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 |
|---|---|---|---|---|---|
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int t[100009], l[100009], r[100009];
int main(void) {
int i, n, m, s, f;
while (scanf("%d %d %d %d", &n, &m, &s, &f) == 4) {
for (i = 1; i <= m; i++) {
scanf("%d %d %d", &t[i], &l[i], &r[i]);
}
int start = 1;
for (i = 1; i <= m; i++, start++) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;
public class B {
static int[] u = new int[1010], a = new int[50];
publi... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
string ans = "";
cin >> n >> m >> s >> f;
int last = 0;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
if (s < f) {
for (int j = 0; j < t - last; j++) {
if (s == f) {
cout << ans << e... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.*;
import java.util.*;
public class Template implements Runnable {
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
void init() throws FileNotFoundException {
try {
in = new BufferedReader(new FileReader("input.txt"));
o... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
cin.sync_with_stdio(0);
int n, m, s, f;
cin >> n >> m >> s >> f;
char pass = (s > f) ? 'L' : 'R';
int t, l, r;
int prev = 0;
for (int i = 0; i < m; i++) {
cin >> t >> l >> r;
if (s == f) continue;
int frees = t - prev - 1;
for (int... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | maxn=100000+200;
ans=""
def prin():
print ans[1:]
opt=[[] for i in range(maxn)]
N,M,S,T=map(int, raw_input().split())
if S<T:
mv=1
chr='R'
else:
mv=-1
chr='L'
opt[0]=[0,1,N]
for i in range(M):
opt[i+1]=list(map(int, raw_input().split()));
opt[M+1]=[99999999,0,0]
for i in range(M+1):
t,l,r... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int i = s, t = 1;
string ans = "";
map<int, pair<int, int>> watching;
for (int i = 1; i <= m; i++) {
int t;
cin >> t;
int l, r;
cin >> l >> r;
watching[t] = {l, r};
}
while (i != f) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int t, l, r;
int d = f - s;
cin >> t >> l >> r;
for (int i = 1;; ++i) {
if (i > t) {
cin >> t >> l >> r;
}
if (d > 0) {
if (i != t || s > r || s + 1 < l) {
s++;
cout <<... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | n,m,s,f=list(map(int,input().split()))
if s>f:
a='L'
c=-1
elif s==f:
a='X'
c=0
else:
a='R'
c=1
l=[]
for i in range(m):
l.append(list(map(int,input().split())))
ans=''
if l[0][0]!=1:
d=l[0][0]-1
while s!=f and d>0:
s=s+c
ans=ans+a
d=d-1
for i in range(m):
... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | n, m, s, f = [int(x) for x in raw_input().strip().split()]
ans = []
prevt = 1
pos = s
movedir = 'R'
if f < s:
movedir = 'L'
togo = (f - pos if movedir == 'R' else pos - f)
done = False
for i in range(m):
t, l, r = [int(x) for x in raw_input().strip().split()]
if done:
continue
freemoves = t ... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class xs
{
public static void main (String[] args) throws Exception{
Scanner in = new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
StringBuilder sb=new StringBuilder();
int n=in.nextInt(),m=in.nextInt(),s=in.nextInt()... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import sys
n,m,s,f=map(int,sys.stdin.readline().split())
L=[]
R=[]
T=[]
for i in range(m):
t,l,r=map(int,sys.stdin.readline().split())
T.append(t)
L.append(l)
R.append(r)
if(f>s):
i=s
step=1
ind=0
Ans=""
while(i!=f):
if(ind>=m or T[ind]!=step):
Ans+="R"
... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int n, m, s, f;
cin >> n >> m >> s >> f;
long long int t[m], l[m], r[m];
for (long long i = 0; i < m; i++) {
cin >> t[i] >> l[i] >> r[i];
}
string str;
long long int po... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, m, s, f;
map<int, int> l;
