prompt string | response string |
|---|---|
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.ArrayList;
import java.io.PrintStream;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Reader;
import java.util.TreeSet;... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
struct one {
int i;
int end;
bool operator<(const one &e) const { return i < e.i; }
};
struct zero {
int i, j;
};
const int inf = 1e9;
class segment {
public:
vector<int> n, t;
segment() { ; }
segment(vector<int> &a) : n(a), t(4 * a.size(), 0) { build(0, 0, a... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
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.util.Arrays;
import java.util.StringTokenizer;
public class CF_275D {
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
import java.math.*;
public class CF483D {
static final int N=100005;
static int l[]=new int[N];
static int r[]=new int[N];
static int v[]=new int[N];
public static void main(String args[]) {
InputReader in=new InputReader(System.in);
PrintWriter ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
int a[111111][33];
int d[111111][33];
int l[111111];
int r[111111];
int x[111111];
int main() {
int t;
int i, j, k, n, m;
scanf("%d%d", &n, &m);
for (i = 1; i <= m; i++) {
scanf("%d%d%d", &l[i], &r[i], &x[i]);
for (t = 0; t < 32; t++)
if ((x[i] >> t) & 1) {
a[l[i]]... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... |
/**
* Date: 20 Nov, 2018
* Link:
*
* @author Prasad-Chaudhari
* @linkedIn: https://www.linkedin.com/in/prasad-chaudhari-841655a6/
* @git: https://github.com/Prasad-Chaudhari
*/
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int p[100005];
int add[100005];
int f[100005];
int first[100005], second[100005], x[100005];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int read() {
int s = 0;
char ch = getchar();
while (ch < '0' || ch > '9') ch = getchar();
while (ch >= '0' && c... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
string A;
int r[1000001], l[1000001], q[1000001], ans[1000001], a[1000001];
struct node {
int x;
void merge(node a, node b) { x = a... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
long long int tree[400000] = {0};
long long int lazy[400000] = {0};
void update(long long int node, long long int a, long long int b,
long long int i, long long int j, long long int val) {
if (lazy[node] != 0) {
tree[node] |= lazy[node];
if (a != b) {
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
long long i, j, k, n, m, x, z, t, f, dp[100007][35], cp[100007][35], l[100007],
r[100007], v[100007];
int main() {
cin >> n >> m;
for (i = 0; i < m; i++) {
cin >> l[i] >> r[i] >> v[i];
for (j = 0; j < 32; j++) {
if ((v[i] & (1LL << j)) != 0) {
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author zodiacLeo
*/
public class Main
{
public static void main(String[] ar... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long maxn = 3e5;
const long long mod = 1e9 + 7;
const long double PI = acos((long double)-1);
long long pw(long long a, long long b, long long md = mod) {
long long res = 1;
while (b) {
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int N = 100001, B = 30;
int l[N], r[N], q[N], a[N], t[4 * N], sum[N];
inline void build(int v, int l, int r) {
if (l + 1 == r) {
t[v] = a[l];
return;
}
int mid = (l + r) >> 1;
build(v * 2, l, mid);
build(v * 2 + 1, mid, r);
t[v] = t[v * 2] & t[v * ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int R = 1e9 + 7;
template <typename T>
T last(vector<T>& v) {
return v[v.size() - 1];
}
template <typename T>
void print(vector<T>& v) {
for (int i = 0; i < v.size(); i++) {
cout << v[i] << ' ';
}
cout << endl;
}
template <typename T1, typename T2>
void print(ve... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
import java.util.List;
public class B {
private static StringTokenizer st;
private static BufferedReader br;
public static int MOD = 1000000007;
public static long tenFive = 100000;
public static void print(Object x) {
System.out.println(x + "");
}... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int n, m;
int l[100005], r[100... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | //package codeforces;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.uti... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int max_n = 100010;
int arr[31][max_n];
long long q[3][max_n];
int sum[31][max_n];
