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stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
983k
|
|---|---|---|---|---|---|---|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
ifstream fin("input.txt");
ofstream fout("output.txt");
long long fast_exp(long long base, long long exp) {
long long res = 1;
while (exp > 0) {
if (exp % 2 == 1) res = (res * base) % 1000000007;
base = (base * base) % 1000000007;
exp /= 2;
}
return res % 1000000007;
}
int palindromecheck(string s) {
int n = s.size();
for (int i = 0; i < n / 2; ++i) {
if (s[i] != s[n - i - 1]) {
return 0;
}
}
return 1;
}
int gcd(int a, int b) {
while (a && b) a > b ? a %= b : b %= a;
return a + b;
}
int val(char c) {
if (c >= '0' && c <= '9')
return (int)c - '0';
else
return (int)c - 'A' + 10;
}
long long pows(int a, int b) {
long long res = 1;
for (int i = 0; i < b; ++i) {
res *= a;
}
return res;
}
long long logx(long long base, long long num) {
int cnt = 0;
while (num != 1) {
num /= base;
++cnt;
}
return cnt;
}
long long divisibles(long long a, long long b, long long m) {
if (a % m == 0)
return (b / m) - (a / m) + 1;
else
return (b / m) - (a / m);
}
string bitstring(int n, int size) {
string s;
while (n) {
s += (n % 2) + '0';
n /= 2;
}
while (s.size() < size) {
s += '0';
}
reverse(s.begin(), s.end());
return s;
}
vector<int> root(200001, 0);
vector<int> size(200001, 1);
int find(int x) {
while (x != root[x]) x = root[x];
return x;
}
bool same(int a, int b) { return find(a) == find(b); }
void unite(int a, int b) {
a = find(a);
b = find(b);
if (size[a] < size[b]) swap(a, b);
size[a] += size[b];
root[b] = a;
}
vector<int> vis(200001, 0);
vector<int> adj[200001];
int main() {
std::ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
int n;
long long l, r;
cin >> n >> l >> r;
vector<long long> vec(n + 1);
long long sum = 0;
int j = 1;
for (int i = n; i >= 1; --i) {
sum += (i - 1) * 2;
vec[j] = sum;
if (i == 1) ++vec[j];
++j;
}
long long idxl = lower_bound(vec.begin(), vec.end(), l) - vec.begin();
long long idxr = lower_bound(vec.begin(), vec.end(), r) - vec.begin();
if (vec[idxl] > l && idxl > 1) --idxl;
vector<pair<long long, long long>> ans;
int temp = vec[idxl];
if (idxl == 1) temp = 1;
long long start;
if (idxl != 1) start = vec[idxl - 1] + 1;
for (int i = idxl; i <= idxr; ++i) {
for (int j = i + 1; j <= n; ++j) {
ans.push_back(make_pair(i, start++));
ans.push_back(make_pair(j, start++));
}
if (i + 1 > n) {
ans.push_back(make_pair(1, start));
}
}
long long begin;
if (idxl == 1) {
begin = l - 1;
} else {
for (int i = 0; i < ans.size(); ++i) {
if (ans[i].second == l) {
begin = i;
break;
}
}
}
for (int i = begin; i <= begin + (r - l); ++i) cout << ans[i].first << " ";
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
long long n;
long long l, r;
long long i;
cin >> n >> l >> r;
long long temp = l;
long long b = 1;
while (temp > 2 * (n - b) && b < n) {
temp = temp - 2 * (n - b);
b++;
}
for (i = l; i <= r && i < n * (n - 1) + 1; i++) {
if (temp % 2 == 1) {
printf("%I64d ", b);
temp++;
} else {
printf("%I64d ", temp / 2 + b);
if (temp / 2 + b == n) {
temp = 1;
b++;
} else
temp++;
}
}
if (r == n * (n - 1) + 1)
printf("1\n");
else
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t, i, n, cnt, od, l, r, ev, g;
cin >> t;
while (t--) {
cin >> n >> l >> r;
g = 0;
if (r == (n * (n - 1)) + 1) {
g = 500;
}
i = 0;
while (i <= n - 1) {
if (2 * n * (i) - (i) * (i + 1) < l) {
i++;
} else {
i--;
break;
}
}
cnt = 2 * n * i - (i) * (i + 1) + 1;
od = i + 1;
ev = i + 2;
if (g) {
r--;
}
while (cnt <= r) {
if (cnt >= l) {
if (cnt % 2 == 0) {
cout << ev << " ";
ev++;
} else {
cout << od << " ";
}
cnt++;
} else {
if (cnt % 2 == 0) {
ev++;
} else {
;
}
cnt++;
}
if (ev > n) {
od++;
ev = od + 1;
}
}
if (g == 500) {
cout << 1 << " ";
}
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int mods = 998244353;
const int maxn = 1e5 + 10;
const int N = 1e5 + 10;
const int E = 2e5 + 10;
long long n, l, r;
long long k[maxn];
vector<int> ans;
int main() {
int T;
cin >> T;
while (T--) {
cin >> n >> l >> r;
k[1] = 1;
for (int i = 2; i <= (n); ++i) {
k[i] = k[i - 1] + 2 * (n - i + 1);