map<int, int> r;
int main() {
cin >> n >> m >> s >> f;
for (int i = 0; i < m; i++) {
int x, y, t;
cin >> t >> x >> y;
l[t] = x;
r[t] = y;
}
for (int i = 1; s < f; i++)
if (s + 1 < l[i] |... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int arr[100009][3];
int main() {
int n, m, s, f, i;
vector<char> V;
scanf("%d%d%d%d", &n, &m, &s, &f);
for (i = 1; i <= m; i++) {
scanf("%d%d%d", &arr[i][0], &arr[i][1], &arr[i][2]);
}
int count = 1, ind = 1, curr = s;
if (s < f) {
while (true) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class ContestTemplate {
public static void main(String[] args) throws IOException {
rd = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(System.out);
st = new StringTokenizer(rd.readLine());
n = Integer.parseInt(st.nextToken());
m = Int... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class XeniaSpies{
public static void main(String[] argv) throws Exception{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] words=br.readLine().trim().split(" ");
int n=Integer.parseInt(words[0]), m=Integer.p... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f;
int main() {
cin >> n >> m >> s >> f;
int tim = 1;
for (int i = 0; i < m; i++) {
int t, l, r;
scanf("%d%d%d", &t, &l, &r);
while (tim < t && s != f) {
if (s < f)
cout << "R";
else
cout << "L";
if (s < f)
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;
public class Xenia2 {
static StringBuilder result = new StringBuilder();
static int current;
static boolean right;
public static... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import sys
def start():
fin=sys.stdin
n,m,s,f=map(int,fin.readline().split())
res=""
cur=s
if s<f:
i=0
for _ in range(m):
t,l,r=map(int,fin.readline().split())
while i+1<t:
res+="R"
cur+=1
i+=1
if... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct W {
int t, l, r;
} w[100010];
int n, m, s, f;
int main() {
while (cin >> n >> m >> s >> f) {
for (int i = (0); i < (m); i++) {
cin >> w[i].t >> w[i].l >> w[i].r;
}
int t = 1, p = 0;
int dir = f > s ? 1 : -1;
while (s != f) {
if (p ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | """http://codeforces.com/problemset/problem/342/B"""
def solve(s, f, t):
res = ''
step = 1 if s < f else -1
i = current = 0
while s != f:
current += 1
ti, l, r = t[i] if i < len(t) else (-1, -1, -1)
if ti == current:
i += 1
if l <= s <= r or l <= s + step... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
struct tn {
int t, l, r;
};
tn lk[111111];
bool in(int loc, int l, int r) {
if (loc >= l && loc <= r) return true;
return false;
}
int absv(int i1) { return i1 < 0 ? -i1 : i1; }
int main() {
int n, m, s, f;
scanf("%d%d%d%d", &n, &m, &s, &f);
int now = s;
int to;
char toc;
if (... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
public class B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStre... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int M = 110000;
int t[M], l[M], r[M];
char ans[2 * M];
int main() {
int i;
int n, m, s, f;
scanf("%d %d %d %d", &n, &m, &s, &f);
for (i = 0; i < m; i++) scanf("%d %d %d", &t[i], &l[i], &r[i]);
int k = 0, cr = 0, ds;
while (s != f) {
if (s < f)
ds... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
map<int, int> m;
void primeFactors(int n) {
while (n % 2 == 0) {
m[2]++;
n = n / 2;
}
for (int i = 3; i <= sqrt(n); i = i + 2) {
while (n % i == 0) {
m[i]++;
n = n / i;
}
}
if (n > 2) m[n]++;
}
long long gcd(long long a, long long b) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct inter {
int l, r, t;
};
int n, m, s, f;
vector<inter> V;
bool pro(inter a, inter b) { return a.t < b.t; }
int main() {
scanf("%d %d %d %d", &n, &m, &s, &f);
for (int i = 0; i < m; i++) {
inter p;
scanf("%d %d %d", &p.t, &p.l, &p.r);