int main() {
int n, m;
cin >> n >> m;
for (int i = 0; i < m; i++) {
long long r, l, val;
cin >> l >> r >> val;
q[0][i] = l;
q[1][i] = r;
q[2][i] = val;
for... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100050;
int a[maxn][32];
int l[maxn], r[maxn], q[maxn];
int cnt[maxn][32];
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < m; ++i) scanf("%d%d%d", &l[i], &r[i], &q[i]);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < 30; ++j)... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 10 + 100000;
const int INF = int(1e9);
int n, m;
int l[MAXN], r[MAXN], q[MAXN];
void input() {
({
int neg = 0;
n = 0;
char ch;
for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
if (ch == '-') neg = 1;
n = ch - 48;
fo... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int dr[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dc[] = {0, 0, 1, -1, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const int INF = 1e9;
const int N = 1e5 + 5;
const long double pi = acos(-1);
const double eps = 1e-9;
struct query {
int l, r, q;
};
bool cmp(query x, quer... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 7;
const int MXN = (int)1e5 + 7;
int n, m;
int a[MXN][31], s[MXN][31], ps[MXN][31];
int l[MXN], r[MXN], q[MXN];
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> m;
for (int i = 1; i <= m; i++) {
cin >> l[i] >> r[i] >> q[i];
for... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | # Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fractio... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
struct pot {
int x, y, d[31];
};
pot a[200020];
int dig[200020][30], sum[200020][30], n, ans[200020], m, k, l;
bool f;
bool cmp(pot u, pot v) { return (u.x < v.x); }
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &a[i].x, &a[i].... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int n, m;
long long lo[4 * 100100 + 1], hi[4 * 100100 + 1], tree[4 * 100100 + 1],
delta[4 * 100100 + 1];
void init(int i, int l, int r) {
lo[i] = l;
hi[i] = r;
if (l == r) return;
int m = (l + r) / 2;
init(2 * i, l, m);
init(2 * i + 1, m + 1, r);
}
void prop... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mn = 1e5 + 1;
int n, m, tree[mn * 4][30];
struct query {
int l, r, k;
} queries[mn];
void update(int id, int l, int r, int x, int y, int type) {
if (l > y || r < x) {
return;
}
if (l >=... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
public static v... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
const int MAXN = 1e5 + 10;
const int MOD = 2333;
const double eps = 1e-6;
using namespace std;
int bit[MAXN << 2], AND[MAXN << 2], l[MAXN], r[MAXN], q[MAXN];
void pushdown(int rt) {
if (bit[rt]) {
bit[rt << 1] |= bit[rt];
bit[rt << 1 | 1] |= bit[rt];
AND[rt << 1] |= bit[rt];
A... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
struct SegTree {
int n;
vector<unsigned int> t1, t2;
SegTree(int n_) {
n = 1;
while (n < n_) n <<= 1;
t1.resize(2 * n - 1, 0);
t2.resize(2 * n - 1, 0);
}
void update(int k) {
t1[k] = (t1[2 * k + 1] | t2[2 * k + 1]) & (t1[2 * k + 2] | t2[2 * k +... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 3e5 + 100;
const long long mod = 1e9 + 7;
const long long INF = 1e9 + 7;
const long long mlog = 22;
const long long SQ = 400;
long long n, m;
long long seg[maxn * 4];
long long l[maxn], r[maxn], q[maxn];
long long a[maxn];
long long lg[maxn];
long lon... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int n, m;
int a[100005];
int l[100005], r[100005], q[100005];
int sgn[100005], pr[100005];
bool ok[100005];
void die() {
cout << "NO\n";
exit(0);
}
int main() {
ios_base::sync_with_stdio(0);
cin >> n >> m;
for (int i = 1; i <= m; i++) cin >> l[i] >> r[i] >> q[i];
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.util.*;
import java.math.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
public class Main{
// ArrayList<Integer> lis = new ArrayList<Integer>();
// ArrayList<String> lis = new ArrayList<String>();
// PriorityQueue<P> que... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws NumberFormatException, IOExceptio... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
static int[] tree;
static int[] lazy;
static void refresh(int idx) {
if (lazy[idx] != 0) {
tree[idx] |= lazy[idx];
if (idx * 2 < tree.... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class D {
static int mo7ayed = (1 << 31) - 1;