}
long long bo = n;
for (int i = 1; i <= (n); ++i) {
if (k[i] > l) {
bo = i - 1;
break;
}
}
if (bo >= n) {
printf("1\n");
continue;
}
long long pc = k[bo];
long long tot = 1;
long long tmp;
ans.clear();
while (pc <= r) {
if (bo == n) {
ans.push_back(1);
break;
}
if (pc % 2 == 1)
tmp = bo;
else {
tmp = bo + tot;
tot++;
}
if (pc >= l) ans.push_back(tmp);
pc++;
if (tmp == n) {
bo++;
tot = 1;
}
}
for (int i = 0; i <= (ans.size() - 1); ++i) {
cout << ans[i] << " ";
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
const int LM = 3e5 + 4;
ll N;
ll L, R;
ll part[LM];
int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%lld%lld%lld", &N, &L, &R);
bool e = 0;
if (R == N * (N - 1) + 1) {
e = 1;
R--;
}
for (int i = 1; i <= N - 1; i++) part[i] = part[i - 1] + (N - i) * 2;
for (int i = 1; i <= N; i++) {
if (L <= part[i] && R > part[i - 1]) {
int v, last = i;
for (int j = 1; j <= (N - i) * 2; j++) {
if (j & 1)
v = i;
else
v = ++last;
if (L <= part[i - 1] + j && part[i - 1] + j <= R) printf("%d ", v);
}
}
}
if (e) printf("1 ");
puts("");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7, MAX = 1e5 + 5;
long long powN(long long a, long long p) {
if (p == 0) return 1;
long long z = powN(a, p / 2);
z = (z * z) % MOD;
if (p % 2) z = (z * a) % MOD;
return z;
}
vector<bool> is_prime(MAX + 1, true);
void Sieve() {
is_prime[0] = is_prime[1] = false;
int i, j;
for (i = 2; i * i <= MAX; i++) {
if (is_prime[i]) {
for (j = i * i; j <= MAX; j += i) {
is_prime[j] = false;
}
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
long long n;
long long l, r;
cin >> n >> l >> r;
long long st = n - 1;
for (long long k = 1; k <= n; k++) {
if ((n * 2 - k - 1) * k >= l) {
st = k - 1;
break;
}
}
long long ex = (2 * n - st - 1) * (st);
l -= ex;
r -= ex;
st++;
vector<int> V = {0};
long long some = st + 1;
long long tot = 0;
for (int i = 1; i <= r; i++) {
if (st == n) {
V.push_back(1);
break;
}
if (i % 2)
V.push_back(st);
else {
V.push_back(some);
some++;
}
if (i == tot + 2 * (n - st)) {
tot += 2 * (n - st);
st++;
some = st + 1;
}
}
for (int i = l; i <= r; i++) {
printf("%d ", V[i]);
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
Scanner in = new Scanner(inputStream);
PrintWriter out = new PrintWriter(outputStream);
DMinimumEulerCycle solver = new DMinimumEulerCycle();
solver.solve(1, in, out);
out.close();
}
static class DMinimumEulerCycle {
public void solve(int testNumber, Scanner sc, PrintWriter pw) {
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long a = sc.nextLong();
long b = sc.nextLong();
long[] arr = new long[n];
arr[0] = 2 * n - 2;
for (int i = 1; i < n - 1; i++) arr[i] = arr[i - 1] + 2 * (n - i - 1);
arr[n - 1] = arr[n - 2] + 1;
int in = 0;
int la = 0;
int l = 0;
int h = n - 1;
while (l <= h) {
int mid = (l + h) / 2;
if (arr[mid] >= a) {
in = mid;
h = mid - 1;
} else {
l = mid + 1;
}
}
l = 0;
h = n - 1;
while (l <= h) {
int mid = (l + h) / 2;
if (arr[mid] >= b) {
la = mid;
h = mid - 1;
} else {
l = mid + 1;
}
}
for (int i = in; i <= la; i++) {
int[] tmp = new int[2 * (n - i) - 2];
int idx = 0;
if (i == n - 1) pw.print(1);
else {
for (int j = i + 2; j <= n; j++) {
tmp[idx] = i + 1;
if (idx + 1 < tmp.length) tmp[idx + 1] = j;
idx += 2;
}
if (i == in) {
if (i == la) {
int s = (int) (a - (i == 0 ? 0 : arr[i - 1]) - 1);
int e = (int) (b - (i == 0 ? 0 : arr[i - 1]));
for (int j = s; j < e; j++) pw.print(tmp[j] + " ");
} else {
int s = (int) (a - (i == 0 ? 0 : arr[i - 1]) - 1);
for (int j = s; j < tmp.length; j++) pw.print(tmp[j] + " ");
}
} else if (i == la) {
int e = (int) (b - (i == 0 ? 0 : arr[i - 1]));
for (int j = 0; j < e; j++) pw.print(tmp[j] + " ");
} else {
for (int j = 0; j < tmp.length; j++) pw.print(tmp[j] + " ");
}
}
}
pw.println();
}
pw.flush();
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
CORRECT
|
java
|
//package com.prituladima.codeforce.contest1334;
import java.io.*;
import java.util.*;
import java.util.function.Predicate;
import static java.util.Arrays.stream;
import static java.util.stream.IntStream.range;
/**
* Don't confuse variables in inner cycles. Don't call variable like (i j k g). Delegate methods.
