <= m; i++) scanf("%d%d%d", f + i, l + i, r + i);
int dir = to - from;
int step = 1;
int x = 0;
f[0] = 0;
l[0] = r[0... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
int main() {
long long int n, m, s, f, j = 0;
cin >> n >> m >> s >> f;
vector<long long int> l, r, t;
for (long long int i = 0; i < m; i++) {
long long int te, le, re;
cin >> te >> le >> re;
t.push_back(te);
l.push_back(l... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100010;
int TT, NP;
int L, R, N, M, F, T;
int K[MAXN][3];
int main() {
cin >> N >> M >> F >> T;
memset(K, -1, sizeof(K));
for (int i = 1; i <= M; i++) cin >> K[i][0] >> K[i][1] >> K[i][2];
TT = 0;
NP = F;
int index = 1;
if (F < T)
while (N... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.Scanner;
public class XeniaAndSpies {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int n;
int m,s,f;
n = sc.nextInt();
m = sc.nextInt();
s = sc.nextInt();
f = sc.nextInt();
int[]... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
long long dp[100000 + 25];
struct node {
int iz, der;
long long tiempo;
node() {}
node(long long t, int i, int d) {
tiempo = t;
iz = i;
der = d;
}
};
struct otro {
int proct, pos;
long long tiemp;
otro() {}
otro(int ps, long long time, int timp... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arra... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | /*package com.fancita.codeforces;*/
import java.io.*;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
/**
* Created by ashutosh on 26/12/16.
*/
public class b199 extends FastIO {
public static void main(String[] args) {
int n = reader.readInt();
int ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, m, s, f;
cin >> n >> m >> s >> f;
long long int i, j;
long long int t[m], l[m], r[m];
long long int curr = s;
long long int step = 0;
for (i = 0; i < m; i++) {... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;
p... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int step = 1;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
if (s == f) return false;
while (step < t) {
if (f > s) {
s++;
cout << "R";
} else if (f < ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import sys
spies, turns, note, final = map(int, input().split())
if note < final:
next = 1
else:
next = -1
answer = []
def pass_note(current_turn, next_watch, left, right, note):
# print(f'\n{current_turn}), next watch: {next_watch}, left: {left}, right: {right} has note: {note}, goal: {final}')
if curr... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
scanf("%d", &n);
;
scanf("%d", &m);
;
scanf("%d", &s);
;
scanf("%d", &f);
;
vector<pair<int, pair<int, int> > > a(m);
for (int i = 0; i < m; i++) {
scanf("%d", &a[i].first);
;
scanf("%d", &a[i].second.first);
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long pos;
long long n, m, s, f;
cin >> n >> m >> s >> f;
map<long long, pair<long long, long long> > mp;
for (long long i = 0; i < m; i++) {
int x, y, z;
cin >> x >> y >> z;
mp.insert(make_pair(x, make_pair(y, z)));
}
if (s < f) {... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.NoSuchElementException;
import java.io.Writer;
import java.math.BigInteger;
import java.io.InputStream;
/**
*... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int s, f, n, m, l, r, dir, now, i, t, dif, k, h = 0, g;
cin >> n >> m >> s >> f;
now = s;
if (f > s)
dir = 1;
else
dir = -1;
char j;
if (dir == 1)
j = 'R';
else
j = 'L';
i = 1;
while (now != f && h < m) {
scanf("%d %d %d"... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.lang.Math;
public class Account {
public static void main (String[] args) {
Scanner sc= new Scanner(System.in) ;
int n= sc.nextInt() ;
i... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
long n, m, s, f, t, l, r, tAnt;
long pos;
char dirout;
int dir;
cin >> n >> m >> s >> f;
pos = s;
if (s > f) {
dir = -1;
dirout = 'L';
} else {
dir = 1;
dirout = 'R';
}
tAnt = 0;
for (int i = 0; i < m; i++) {
cin >> t >... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.Scanner;
public class B {
public static void main(String[] args) {
B solver = new B();
solver.solve();
}
private void solve() {
Scanner sc = new Scanner(System.in);
// sc = new Scanner("3 5 1 3\n" +
// "1 1 2\n" +
// "2 2 3\n" ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
void print_steps(int rtime, int steps, bool swap) {
if (swap) {
cout << string(steps, 'L');
cout << string(rtime - steps, 'X');
} else {
cout << string(steps, 'R');
cout << string(rtime - steps, 'X');
}
}
int main() {
int n, m, si, fi, s, f, t, ct = ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... |
if __name__=='__main__':
inp = input()
arr = inp.split(' ')
n = int(arr[0])
m = int(arr[1])
s = int(arr[2])
f = int(arr[3])
ans = ""
ch = 'L'
inc = -1
if s<f:
ch='R'
inc = 1