V.push_back(p);
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct T {
int l, r;
int t;
};
T p[100010];
int main() {
long long n, m, s, f;
cin >> n >> m >> s >> f;
s--;
f--;
int last = 0;
for (int i = 0; i < m; i++) {
cin >> p[i].t >> p[i].l >> p[i].r;
p[i].l--;
p[i].r--;
}
if (s < f)
for (int i =... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int spys, steps, start, finish, l, r, t;
int counter;
string res;
map<int, pair<int, int> > step;
bool left;
while (cin >> spys >> steps >> start >> finish) {
res.clear();
step.clear();
counter = 1;
for (int i = 0; i < steps; i++) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
int L[100010 * 2], R[100010 * 2];
int main() {
int n, m, x, y;
int t, l, r;
scanf("%d%d%d%d", &n, &m, &x, &y);
for (int i = 0; i < m; i++) {
scanf("%d%d%d", &t, &l, &r);
if (t <= n * 2) {
L[t] = l, R[t] = r;
}
}
for (in... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, u, v, prev = 1, move;
cin >> n >> m >> u >> v;
string output;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
move = t - prev;
if (u > v) {
move = min(u - v, move);
output += string(move, 'L');
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import sys
a = [int(i) for i in raw_input().split()]
(n, m, s, f) = (a[0], a[1], a[2], a[3])
w = []
for i in range(m):
w.append([int(i) for i in raw_input().split()])
def isWarching(i, st, cw):
if cw >= len(w): return False
if w[cw][0] == st:
return w[cw][1] <= i <= w[cw][2]
cw += 1
if cw >= len(w): return Fa... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long p) {
int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long PO(long long a, long long n) {
long long val = 1;
for (long long... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
map<int, int> L, R;
int main() {
int n, m, s, t;
while (scanf("%d%d%d%d", &n, &m, &s, &t) == 4) {
L.clear(), R.clear();
for (int i = 0; i < (m); i++) {
int a, b, x;
scanf("%d%d%d", &x, &a, &b);
L[x] = a, R[x] = b;
}
int dx = (s < t ? 1 ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | n, m, s, f = map(int, input().split())
data = {}
for _ in range(m):
t, l, r = map(int, input().split())
data[t] = (l, r)
k = 1
curr = s
while curr != f:
if k in data:
if data[k][0] <= curr <= data[k][1]:
print('X', end='')
else:
if 1 < curr < n:
if cur... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
class Triplet {
public:
long long x;
long long y;
long long gcd;
};
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
Triplet extendedEuclid(long long a, long long b) {
if (b == 0) {
Triplet ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
int l[100005], r[100005], t[100005];
int main() {
int n, m, s, f, step = 1;
scanf("%d%d%d%d", &n, &m, &s, &f);
for (int i = 0; i < m; i++) scanf("%d%d%d", &t[i], &l[i], &r[i]);
for (int i = 0; i < m || s != f; i++) {
if (t[i] != step++) {
if (s < f) {
printf("R");
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int s, f, n, m;
int t[100300], x[100300], y[100300];
int main() {
scanf("%d %d %d %d", &n, &m, &s, &f);
for (int i = 0; i < m; i++) scanf("%d %d %d", &t[i], &x[i], &y[i]);
string res = "";
int next = 0;
int tiempo = 1;
int cur = 0;
char ch;
while (s != f) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int nei(int now, int s, int f) {
if (s < f)
return now + 1;
else
return now - 1;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);
int n, m, s, f;
cin >> n >> m >> s >> f;
int now = s;
int t_now = 1;
for (int i = 0; i < m; i++) {
int t, l,... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
const long double pi = 3.14159265359;
const long long N = 1e9 + 7;
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n = 0, s, f, e = 1, a, b, c;
cin >> a >> b >> s >> f;