* -Xmx64m maximum heap size allocation
* 90% errors is copy-paste, wrong indexes and TOO MUCH variables
*/
public class SolutionD {
private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
private static final boolean MULTI_TEST = true;
private static final int BITS = 31;
private static final int MODULO = (int) 1e9 + 7;
private static final int INF = (int) 1e7 + 7;
private static final Random rand = new Random(1278367);
private static final String yes = "YES", no = "NO";
private static final int MAXN = 2 * (int) 10e5 + 10;
private boolean[] used = new boolean[MAXN];
private void solveAll() {
int t = MULTI_TEST ? nextInt() : 1;
while (t-- > 0) {
solve();
}
}
private long pathLen(long n) {
return n * (n - 1) + 1;
}
private long sum(long n, long k) {
return n * k - k * (k + 1) / 2;
}
private void solve() {
long n = nextInt();
long L = nextLong();
long R = nextLong();
boolean tillEnd = false;
if (R == pathLen(n)) {
R--;
tillEnd = true;
}
long curPointer = 1;
long lev = 1;
for (; curPointer + 2*(n - lev) < L; lev++) {
curPointer += 2*(n - lev);
}
// lev = upperBound();
// debug("curPointer = " + curPointer);
// debug("lev = " + lev);
boolean trigger = true;
outer:
for (; lev < n; lev++) {
for (long lev2 = lev + 1; lev2 <= n; curPointer++) {
// if (curFrom == L && curFrom + 1 == R) {
// print(lev);
// print(' ');
// print(lev2);
// print(' ');
//
// }
if (curPointer < L) {
if (trigger) {
// print(lev);
// printSpace();
} else {
// print(lev2);
// printSpace();
lev2++;
}
trigger = !trigger;
} else if (R < curPointer) {
break outer;
} else {
if (trigger) {
print(lev);
printSpace();
} else {
print(lev2);
printSpace();
lev2++;
}
trigger = !trigger;
}
}
}
if (tillEnd) {
print(1);
}
println();
// }
//
//1.
/**
* n vertexes
* n - 1
* n - 2
* n - 3
*
* r - l + 1;
* 1 2
* 1 3
* 2 3
* 1
*
*/
// long k;
// if ((L + 1) / 2 <= n - 1) {
// k = 0;
// } else {
// long lowK = 0, highK = n;
// while (highK - lowK > 1) {
// long mid = lowK + (highK - lowK) / 2;
// long sum = n * mid - mid * (mid + 1) / 2;
// if (sum < (L + 1) / 2) {//row number
// lowK = mid;
// } else {
// highK = mid;
// }
// }
// k = lowK;
// }
// debug("k = " + k);
// long P = n * k - k * (k + 1) / 2 + 1;
// long diff = (L + 1) / 2 - P;
// if (L % 2 == 0) {
// print((k + 2 + diff) + " ");
// L++;
// diff++;
// }
// boolean oneMore = false;
// boolean last = false;
// if (R == n * (n - 1) + 1) {
// last = true;
// R--;
// } else if (R % 2 == 1) {
// oneMore = true;
// R--;
// }
// while (L / 2 <= R / 2) {
// print((k + 1) + " " + (k + 2 + diff) + " ");
// if ((k + 2 + diff) == n) {
// k++;
// diff = 0;
// }
// L++;
// L++;
// diff++;
// }
//
//
// if (last) {
// println(1);
// } else if (oneMore) {
// println((k + 1));
// } else {
// println();
// }
}
private boolean inRange(int val, int fromInclusive, int toExclusive) {
return fromInclusive <= val && val < toExclusive;
}
private int nextRandInt(int minInclusive, int maxExclusive) {
return rand.nextInt(maxExclusive - minInclusive) + minInclusive;
}
/**
* Graph traverses
*/
private void dfs(int from, boolean[] used, Graph graph) {
used[from] = true;
for (int to : graph.get(from)) {
if (!used[to]) {
dfs(to, used, graph);
}
}
}
private void dfsTree(int from, int parent, Tree tree) {
for (int to : tree.get(from)) {
if (parent != to) {
dfsTree(to, from, tree);
}
}
}
/**
* Binary searches.