public static void main(String[] args) throws Exception
{
Scanner sc = new... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class CF_275_Div1_InterestingArray {
public static void main(String[] args) throws IOException {
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
inline void ri(int& x) { scanf("%d", &x); }
class FenwickTree {
private:
vector<int> ft;
void update(int pos, int val) {
while (pos < ft.size()) {
ft[pos] += val;
pos += pos & (-pos);
}
}
public:
FenwickTree() {}
FenwickTree(int n) { ft.assi... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
using namespace std;
long long dp[100005][35];
long long val[100005][35];
long long a[100005][35];
struct node {
long long l, r, q;
};
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100, B = 30;
int L[N], R[N], X[N];
int A[N], res[N];
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, q;
cin >> n >> q;
for (int i = 1; i <= q; i++) {
cin >> L[i] >> R[i] >> X[i];
}
for (int bit = B - 1; bit >= 0; bit--) {
m... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
template <class T>
T pmax(T a, T b) {
return max(a, b);
}
const long long int mod = 1e9 + 7;
const long long int Maa = 3e5;
vector<long long int> edge[Maa];
long long int n, m, l[Maa], r[Maa], q[Maa], A[Maa], p[Maa], c[Maa];
int main() {
cin >> n >> m;
for (int i = 0;... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXB = 30;
int ls[MAXN], rs[MAXN], qs[MAXN], out[MAXN];
int res[MAXN];
int main(void) {
int n, m;
scanf("%d", &(n));
scanf("%d", &(m));
vector<pair<int, int> > intv[MAXN];
for (int i = 0; i < m; i++) {
scanf("%d", &(ls[i]));... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
public class InterestingArray {
public static void main(String[] args) throws Exception{
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(f.readLine());
int n = Integer.parseInt(st.nextToken());
int ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int arr[100002][30], res[100002][30], cnt[100002][30];
int l[100002], r[100002], q[100002];
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i++) {
scanf("%d%d%d", &l[i], &r[i], &q[i]);
for (int j = 0; j < 30; j++) {
if (q[i] & (1 << ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt(), m = s.nextInt();
Condition[] c = new Condition[m];
for (int i = 0; i < m; i++) {
c[i] = new Condition(s.nextInt() -... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const char lend = '\n';
const int N = 100000, M = 30;
int vet[N][M], n, m, l[N], r[N], x[N];
void first(int b) {
set<int> s;
for (int i = 0; i < n; ++i) s.insert(i);
for (int i = 0; i < m; ++i)
if (x[i] >> b & 1) {
auto it = s.lower_bound(l[i]);
while ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int N = 200010;
int from[N], to[N], num[N];
int s[N][30];
int a[N][30];
int sum[N][30];
void tree(int n, int m) {
for (int i = 0; i <= n; i++) {
for (int j = 0; j < 30; j++) {
s[i][j] = 0;
}
}
for (int k = 0; k < m; k++) {
scanf("%d %d %d", fro... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
vector<pair<pair<int, int>, int> > vec;
int tree[4 * 100011];
int arr[30][100011];
int n, k;
int ans[100011];
int build(int node, int b, int e) {
if (b == e) {
return tree[node] = ans[b];
}
int mid = (b + e) >> 1;
int l = node << 1;
int r = l + 1;
int p = bu... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
mt19937 rnd(239);
const int N = 100001;
int a[2 * N];
int t[2 * N];
void prop(int i) {
t[i] |= t[i >> 1];
t[i ^ 1] |= t[i >> 1];
}
void f(int ls, int rs, int x) {
for (int l = ls + N, r = rs + N; l <= r; l = l + 1 >> 1, r = r - 1 >> 1) {
if (l & 1) {
t[l] |=... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
class SGTreeLazy {
long long int size;
vector<long long int> arr;
public:
vector<long long int> result;
SGTreeLazy(long long int s) {
size = s;
arr.resize(4 * s, 0);
}
void build(long long int x, vector<long long int> &a, long long int l,
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
const int maxm = 200010;
const int inf = 1e9;
int n, m;
int a[maxn];
int b[maxn];
int r1[maxn];
int r2[maxn];
int r3[maxn];
struct Node {
int l;
int r;
int val;
int re;
} Tree[maxn << 2];
void BuildTree(int left, int right, int rt) {
Tree[... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int N = 111111;
int req[N][32], c[N][32];
int s[32];
int ans[N];
int main() {
int n, m;
scanf("%d%d", &n, &m);
vector<pair<int, pair<int, int> > > z;