bool g, r = s < f;
while (s != f) {
c = (r ? s + 1 : s - ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | def seen(i, l, r):
return (i >= l and i <= r)
n, m, src, dst = map(int, raw_input().split())
steps = [map(int, raw_input().split()) for i in range(m)]
curr = 1
idx = 0
dx = 1
ans = []
sdx = "R"
if src > dst:
dx = -1
sdx = "L"
while src != dst:
if idx < len(steps) and curr == steps[idx][0]:
... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int step = 0;
int flag = 0;
while (m) {
if (s == f) return;
int t, l, r;
if (flag == 0) {
cin >> t >> l >> r;
m--;
}
if (s != f) {
step++;
if (step == t) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.math.*;
import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
//FileInputStream cao = new FileInputStream("injava.txt");
Reader.init(System.in);
StringBuilder ans = new StringBuilder();
int n = Reader.nextInt();
int m = Reader.n... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100001;
int n, m, s, f, t[maxn], l[maxn], r[maxn];
int main() {
scanf("%d%d%d%d", &n, &m, &s, &f);
for (int i = 1; i <= m; i++) scanf("%d%d%d", &t[i], &l[i], &r[i]);
if (s < f) {
for (int i = 1, j = 1; i < maxn << 1; i++) {
if (s == f) break... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class Main implements Runnable {
public void solve() throws IOException {
int N = nextInt();
int M = nextInt();
int src = nextInt();
int tgt = nextInt();
int time = 0;
... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
inline void boost() {
ios_base::sync_with_stdio();
cin.tie(0);
cout.tie(0);
}
const long long MAXN = 1e5 + 123;
const long long inf = 1e9 + 123;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
map<int, pair<int, int> > mm;
int main() {
boost();
int n, m... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
scanf("%d%d%d%d", &n, &m, &s, &f);
int i;
int x[m][3];
for (i = 0; i < m; i++) {
scanf("%d%d%d", &x[i][0], &x[i][1], &x[i][2]);
}
int step = 1;
int add;
int current = s;
if (s < f) {
add = 1;
} else {
add = -1;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main(int argc, char* argv[]) {
int n = 3, m = 5, s = 1, f = 3;
cin >> n >> m >> s >> f;
int l, r;
char c = 'L';
int d = -1;
if (s < f) {
c = 'R';
d = 1;
}
int cur = s, step = 0, km = 0;
int next = -1;
if (m > 0) {
cin >> next >> l >> r;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | R = lambda : map(int , raw_input().split())
n,m,s,f = R()
out = ""
prev_t = 1
sign = 1 if s < f else -1
dir = "R" if sign == 1 else "L"
for i in xrange(m):
t, l, r = R()
while t > prev_t and s != f:
out += dir
s += sign
prev_t += 1
if s == f:
break
if l <= s <= r or l <= s + sign <= r:
out += "X"
else:
... | PYTHON |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.HashMap;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int s = sc.nextInt();
int f = sc.nextInt();
StringBuilder res = new StringBuilder();
HashMap<Integer, Integ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
i... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m, s, f;
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> m >> s >> f;
vector<pair<int, pair<int, int> > > v(m + 1);
for (long long int i = 0; i < int(m); ++i) {
cin >> v[i].first >> v[i].second.first >> v[i].second.second;
}
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
int l[100009], r[100009], t[100009];
int main() {
int n, m, s, f;
int i, j;
while (scanf("%d%d%d%d", &n, &m, &s, &f) != -1) {
t[0] = 0;
for (i = 1; i <= m; i++) {
scanf("%d%d%d", &t[i], &l[i], &r[i]);
}
int pos = s;
if (s < f) {
int time = 1, d = 1;
f... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | n,m,s,f=map(int,input().split())
t=dict();
for i in range(m):
t1,l1,r1=map(int,input().split())