*/
private long upperBound(long inclusiveLeft, long exclusiveRight, Predicate<Long> predicate) {
while (exclusiveRight - inclusiveLeft > 1) {
long middle = inclusiveLeft + (exclusiveRight - inclusiveLeft) / 2;
if (predicate.test(middle)) {
inclusiveLeft = middle;
} else {
exclusiveRight = middle;
}
}
return inclusiveLeft;
}
private int lowerBound(int exclusiveLeft, int inclusiveRight, Predicate<Integer> predicate) {
while (inclusiveRight - exclusiveLeft > 1) {
int middle = exclusiveLeft + (inclusiveRight - exclusiveLeft) / 2;
if (predicate.test(middle)) {
inclusiveRight = middle;
} else {
exclusiveLeft = middle;
}
}
return inclusiveRight;
}
private int minAns(int lev) {
char[] tabs = new char[lev];
Arrays.fill(tabs, '\t');
debug(new StringBuilder().append(tabs).append(" ").append(lev));
return 0;
}
public static void main(String[] args) {
new SolutionD().run();
}
private BufferedReader reader;
private PrintWriter writer;
private StringTokenizer tokenizer;
private void run() {
try (BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
PrintWriter writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)))) {
this.writer = writer;
this.reader = reader;
solveAll();
} catch (Exception e) {
e.printStackTrace();
System.exit(1);
}
}
/**
* Base types: Strings, int, long, double
*/
private String nextToken() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
private int nextInt() {
return Integer.parseInt(nextToken());
}
private long nextLong() {
return Long.parseLong(nextToken());
}
private double nextDouble() {
return Double.parseDouble(nextToken());
}
/**
* Primitives 1D arrays: char, int, long, double
*/
private char[] nextCharArray() {
return nextToken().toCharArray();
}
private int[] nextIntArray(int size) {
return stream(new int[size]).map(c -> nextInt()).toArray();
}
private long[] nextLongArray(int size) {
return stream(new long[size]).map(c -> nextLong()).toArray();
}
private double[] nextDoubleArray(int size) {
return stream(new double[size]).map(c -> nextDouble()).toArray();
}
private String[] nextStringArray(int size) {
return range(0, size).mapToObj(i -> nextToken()).toArray(String[]::new);
}
/**
* Primitives 2D arrays: char, int, long, double
*/
private char[][] nextCharMatrix(int n) {
return range(0, n).mapToObj(i -> nextToken().toCharArray()).toArray(char[][]::new);
}
private int[][] nextIntMatrix(final int n, final int m) {
return range(0, n).mapToObj(i -> nextIntArray(m)).toArray(int[][]::new);
}
private long[][] nextLongMatrix(final int n, final int m) {
return range(0, n).mapToObj(i -> nextLongArray(m)).toArray(long[][]::new);
}
private double[][] nextDoubleMatrix(final int n, final int m) {
return range(0, n).mapToObj(i -> nextDoubleArray(m)).toArray(double[][]::new);
}
/**
* Graphs
*/
private static class Graph extends HashMap<Integer, Collection<Integer>> {
}
private static class Tree extends Graph {
}
private Graph nextGraph(int amountOfVertexes, int amountOfEdges, boolean isDirected) {
Graph graph = new Graph();
for (int i = 1; i <= amountOfVertexes; i++) {
graph.put(i, new HashSet<>());
}
for (int i = 1; i <= amountOfEdges; i++) {
int from = nextInt();
int to = nextInt();
graph.get(from).add(to);
if (!isDirected) {
graph.get(to).add(from);
}
}
return graph;
}
private Tree nextTree(int amountOfVertexes, boolean isDirected) {
Tree tree = new Tree();
for (int i = 1; i <= amountOfVertexes; i++) {
tree.put(i, new HashSet<>());
}
for (int i = 1; i <= amountOfVertexes - 1; i++) {
int from = nextInt();
int to = nextInt();
tree.get(from).add(to);
if (!isDirected) {
tree.get(to).add(from);
}
}
return tree;
}
/**
* Output
*/
private void printf(String format, Object... args) {
writer.printf(format, args);
}
private void print(Object o) {
writer.print(o);
}
private void printSpace() {
writer.print(' ');
}
private void println() {
writer.println();
}
private void println(Object o) {
writer.println(o);
}
private void flush() {
writer.flush();
}
/**
* Utils
*/
private boolean isValidIndex(int ind, int n) {
return 0 <= ind && ind < n;
}
private void printSeparator() {
if (ONLINE_JUDGE) return;
println("--------------Answer-----------------");
}
private void debug(Object o) {
if (ONLINE_JUDGE) return;
println(o);
}
private void debug(int[] array) {
if (ONLINE_JUDGE) return;