for (int i = 0; i < m; ++i) {
int l, r, q;
scanf("%d%d%d", &l, &r, &q);
for (int j = 0; j < 30; ++... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
vector<int> seg1;
void update(int st, int en, int i, int val, int l, int r) {
if (r < st || en < l) return;
if (l <= st && en <= r) {
seg1[i] |= val;
return;
}
int mid = (st + en) / 2;
update(st, mid, (2 * i), val, l, r);
update(mid + 1, en, (2 * i) + 1,... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
inline long long gcd(long long a, long long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
inline long long power(long long a, long long n, long long m) {
if (n == 0) return 1;
long long... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int lim = 1e5 + 5;
int arr[lim][31], qur[lim][3], fin[lim], seg[4 * lim], q, n;
void make() {
cin >> n >> q;
for (int i = 0; i < q; ++i) {
cin >> qur[i][0] >> qur[i][1] >> qur[i][2];
for (int j = 0; j < 31; ++j)
if ((qur[i][2] >> j) & 1) arr[qur[i][0... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
int n;
class bit {
public:
int s[200005];
void up(int p, int x) {
while (p <= n) {
s[p] += x;
p += p & -p;
}
}
int get(int p) {
int re = 0;
while (p) {
re = re + s[p];
p -= p & -p;
}
retu... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import static java.lang.System.in;
import static java.lang.System.out;
import java.io.*;
import java.util.*;
public class B482bitSegmentTree {
static final double EPS = 1e-10;
static final int INF = (1 << 30)-1;
static final double PI = Math.PI;
public static Scanner sc = new Scanner(in);
StringBuilder sb = new... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Mahmoud Aladdin <aladdin3>
*/
public ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int dx[100005][30];
int n, q, s[100005], e[100005], x[100005];
int main() {
scanf("%d %d", &n, &q);
for (int i = 0; i < q; i++) {
scanf("%d %d %d", &s[i], &e[i], &x[i]);
for (int j = 0; j < 30; j++) {
if ((x[i] >> j) & 1) {
dx[s[i]][j]++;
d... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.util.*;
import java.io.*;
import java.math.*;
public class Solution{
public static int n,m;
public static int[] l;
public static int[] r;
public static int[] q;
public static int[][] a;
public static int[][] beg;
public static int[][] end;
public static int k;
public stat... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5 + 10;
const int MOD = (int)1e9 + 7;
int rs[MAXN * 4], vl[MAXN * 4];
void update(int l, int r, int x, int y, int z, int rt) {
if (l == x && r == y) {
vl[rt] |= z;
return;
}
int mid = l + r >> 1;
if (mid >= y)
update(l, mid, x, y,... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
const long long maxn = 1e5 + 9, modn = 1e9 + 7;
using namespace std;
long long n, m, l[maxn], r[maxn], q[maxn], s[4 * maxn], a[maxn];
void query(long long x, long long y, long long q, long long v = 1,
long long l = 0, long long r = n) {
if (x >= r || y <= l) return;
if (l >= x &&... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
struct dat {
int L, R, V;
} a[100005];
int N, M;
bool cmp(dat a, dat b) {
if (a.L != b.L) return a.L < b.L;
if (a.R != b.R) return a.R < b.R;
return a.V < b.V;
}
int ans[100005], s[100005];
void build(int k) {
int last = 0;
for (int i = (1), _b = (M); i <= _b; i... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int N = 100010, D = 30;
int n, m, c[D], a[N];
int l[N], r[N], q[N], s[N][D];
vector<int> p[N], k[N];
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= m; i++) {
scanf("%d %d %d", &l[i], &r[i], &q[i]);
p[l[i]].push_back(q[i]);
k[r[i]].push_back(... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int ar[101000], a, b, c, n, m;
vector<pair<pair<int, int>, int> > V;
vector<pair<int, pair<int, int> > > A;
int ans[101000];
bool cari(int k) {
A.clear();
for (int i = 0; i < V.size(); i++)
if ((1 << k) & (V[i].second))
A.push_back(make_pair(1, V[i].first));
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Rubanenko
*/
public class M... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
struct sequence {
int t[100010], s[100010];
inline void insert(int l, int r) {
t[l]++;
t[r + 1]--;
}
inline void modify(int n) {
for (int i = 1; i <= n; i++) t[i] += t[i - 1];
for (int i = 1; i <= n; i++) t[i] = t[i] > 0;
for (int i = 1; i <= n; i++) s[i] = s[i - 1] ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
if (x == 0) return y;