t[t1]=(l1,r1);
pos=s;i=1;
while(1):
if(pos==f):
break
if i in t:
if t[i][0] <= pos<=t[i][1]:
print('X',end='')
i+=1
continue
elif(f-pos>0 and t[i][0... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | /*
Aman Agarwal
algo.java
*/
import java.util.*;
import java.io.*;
public class B342
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.Scanner;
public class XeniaAndSpies {
@SuppressWarnings("unused")
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int spies = sc.nextInt();
int steps = sc.nextInt();
int start = sc.nextInt();
int finish = sc.nextInt();
int mat[][] = new int[3][100001];
in... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int INF = 1901486712;
const int MOD = 1000000007;
const double PI = acos(-1);
const double EPS = 1E-9;
bool between(int x, int l, int r) { return (l <= x && x <= r); }
string tostring(int x) {
char dum[20];
sprintf(dum, "%d", x);
string ret(dum);
return ret;
}... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | //B. Xenia and Spies
//time limit per test2 seconds
//memory limit per test256 megabytes
//inputstandard input
//outputstandard output
//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 ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, l, r, mi;
int curr, prev = 1;
scanf("%d %d %d %d", &n, &m, &s, &f), curr = s;
for (int i = 1; i <= m && curr != f; i++) {
scanf("%d %d %d", &mi, &l, &r);
for (int j = 0; j < mi - prev && curr != f; j++)
if (s < f)
c... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
struct Node {
int times, l, r;
} node[100005];
int main() {
int n, m, s, f;
int flag = 0;
scanf("%d %d %d %d", &n, &m, &s, &f);
int now = 0;
for (int i = 0; i < m; i++) {
scanf("%d %d %d", &node[i].times, &node[i].l, &node[i].r);
}
if (s <= f) flag = 1;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long mode = pow(10, 9) + 7;
bool cmp(pair<long double, int> x, pair<long double, int> y) {
return x.first < y.first;
}
const long long maxn = 2 * 1e5 + 5;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
long long n, m, s, f;
cin >> n >> m >> s >> f;
map<l... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class BX {
/*
*/
public static BR in;
public static LR lin;
public static PrintWriter out;
public static PrintWriter fout;
static {
try {
in = new BR();
out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(S... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 |
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import static java.lang.Math.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, s, f, t, l, r, i, j = 1;
string st = "";
cin >> n >> m >> s >> f;
for (i = 1; i <= m; i++) {
cin >> t >> l >> r;
if (s == f) {
} else if (s < f) {
while (j <= t) {
if (s == f) {
break;
}
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.*;
import java.util.*;
public class Main
{
public static void main(String args[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
PrintWriter pr=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
//System.ou... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int M = 100005;
int n, m, s, f;
struct interval {
int l, r, ti;
void read() { scanf("%d %d %d", &ti, &l, &r); }
bool judge(int x) {
if (x >= l && x <= r)
return false;
else
return true;
}
} q[M];
void solve() {
int dir = s < f ? 1 : -1;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
long t[100001];
int l[100001], r[100001];
for (int i = 0; i < m; i++) {
cin >> t[i] >> l[i] >> r[i];
}
t[m] = 1000000001;
int pos = 1;
int tps = s;
int t... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | /**
* Created with IntelliJ IDEA.
* User: den
* Date: 9/7/13
* Time: 11:35 AM
* To change this template use File | Settings | File Templates.