for (int i = 0; i < array.length; i++) {
print(array[i]);
print(' ');
}
println();
}
private void debug(int[][] matrix) {
if (ONLINE_JUDGE) return;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
print(matrix[i][j]);
print(' ');
}
println();
}
}
private void debug(char[][] matrix) {
if (ONLINE_JUDGE) return;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
print(matrix[i][j]);
print(' ');
}
println();
}
}
public static double maxn(double req, double... opt) {
double max = req;
for (double value : opt) max = Math.max(max, value);
return max;
}
public static double minn(double req, double... opt) {
double min = req;
for (double value : opt) min = Math.min(min, value);
return min;
}
public static double sumn(double... a) {
return stream(a).sum();
}
public static int maxn(int req, int... opt) {
int max = req;
for (int value : opt) max = Math.max(max, value);
return max;
}
public static int minn(int req, int... opt) {
int min = req;
for (int value : opt) min = Math.min(min, value);
return min;
}
public static int sumn(int... a) {
return stream(a).sum();
}
public static long maxn(long req, long... opt) {
long max = req;
for (long value : opt) max = Math.max(max, value);
return max;
}
public static long minn(long req, long... opt) {
long min = req;
for (long value : opt) min = Math.min(min, value);
return min;
}
public static long sumn(long... a) {
return stream(a).sum();
}
public static void sort(long[] array) {
shuffle(array);
Arrays.sort(array);
}
public static void shuffle(long[] array) {
Random random = new Random();
for (int i = 0, j; i < array.length; i++) {
j = i + random.nextInt(array.length - i);
long buf = array[j];
array[j] = array[i];
array[i] = buf;
}
}
public static void sort(int[] array) {
shuffle(array);
Arrays.sort(array);
}
public static void shuffle(int[] array) {
Random random = new Random();
for (int i = 0, j; i < array.length; i++) {
j = i + random.nextInt(array.length - i);
int buf = array[j];
array[j] = array[i];
array[i] = buf;
}
}
public static Map<Double, Integer> multiSet(double[] arr) {
Map<Double, Integer> multiSet = new HashMap<>();
for (int i = 0; i < arr.length; i++) multiSet.put(arr[i], multiSet.getOrDefault(arr[i], 0) + 1);
return multiSet;
}
public static Map<Integer, Integer> multiSet(int[] arr) {
Map<Integer, Integer> multiSet = new HashMap<>();
for (int i = 0; i < arr.length; i++) multiSet.put(arr[i], multiSet.getOrDefault(arr[i], 0) + 1);
return multiSet;
}
public static Map<Long, Integer> multiSet(long[] arr) {
Map<Long, Integer> multiSet = new HashMap<>();
for (int i = 0; i < arr.length; i++) multiSet.put(arr[i], multiSet.getOrDefault(arr[i], 0) + 1);
return multiSet;
}
public static int[] calculatePrefixSum(int[] a) {
int[] pref = new int[a.length];
pref[0] = a[0];
for (int i = 1; i < a.length; i++) pref[i] = pref[i - 1] + a[i];
return pref;
}
public static int[] calculateSuffixSum(int[] a) {
int[] suff = new int[a.length];
suff[a.length - 1] = a[a.length - 1];
for (int i = a.length - 2; i >= 0; i--) suff[i] = suff[i + 1] + a[i];
return suff;
}
public static long[] calculatePrefixSum(long[] a) {
long[] pref = new long[a.length];
pref[0] = a[0];
for (int i = 1; i < a.length; i++) pref[i] = pref[i - 1] + a[i];
return pref;
}
public static long[] calculateSuffixSum(long[] a) {
long[] suff = new long[a.length];
suff[a.length - 1] = a[a.length - 1];
for (int i = a.length - 2; i >= 0; i--) suff[i] = suff[i + 1] + a[i];
return suff;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
/**
* @author Mubtasim Shahriar
*/
public class MinEu {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader sc = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
int t = sc.nextInt();
// int t = 1;
while(t--!=0) {
solver.solve(sc, out);
}
out.close();
}
static class Solver {
public void solve(InputReader sc, PrintWriter out) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
long[] cnt = new long[n+1];
for(int i = 1; i <= n; i++) {
cnt[i] = ((long)(n-i))*2l;
}
long[] sum = new long[n+1];
sum[1] = cnt[1];
for(int i = 2; i <= n; i++) {
sum[i] = sum[i-1]+cnt[i];
}
int idx = 0;
for(int i = 1; i <= n; i++) {
if(sum[i]>=l) {
idx = i;
break;
}
}
if(idx==0) {
out.println(1);
return;
}
long from = l-sum[idx-1];
// System.out.println(sum[idx-1]);
ArrayList<Long> ans = new ArrayList();