return gcd(y % x, x);
}
long long powmod(long long x, long long y, long long m) {
if (y == 0) return 1;
long long p = powmod(x, y / 2, m) % m;
p = (p * p) % m;
return (y % 2 == 0) ? p : (x * p) % m;... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner in = new Scanner(new BufferedInputStream(System.in));
PrintWriter out = new PrintWriter(new BufferedOutputStream(Syste... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... |
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[]args) throws IOException {
Scanner sc = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
StringBuilder sb = new StringBuilder();
i... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
class Node {
public:
int left;
int right;
int value;
};
class segTree {
public:
Node *tree;
int length;
segTree(int length) {
this->length = length;
tree = new Node[length * 4];
createTree(0, 0, length - 1);
}
void createTree(int root, int left... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
public class Main {
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const long long INF = (1ll << 32) - 1;
long long sum[maxn << 2], add[maxn << 2];
int n, m;
struct Node {
int l, r;
long long q;
} nd[maxn];
void Pushup(int root);
void Pushdown(int root);
void update(long long q, int L, int R, int left, int rig... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int mod = (int)1e+9 + 7;
const double pi = acos(-1.0);
unsigned int t[32][100100];
bool x[100100];
unsigned int z[2 * 131072], k = 131072;
struct zp {
unsigned int l, r, q;
} zap[100100];
unsigned int getand(unsigned int l, unsigned int r) {... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m;
pair<pair<int, int>, int> a[101010];
cin >> n >> m;
int x, y, z;
for (int i = 1; i <= m; i++) {
cin >> x >> y >> z;
pair<int, int> p = make_pair(x, y);
pair<pair<int, int>, int> q = make_pair(p, z);
a[i] = q;
}
bool ok = ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
struct _ {
ios_base::Init i;
_() {
cin.sync_with_stdio(0);
cin.tie(0);
}
} _;
int n, m;
const int N = 100 * 1000 + 1;
int l[N], r[N], q[N], sum[N];
int MAXBIT = 30;
int main() {
cin >> n >> m;
for (int i = 0; i < m; ++i) {
cin >> l[i] >> r[i] >> q[i];
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
public class Main {
static PrintWriter pw;
static Scanner sc;
static int[] a, b, dp[];
static long ceildiv(long x, long y){ return (x+y-1)/y;}
public static void main(String[] args) throws Exception {
sc = new Scanner(System.in);
pw = new Print... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | /**
* Created by Aminul on 10/28/2017.
*/
import java.io.*;
import java.util.*;
public class CF482B {
public static void main(String[] args) throws Exception {
FastReader in = new FastReader(System.in);
PrintWriter pw = new PrintWriter(System.out);
int n = in.nextInt(), k = in.nextInt();... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
import java.math.*;
public class CF483D {
static final int N=100005;
static int l[]=new int[N];
static int r[]=new int[N];
static int v[]=new int[N];
public static void main(String args[]) {
InputReader in=new InputReader(System.in);
PrintWriter ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int a[100009][30];
int b[100009][30];
int c[30];
void solve() {
int n, m;
cin >> n >> m;
int mark[m][3];
c[0] = 1;
for (int i = 0; i < m; i++) {
cin >> mark[i][0] >> mark[i][1] >> mark[i][2];
int ele = mark[i][2];
int ind = 0;
int x = mark[i][0];
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
const int N = 100100;
int n, m, l[N], r[N], q[N], c[N], s[N], a[N], f;
int go(int b) {
memset(c, 0, sizeof(c));
for (int i = 0; i < m; ++i)
if (q[i] & b) ++c[l[i]], --c[r[i] + 1];
for (int i = 1; i <= n; ++i)
c[i] += c[i - 1], s[i] = s[i - 1] + (c[i] < 1), a[i] |= c[i] ? b : 0;
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int INF = (1LL << 31) - 1;
const long long int LINF = (1LL << 62) - 1;
const int NMAX = 100000 + 5;
const int MMAX = 100000 + 5;
const int KMAX = 100000 + 5;
const int PMAX = 100000 + 5;
const int LMAX = 100000 + 5;
const int VMAX = 100000 + 5;
int N, M;
int L[NMAX];
... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.*;
import java.util.*;
public class Main {
// main
public static void main(String [] args) throws IOException {
// makes the reader and writer
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline T poww(T b, T p) {