tm = 1
done = False
for i in range(m):
inp = input()
if done:... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
char str[10000001];
int k = 0;
int main() {
int n, m, s, f;
scanf("%d%d%d%d", &n, &m, &s, &f);
int ptr = s;
int tmp = 1;
for (int i = 0; i < m; i++) {
int t, l, r;
scanf("%d%d%d", &t, &l, &r);
if (ptr == f) continue;
while (tmp < t) {
if (s < f) {
ptr++;
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
int main() {
int q, w, e, r;
int flag = 0;
int a, b, sum, i, j;
int ans = 0;
int n, m, s, now;
while (~scanf("%d%d%d%d", &n, &m, &s, &e)) {
now = s;
i = 1;
while (m--) {
scanf("%d%d%d", &q, &a, &b);
while (i - 1 != q) {
if (now == e) break;
if... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | I=lambda:map(int, raw_input().split())
n, m, s, f = I()
d = {}
for _ in xrange(m):
t, l, r = I()
d[t] = (l, r)
W = 'R' if s < f else 'L'
move = 1 if s < f else -1
i = 1
ans = ''
while True:
if s == f:
break;
if i in d:
if (d[i][0] -1 <= s <= d[i][1] and W == 'R') or (d[i][0] <=... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
char ans[1000009];
int tot;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int t[100009], l[100009], r[100009];
for (int i = 1; i <= (m); i++) cin >> t[i] >> l[i] >> r[i];
tot = 0;
int p = 1;
while (1) {
++tot;
if (t[p] == tot) {
if (s >= ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.Scanner;
public class XeniaAndSpies {
public static void main(String[] args) throws Exception {
Scanner in = new Scanner(System.in);
int n = in.nextInt(), m = in.nextInt(), s = in.nextInt(), f = in
.nextInt();
char move = (s < f) ? 'R' : 'L';
int start = s, it = 0;
StringBuilder ans = n... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int t[100010], l[100010], r[100010];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int i, n, m, s, f, j;
cin >> n >> m >> s >> f;
for ((i) = 0; (i) < (int)(m); (i)++) {
cin >> t[i] >> l[i] >> r[i];
}
i = 0;
for (j = 1;; j++) {
if (s == f) {
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f, t[110010], l[110010], r[110010], nxt;
char c;
int main() {
cin >> n >> m >> s >> f;
for (int i = 1; i <= m; i++) cin >> t[i] >> l[i] >> r[i];
t[m + 1] = 2 * (int)1e9;
if (f < s)
c = 'L';
else
c = 'R';
int add = f < s ? -1 : 1;
for (int ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | n,m,s,f = map(int,raw_input().split())
flag = 1
if s > f :
flag = -1
res = ""
pre_t = 0
for i in range(m) :
t,l,r = map(int,raw_input().split())
if s == f:
break
if t != (pre_t + 1) :
num = 0
if ( t - pre_t - 1) > abs(f-s):
num = abs(f-s)
else :
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class Main{
static StringBuilder c;
static class pair{
int form,to;
public pair(int from,int to) {
this.form=from;this.to=to... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, s, f;
cin >> n >> m >> s >> f;
map<int, pair<int, int> > mop;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
mop[t] = {l, r};
}
string ans;
while (s... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
map<int, pair<int, int> > w;
int main() {
int n, m, s, f, temp;
cin >> n >> m >> s >> f;
for (int i = 0; i < m; i++) {
cin >> temp;
cin >> w[temp].first >> w[temp].second;
}
int pasos = 1, pos = s;
while (pos != f) {
if (s < f) {
while (w[pasos... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;
public class XeniaAndSpies {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
sc.nextInt();
int M = sc.nextInt();
int S = sc.nextInt();
int F = sc.nextInt... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... |
import java.util.HashMap;
import java.util.Scanner;
public class Problem1 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n, m, s, f;
n = scanner.nextInt();
m = scanner.nextInt();
s = scanner.nextInt();
f = scanner.nextInt();... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int inf = 0x7FFFFFFF;
const int maxn = 1000000;
string ans = "";
int n, m, s, f;
int t, l, r, w = 1;
bool ok = false;
void work() {
if (s == f) {
ok = true;
return;
}
if (s < f)
++s, ans += 'R';
else
--s, ans += 'L';
}
int main() {
cin >> n >... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | //package codeforces.train;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class ProblemB {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(new OutputStreamWriter(System.out));
long[] ttl = null;
int txl = 0;... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | n,m,s,f=map(int,input().split())
p=s
d=-1
c='L'
if s<f:
d=1
c='R'
t=1
ts={}
ans=""
for _ in range(m):
x,y,z=map(int,input().split())
ts[x]=(y,z)
while(p!=f):
if t in ts:
(l,r)=ts[t]
if l<=p<=r or l<=p+d<=r:
ans+='X'
else:
p+=d
ans+=c
el... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.HashMap;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelpe... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.Scanner;