*/
import java.io.*;
import java.util.StringTokenizer;
public class TaskB extends Thread {
private void solve() throws IOException {
int n = _int();
... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
public class XeniaAndSpies {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int s = in.nextInt() - 1;
int f = in.nextInt() - 1;
int dir = s < f ? 1 : -1;
StringBuilder sb = new StringBuilder();
int ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class XeniaAndSpies {
public static InputReader in;
public static PrintWriter out;
public static final int MOD = (int) (1e9 + 7);
public static void main(String[] args) {
in = new InputReader(System.in);
out = new PrintWriter(System.out);
int n = in.n... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int main() {
ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
long long int n, i, j, t, m, k, l, x, s, f, la... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int msg, dest, m, n, i, li, ri, ti, step = 1;
cin >> n >> m >> msg >> dest;
for (i = 1; i <= m; i++) {
cin >> ti >> li >> ri;
while (step != ti) {
if (msg == dest) goto end;
if (msg < dest)
cout << 'R', msg++;
else
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long n, m, st, f, i, j;
long long a, b, c;
set<long long> s;
map<long long, pair<long long, long long> > mp;
cin >> n >> m >> st >> f;
for (i = 0; i < m; i++) {
cin >> a... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 |
import java.io.*;
import java.util.*;
/**
*
* @author Do Quoc bao
*/
public class ProblemB {
public static void main(String[] args) throws java.lang.Exception {
in.init(System.in);
StringBuilder kq=new StringBuilder();
int n=in.nextInt(),m=in.nextInt(),s=in.nextInt(),f=in.nextInt(),i,time=0;
char d=' ';if ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, _f, i, j, t, L, R, pre;
string ans = "";
cin >> n >> m >> s >> f;
for (pre = i = 0; i < m; i++) {
cin >> t >> L >> R;
if (s == f) continue;
for (j = 1; j < t - pre; j++) {
if (f > s) {
ans += 'R';
++s;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n, m, s, f;
cin >> n >> m >> s >> f;
map<int, pair<int, int> > mp;
for (int i = 0; i < m; i++) {
int t, l, r;
cin >> t >> l >> r;
mp[t] = {l, r};
};
string ans;
if (s < f) {
for (int step = 1;; step++) {
if (s == f) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
using namespace std;
inline long long in() {
long long x;
scanf("%lld", &x);
return x;
}
int32_t main() {
long long n = in();
long long m = in();
long long s = in();
long long f = in();
lo... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int t[100005], l[100005], r[100005];
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
int c = s;
int last = 0;
for (int i = 0; i < m; i++) cin >> t[i] >> l[i] >> r[i];
string out = "";
int ct = 1;
int k = 0;
while (s != f) {
char ch;
int tt;
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f, t0 = 0;
scanf("%d%d%d%d", &n, &m, &s, &f);
string step;
for (int t, l, r, i = 0; i < m; ++i, t0 = t) {
scanf("%d%d%d", &t, &l, &r);
int dt = t - t0;
if (s < f) {
int d = f - s;
int dx = min(d, dt - 1);
for (... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
map<int, pair<int, int> > mp;
string move = "LRX";
int n, m, s, f, temp, counter = 1;
cin >> n >> m >> s >> f;
for (int i = 0; i < m; i++) {
cin >> temp;
cin >> mp[temp].first;
cin >> mp[temp].second;
}
temp = 1;
if (s > f) {
tem... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int t[100001], L[100001], R[100001];
char dev[300001];
char ff(int dir) {
if (dir == 1) return 'R';
return 'L';
}
bool noesta(int x, int a, int b) {
if (x < a || x > b) return 1;
return 0;
}
int main() {
int n, m, s, f;
while (cin >> n >> m >> s >> f) {
for ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.util.Comparator;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.List;
import java.io.IOException;
import java.util.InputMismatchException;
import java.util.ArrayList;
import java.util.NoSuchEl... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;
public class Main {
public static void main(String[] args) throw... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long f = 0, j, q = 1, i, n;
while (q--) {
long long y, k = 1, x, M, s;
cin >> n >> M >> s >> f;
unordered_map<int, pair<int, int> > m;
for (i = 0; i < M; i++) {
... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, s, f;
cin >> n >> m >> s >> f;