long cntu = 0;
long tmp = from/2;
if(from%2==0) {
ans.add(tmp+idx);
// tmp += idx+1;
tmp++;
} else tmp++;
tmp += idx;
long now = idx;
// System.out.println(tmp);
// System.out.println(ans.size());
while(true) {
if(ans.size()>=r-l+1) break;
boolean ok = false;
// System.out.println(now);
while(tmp<=n) {
ans.add(now);
ans.add(tmp);
ok = true;
tmp++;
}
if(!ok) break;
now++;
tmp = now+1;
}
// System.out.println("HI");
if(ans.size()<r-l+1) ans.add(1l);
long cnn = 0;
for(int i = 0; i < ans.size(); i++) {
out.print(ans.get(i) + " ");
cnn++;
if(cnn==r-l+1) break;
}
// System.out.println("HI");
out.println();
}
}
static class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n){
int[] array=new int[n];
for(int i=0;i<n;++i)array[i]=nextInt();
return array;
}
public int[] nextSortedIntArray(int n){
int array[]=nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n){
int[] array=new int[n];
array[0]=nextInt();
for(int i=1;i<n;++i)array[i]=array[i-1]+nextInt();
return array;
}
public long[] nextLongArray(int n){
long[] array=new long[n];
for(int i=0;i<n;++i)array[i]=nextLong();
return array;
}
public long[] nextSumLongArray(int n){
long[] array=new long[n];
array[0]=nextInt();
for(int i=1;i<n;++i)array[i]=array[i-1]+nextInt();
return array;
}
public long[] nextSortedLongArray(int n){
long array[]=nextLongArray(n);
Arrays.sort(array);
return array;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
#!/usr/bin/env python
# coding: utf-8
# In[48]:
#from __future__ import print_function
#from sys import stdin
# In[52]:
cases = int( input() )
# In[53]:
def ecycle(n,l,r):
cnt = r - l + 1
p = n-1
start = 1
while(l >= 2*p and p>0):
l -= 2*p
p -= 1
start += 1
if(start==n):
start = 1
flag = l%2
nextn = start + 1 + (l-1)//2
while(cnt>0):
cnt-=1
if(flag==1):
print(start,end=" ")
if(flag==0):
print(nextn, end=" ")
nextn += 1
if(nextn>n):
start += 1
nextn = start + 1
flag = 0
if(start==n):
start = 1
flag = 1-flag
# In[54]:
while(cases>0):
n,l,r = map( int, input().split() )
ecycle(n,l,r)
cases -= 1
# In[ ]:
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
/*
*created by Kraken on 02-05-2020 at 14:27
*/
//package com.kraken.cf.practice;
import java.util.*;
import java.io.*;
public class D1334 {
public static void main(String[] args) {
FastReader sc = new FastReader();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long l = sc.nextLong(), r = sc.nextLong();
long[] block = new long[n + 1];
block[1] = 1;
for (int i = 2; i <= n; i++) block[i] = 2 * (i - 1) + block[i - 1];
// System.out.println(Arrays.toString(block));
long left = findBlock(block, l);
long right = findBlock(block, r);
// System.out.printf("left: %d, right: %d\n", left, right);
long curr = left;
ArrayList<Long> path = new ArrayList<>();
while (curr <= right) {
for (int i = 0, j = 2; i < curr - 2; i++, j++) {
path.add(curr);
path.add((long) j);
}
path.add(curr);
path.add((long) 1);
curr++;
}
// System.out.println(path.toString());
StringBuilder sb = new StringBuilder();
long lidx = l - block[(int) (left - 1)] - 1;
if (l == 1) lidx++;
for (int i = 0; i < r - l + 1; i++) {
sb.append(path.get((int) (lidx + i))).append(" ");
}
System.out.println(sb.toString());
}
}
private static int findBlock(long[] a, long key) {
int l = 1, r = a.length - 1;
while (l < r) {
int mid = l + (r - l) / 2;
if (a[mid] >= key) r = mid;
else l = mid + 1;
}
return r;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python2
|
T = input()
for _ in xrange(T):
n, l, r = map(int, raw_input().split())
now = 0
sn = 2
result = []
need = r-l+1
while len(result) < need:
if now+1 >= l:
result.append("1")
result.append(str(sn))
elif now + 2 >= l:
result.append(str(sn))
now += 2
if (sn-2) * 2 + now < l:
now += (sn-2) * 2
sn += 1
continue
for i in xrange(2, sn):
if now+1 >= l:
result.append(str(i))
result.append(str(sn))
elif now + 2 >= l:
result.append(str(sn))
now += 2
sn += 1
#print now, sn, l, r, result
print " ".join(result[:need])
#10 2 5
#1 2 1 3 2 3 1 4 2 4 3 4 1 5 2 5 3 5 4 5
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using ld = long double;
using pll = pair<ll, ll>;
void buff() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
constexpr ll MOD = 1e9 + 7;
inline ll pow_mod(ll a, ll b, ll mod = MOD) {
ll res = 1;
a %= mod;
assert(b >= 0);
for (; b; b >>= 1) {