long long a = 1;
while (p) {
if (p & 1) {
a = (a * b);
}
p >>= 1;
b = (b * b);
}
return a;
}
template <class T>
inline T poww2(T b, int p) {
T a = 1;
while (p) {
if (p & 1) {
a = (a * b... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
const int N = 1000 * 1000;
const int MAXBIT = 30;
int l[N], r[N], q[N], a[N], t[4 * N];
int sum[N];
inline void build(int v, int l, int r) {
if (l + 1 == r) {
t[v] = a[l];
return;
}
int mid = (l + r) >> 1;
build(v * 2, l, mid);
build(v * 2 + 1, mid, r);
t[v] = t[v * 2] & t[v... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
long long int t[100100 * 4], lz[100100 * 4];
pair<pair<int, int>, int> qr[100100];
void update(int node, int st, int en, int l, int r, int q) {
t[node] = t[node] | lz[node];
if (st != en) {
lz[node * 2 + 1] = lz[node] | lz[node * 2 + 1];
lz[node * 2 + 2] = lz[no... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int l[100010], r[100010], n, m;
;
long long val[100010], num[100010][30], a[100010][30];
int main() {
cin >> n >> m;
for (int i = 0; i < m; i++) {
cin >> l[i] >> r[i] >> val[i];
for (int j = 0; j < 30; j++)
if ((val[i] & (1 << j)) != 0) {
a[l[i]][j... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
int a[32][101111], n, m, d[32][111111], b[111111], t[400111];
pair<pair<int, int>, int> p[111111];
int bit(int mask, int i) { return (mask & (1 << i)) > 0; }
void build(int v, int tl, int tr) {
if (tl == tr)
t[v] ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const int MX = 1e5 + 5;
const int INF = 0x3f3f3f3f;
int n, m;
int L[MX], R[MX], Q[MX], A[MX];
bool ok[32][MX << 2];
int cnt[32][MX];
void update(int L, int R, int id, int l, int r, int rt) {
if (L <= l && r <= R) {
ok[id][rt] = 1;
return;
}
int m = (l + r) >> ... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
const long long inf = 998244353;
vector<int> adj[200010], vis(200010, 0), par(200010, 0), dis(200010, 0),
di(200010, 0)... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | // package CodeForces;
import java.io.*;
import java.util.*;
public class Problem_482B {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner();
PrintWriter pw = new PrintWriter(System.out);
int n=sc.nextInt(),q=sc.nextInt();
SegmentTree sgt=new SegmentTree(n);
Triple[]query... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
const long long inf = 998244353;
vector<int> adj[200010], vis(200010, 0), par(200010, 0), dis(200010, 0),
di(200010, 0)... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
int n, m;
int a[100010];
int queryL[100010], queryR[100010], b[100010];
int s[30][100010];
int data[400010];
void build(int p, int L, int R) {
if (L == R) {
data[p] = a[L];
return;
}
int mid = L + R >> 1;
build(p << 1, L, mid);
build(p << 1 | 1, mid + 1, R);
data[p] = data[p... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
import jav... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Solver {
public static void main(String[] Args) t... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
const long long D = 31;
const long long N = 100010;
struct constraint {
long long l, r, q;
constraint(long long _l, long long _r, long long _q) {
l = _l;
r = _r;
q = _q;
}
};
void err() {
cout << "NO" << '\n';
exit(0);
}
long long n, m;
long long a[N][... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int n, m;
int l[100005], r[100005], q[100005];
struct node {
int v, add;
};
node t[4 * 100005];
const int MX = (1LL << 31) - 1;
void Push(int v, int tl, int tm, int tr) {
if (t[v].add != 0) {
t[2 * v].add |= t[v].add;
t[2 * v].v |= t[v].add;
t[2 * v + 1].add... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | import java.util.*;
import java.io.*;
import javafx.util.*;
public class Solution
{
public static void main(String []args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
HashMap<Pair<Integer,Integer> , Long> map = new HashMap<Pair<Integer,Integer> , Long>... |
Problem: We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or stat... | #include <bits/stdc++.h>
using namespace std;
int n, m;
int l[112345], r[112345], q[112345];
int prefix[31][112345], s[31][112345], cnt[31][112345];
int main() {
ios_base::sync_with_stdio(false);
cin >> n >> m;
for (int i = 0; i < m; ++i) cin >> l[i] >> r[i] >> q[i];
for (int i = 0; i < m; ++i) {
for (int b... |
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