import java.io.StreamTokenizer;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintStream;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution i... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
#pragma comment(linker, "/stack:225450978")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long Mod = 1000000007LL, INF = 1e9, LINF = 1e18;
const long double Pi = 3.141592653589793116, EPS = 1e-9,
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import sys
N, M, S, F = map(int, raw_input().split())
T = []
for i in xrange(M):
T.append(map(int, raw_input().split()))
T.append((0, 0, 0))
c = 1
p = 0
while S!=F:
NS = S
if S<F:
NS = S+1
else:
NS = S-1
if c==T[p][0] and (T[p][1]<=S<=T[p][2] or T[p][1]<=NS<=T[p][2]):
sys.stdout.write('X')
else:
if S<F... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class A {
public void solve() throws Exception {
int n = nextInt();
int m = nextInt();
int s = nextInt();
int f = nextInt(... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import javax.print.DocFlavor;
import javax.swing.plaf.basic.BasicInternalFrameTitlePane;
import java.io.*;
import java.math.BigInteger;
import java.nio.Buffer;
import java.sql.BatchUpdateException;
import java.util.*;
import java.util.stream.Stream;
import java.util.Vector;
import java.io.IOException;
import java.nio.f... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int32_t main() {
long long n, m, s, f, t, l, r;
cin >> n >> m >> s >> f;
long long t2 = 0;
for (long long i = 1; i <= m; i++) {
if (s == f) return 0;
cin >> t >> l >> r;
if (t2 + 1 < t) {
for (long long i = t2 + 1; i < t; i++) {
if (s == f)... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws IOException {
Solution sol = new Solution();
sol.run();
}
void out(String ans) {
System.out.println(ans);
System.exit(0);
}
void run() throws IOException {
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
int t, l, r, backT = 0;
cin >> n >> m >> s >> f;
int cur = s;
for (int i = 0; i < m; i++) {
cin >> t >> l >> r;
for (int j = t - backT; j > 0; j--) {
if (cur < f && (j > 1 || (r < cur || cur + 1 < l))) {
cur++;
... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f, t, l, r, t1, px;
char x;
int main() {
cin >> n >> m >> s >> f;
if (f < s)
x = 'L', px = -1;
else
x = 'R', px = 1;
for (int i = 1; i <= m; i++) {
if (s == f) break;
cin >> t >> l >> r;
t1 = t - t1 - 1;
for (int j = 1; j <= t1; ... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, l2 = 1;
string sj = "";
int t1 = 0, t2 = 0, l, r;
bool ch = true;
cin >> n >> m >> s >> f;
if (s > f) ch = false;
int e = 0;
for (int i = 0; i < m; i++) {
t1 = t2;
cin >> t2 >> l >> r;
e = t2 - t1 - 1;
while (e &&... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
public class Main{
public static void main(String []args){
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int m=in.nextInt();
int s=in.nextInt();
int f=in.nextInt();
int[] t=new i... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | n,m,s,f=map(int,raw_input().split())
ar=[]
for i in range(m):
a,b,c=map(int,raw_input().split())
ar.append([a,b,c])
ans=""
curpos=s
if s<f:
i=0
tm=1
while 1:
if curpos==f:
break
elif i>=m:
ans+="R"
curpos+=1
tm+=1
elif tm!=ar[i][0]:
ans+="R"
curpos+=1
tm+=1
elif (curpos<=ar[i][2] and c... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | import java.io.*;
import java.util.*;
public class XeniaAndSpies {
public static void main(String[] args) throws Exception {
BufferedReader k = new BufferedReader(new InputStreamReader(System.in));
// Scanner k = new Scanner(System.in);
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
S... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
static inline bool between(int x, int l, int r) { return x >= l && x <= r; }
int main() {
string actions;
int n, m, s, f;
cin >> n >> m >> s >> f;
map<int, pair<int, int> > stages;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
stages[t... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int inf = INT_MAX / 2;
int n, m, s, f;
struct node {
int l, r, t;
};
node a[maxn];
bool cmp(node a, node b) { return a.t < b.t; }
int main() {
while (scanf("%d%d%d%d", &n, &m, &s, &f) != EOF) {
for (int i = 0; i < m; i++) {
scan... |
Problem: Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.
Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can ... | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f;
int main() {
int i;
scanf("%d%d%d%d", &n, &m, &s, &f);
int now = 1;
int t, l, r;
int flag = s < f ? 1 : -1;
char step = 'L';
if (flag == 1) step = 'R';
for ((i) = (1); (i) <= (m); ++(i)) {
scanf("%d%d%d", &t, &l, &r);
while (s != f &&... |
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