if (b & 1) res = (res * a) % mod;
a = (a * a) % mod;
}
return res;
}
int main() {
buff();
int t;
cin >> t;
for (int i = 0; i < (t); ++i) {
ll n, l, r;
cin >> n >> l >> r;
vector<pair<ll, ll> > c(n);
ll cur_start = 1;
ll cur_len = 2;
for (int j = 1; j < n; ++j) {
c[j] = make_pair(cur_start, cur_start + cur_len - 1ll);
cur_start = (cur_start + cur_len);
cur_len += 2ll;
}
c.push_back(make_pair(c.back().second + 1, c.back().second + 1));
for (int j = 1; j <= n; ++j) {
if (l > c[j].second) continue;
if (r < c[j].first) break;
ll first_good = max(l, c[j].first);
ll last_good = min(r, c[j].second);
for (ll x = first_good; x <= last_good; ++x) {
if (x == c[j].first) {
cout << 1ll << " ";
} else {
ll dist = (x - c[j].first);
if (dist & 1ll) {
cout << j + 1ll << " ";
} else
cout << (dist / 2ll + 1) << " ";
}
}
}
cout << '\n';
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main
{
public static void main(String args[])throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw=new PrintWriter(System.out);
int t=Integer.parseInt(br.readLine());
for(int x=0;x<t;x++)
{
String str[]=br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
long l=Long.parseLong(str[1]);
long r=Long.parseLong(str[2]);
if(l<=(n-1)*2+1)
{
int arr[]=new int[(int)(r-l+1)];
int ind;
if(l%2==0)
{
arr[0]=(int)(l/2)+1;
arr[1]=1;
ind=1;
for(int i=(int)(l/2)+2;i<=n-1&&ind<arr.length-1;i++)
{
if(ind<arr.length-1)
arr[++ind]=i;
if(ind<arr.length-1)
arr[++ind]=1;
}
}
else
{
arr[0]=1;
ind=0;
for(int i=(int)((l+1)/2)+1;i<=n-1&&ind<arr.length-1;i++)
{
if(ind<arr.length-1)
arr[++ind]=i;
if(ind<arr.length-1)
arr[++ind]=1;
}
}
if(ind<arr.length-1)
arr[++ind]=n;
for(int i=2;i<n&&ind<arr.length;i++)
{
for(int j=i+1;j<=n&&ind<arr.length-1;j++)
{
arr[++ind]=i;
if(ind<arr.length-1)
arr[++ind]=j;
}
}
if(ind<arr.length-1)
arr[++ind]=1;
for(int i=0;i<arr.length;i++)
pw.print(arr[i]+" ");
pw.println();
}
else if(l<=(n-1)*2+1+2*(n-2))
{
int arr[]=new int[(int)r-(n-1)*2+5];
int ind=0;
arr[0]=n;
for(int i=2;i<n&&ind<arr.length;i++)
{
for(int j=i+1;j<=n&&ind<arr.length-1;j++)
{
arr[++ind]=i;
if(ind<arr.length-1)
arr[++ind]=j;
}
}
for(int i=0;i<arr.length;i++)
pw.print(arr[i]+" ");
pw.println();
}
else
{
long sum=2*(n-1)+1;
int num=n-2;
int num2=2;
while(sum+2*(num)<l&&num>0)
{
sum=sum+2*(num);
num--;
num2++;
}
int ind=-1;
int arr[]=new int[(int)(r-sum)+5];
// while(sum<=r)
//{
for(int i=num2;i<n&&ind<arr.length;i++)
{
for(int j=i+1;j<=n&&ind<arr.length-1;j++)
{
arr[++ind]=i;
if(ind<arr.length-1)
arr[++ind]=j;
}
}
//}
if(ind<arr.length-1)
arr[++ind]=1;
//for(int i=0;i<arr.length;i++)
//pw.print(arr[i]+" ");
for(int i=(int)(l-sum);i<=(int)(r-sum);i++)
pw.print(arr[i]+" ");
pw.println();
}
}
pw.flush();
pw.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
T = int(input())
for _ in range(T):
n, l, r = map(int, input().split())
size = r - l + 1
st = None
en = None
cur = 0
offset = 0
for i in range(1, n):
if st is None and l < cur + (n-i)*2:
st = i
offset = l - cur - 1
if en is None and r < cur + (n-i)*2:
en = i
cur += (n-i)*2
if st is None:
st = n
if en is None:
en = n
arr = []
for i in range(st, en):
for j in range(i+1, n+1):
arr.append(i)
arr.append(j)
if en == n:
arr.append(1)
else:
i = en
for j in range(i+1, n+1):
arr.append(i)
arr.append(j)
print(' '.join(map(str, arr[offset:offset+size])))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
void solve() {
long long n, l, r;
cin >> n >> l >> r;
long long p = 0;
long long s = 1;
vector<int> bp;
vector<long long> res;
int cu = l;
int a = 1;
int b = n;
int c = 1;
while (a <= b) {
long long cnt = 2 * (b - a);
if (p + cnt >= l) {
vector<long long> list;
for (int i = a + 1; i <= b; i++) {
list.push_back(a);
list.push_back(i);
}
while (cu <= r && cu - 1 - p < list.size()) {
res.push_back(list[cu - 1 - p]);
cu++;
}
}
if (cu > r) break;
p += cnt;
bp.push_back(c);
if (c == a) {
a++;
c = b;
} else {
c = b - 1;
b--;
}
}
long long start = p;
for (int k : bp) res.push_back(k);
int sp = 0;
if (start < l) {
sp = l - start - 1;
}
for (int i = 0; i <= r - l; i++) {
if (i != 0) cout << " ";
cout << res[i + sp];
}
cout << "\n";
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.precision(10);
int T = 1;
cin >> T;
for (int i = 1; i <= T; i++) {
solve();
}
cout.flush();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