char move;
int inc;
if (s < f)
move = 'R', inc = 1;
else
move = 'L', inc = -1;
int cur = s, i = 1, cnt = 0;
while (cur != f) {
int t, l, r;
if (cnt < m) {
cin >> t >> l >> r;
c... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
int main() {
int n, m, s, f, t, l, r, cur, inc, i, temp;
char ch;
scanf("%d %d %d %d", &n, &m, &s, &f);
cur = s;
if (s < f) {
inc = 1;
ch = 'R';
} else {
inc = -1;
ch = 'L';
}
temp = 0;
for (i = 0; i < m; i++) {
scanf("%d %d %d", &t, &l, &r);
while (++t... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | n, m, s, f = map(int, input().split())
ans, p = '', 1
if s < f:
for i in range(m):
t, l, r = map(int, input().split())
if t > p:
if t - p < f - s:
ans += 'R' * (t - p)
s += t - p
else:
ans += 'R' * (f - s)
s = f... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class B {
public static void main(String[] args) throws Exception {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(bf.readLine());
int n = ... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.StringTokenizer;
import java.io.*;
public class XeniaAndSpies
{
public static void main(String[]args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer tt = new StringTokenizer(br.readLine());
int n = Integer.parseInt(t... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f, idx_spy, step = 1, idx_xen = 0;
vector<int> tim(100001), lef(100001), righ(100001);
string ans;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> m >> s >> f;
for (int i = 0; i < m; i++) cin >> tim[i] >> lef[i] >> righ[i];
tim[m] =... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
public class cf342B{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int s = sc.nextInt();
int f = sc.nextInt();
char toApp = 'L';
int toAdd = -1;
StringBuilder ans = new StringBuilder();
if(s < f)
... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | import java.util.*;
import java.io.*;
public class XeniaSpies324B
{
public static class MyFasterScanner
{
BufferedReader br;
StringTokenizer st;
public MyFasterScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !s... | JAVA |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, m, s, f, l, r, T = 1, t, shift;
char C;
char res[1111111];
int resl;
void read_input() { scanf("%d%d%d%d", &n, &m, &s, &f); }
void write_output() { printf("%s\n", res); }
void solve() {
if (s < f) {
C = 'R';
shift = 1;
} else {
C = 'L';
shift = -1... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int i, j, k, l, x, y, z, m, n, ans, dir, s, f, current, p, q;
pair<int, int> a;
pair<int, pair<int, int> > spies[300000];
int main() {
scanf("%d %d %d %d", &n, &m, &s, &f);
for (i = 0; i < m; i++) {
scanf("%d %d %d", &x, &y, &z);
a = make_pair(y, z);
spies[i... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | def geto(a, b):
return max(0, min(a[1], b[1]) - max(a[0], b[0])+1)
n,m,s,f = map(int, input().split())
di = {}
for _ in range(m):
t, l, r = map(int, input().split())
di[t] = [l, r]
t = 1
ans = []
while s != f:
if f > s:
inte = [s, s+1]
if t in di and geto(inte, di[t]): ans += ['X']
... | PYTHON3 |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
const int N = 1e5 + 10;
using namespace std;
map<int, pair<int, int> > mp;
int main() {
int n, m, s, f;
scanf("%d%d%d%d", &n, &m, &s, &f);
while (m--) {
int t, x, y;
scanf("%d%d%d", &t, &x, &y);
mp[t] = make_pair(x, y);
}
char c;
int mv;
if (s < f)
c = 'R', mv = 1;... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
bool sortbysec(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
return (a.second < b.second);
}
pair<long long int, long long int> call(long long int n) {
long long int i = 0;
long long int k = n;
while (n % ... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n, m, s, f, t[N], l[N], r[N];
int main() {
scanf("%d%d%d%d", &n, &m, &s, &f);
for (int i = 0; i < m; ++i) {
scanf("%d%d%d", t + i, l + i, r + i);
}
t[m] = -1;
int timer = 0, pos = 0;
string ans;
while (s != f) {
char ch;
i... | CPP |
342_B. Xenia and Spies | 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 pass the ... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int t, l, r, n, m, s, f;
cin >> n >> m >> s >> f;
int cnt = 1;
for (int i = 0; i < m; i++) {
cin >> t >> l >> r;
while (t > cnt && s != f) {
cnt++;
if (s < f)
s++, cout << "R";
else
s--, cout << "L";
}
i... | CPP |
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