for (int tc = 0; tc < (t); tc += 1) {
long long n, l, r;
cin >> n >> l >> r;
l--;
long long i = 1;
long long ptr = 0;
while (ptr + (n - i) * 2 <= l) {
ptr += (n - i) * 2;
i++;
}
long long j = i;
j += (l - ptr) / 2;
if (l % 2 == 1) {
cout << j + 1 << " ";
j++;
l++;
}
if (j != n) {
j++;
} else {
i++;
j = i + 1;
}
while (r > l) {
if (r - l == 1) {
cout << i;
break;
}
cout << i << " " << j << " ";
if (j != n) {
j++;
} else {
i++;
j = i + 1;
}
r -= 2;
}
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Sol4{
public static void main(String[] args) throws IOException{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
long n = sc.nextInt();
long l = Long.parseLong(sc.next());
long r = Long.parseLong(sc.next());
long idx = 1;
long cnt = 2*(n-idx);
while(cnt+1<l) {
idx++;
cnt +=(long)2*(n-idx);
}
cnt++;
cnt-=2*(n-idx)-1;
long ix = idx+1;
while(cnt<r) {
if(ix == n+1) {
idx++;
ix = idx+1;
}
if(cnt%2==0) {
if(cnt>=l)System.out.print(ix + " ");
ix++;
}else {
if(cnt>=l)System.out.print(idx + " ");
}
cnt++;
}
if(r == ((n)*(n-1)+1))System.out.println(1);
else {
if(ix == n+1) {
idx++;
ix = idx+1;
}
if(cnt%2==0) {
if(cnt>=l)System.out.println(ix);
}else {
if(cnt>=l)System.out.println(idx);
}
}
}
sc.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.Scanner;
public class ProblemD {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int a=0;a<t;a++) {
int n = s.nextInt();
long l = s.nextLong();
long r = s.nextLong();
long[] arr = new long[n];
long sum = 0;
long val = 2*(n-1);
for(int i=0;i<n-1;i++) {
sum += val;
val -= 2;
arr[i] = sum;
}
arr[n-1] = arr[n-2] + 1;
// for(int i=0;i<n;i++)
// System.out.println(arr[i]);
int index = upperBound(arr, l);
System.out.println(index);
print(arr, l, r, index);
System.out.println();
}
}
public static void print(long[] arr, long l, long r, int index) {
int n = arr.length;
if(index == n-1) {
System.out.print(1+" ");
return;
}
long val1 = index + 1, val2 = 0;
long end = arr[index];
if(l%2 == 0) {
val2 = n - (end-l)/2;
}
else {
val2 = n - (end-l-1)/2;
}
for(long i=l;l <= Math.min(end, r);l++) {
if(l%2 == 1)
System.out.print(val1+" ");
else {
System.out.print(val2+" ");
val2++;
}
}
if(end < r)
print(arr, end + 1, r, index + 1);
}
public static int upperBound(long[] arr, long v) {
if(v > arr[arr.length-1])
return -1;
if(arr[0] > v)
return 0;
int low = 0, high = arr.length - 1;
while(low < high) {
if(low == high - 1) {
if(arr[low] >= v)
high = low;
else
low = high;
break;
}
int mid = (low + high)/2;
if(arr[mid] >= v)
high = mid;
else
low = mid + 1;
}
return low;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author bhavy seth
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskD solver = new TaskD();
solver.solve(1, in, out);
out.close();
}
static class TaskD {
public void solve(int testNumber, InputReader sc, PrintWriter out) {
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
int start = 0;
long count = 1;
if (count == l)
start = 1;
else {
for (int i = 2; i <= n; i++) {
int smalller = Math.max(0, (i - 2) * 2) + 2;
if (l <= count + smalller) {
start = i;
break;
} else {
count += smalller;
}
}
}
if (start == 2 && l == 2) {
out.print(2 + " ");
count++;
} else if (start == 1) {
out.print(1 + " ");
} else
count++;
for (int i = 2; i < start; i++) {
if (count >= l && count < r) {
out.print(i + " ");
count++;
}
if (count >= l && count < r) {
out.print(start + " ");
count++;
}
if (count >= r) {
break;
}
}
if (count < r && start != 1) {
count++;
out.print(1 + " ");
}
if (count < r) {
for (int i = start + 1; i <= n; i++) {
for (int j = 2; j < i; j++) {
if (count < r && j == 2) {
out.print(i + " ");
count++;
}
if (count >= l && count < r) {
out.print(j + " ");
count++;
}
if (count >= l && count < r) {
out.print(i + " ");
count++;
}
if (count >= r) {
break;
}
}
if (i == 2) {
out.print(i + " ");
count++;
}
if (count < r) {
out.print(1 + " ");
count++;
}
if (count >= r)
break;
}
}
out.println();
}
}
}
static class InputReader {
BufferedReader br;
StringTokenizer st;
public InputReader(InputStream inputStream) {
br = new BufferedReader(new
InputStreamReader(inputStream));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python2
|
import sys
from math import sqrt, floor
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
def main():
T = int(input())
while T:
n, l, r = get_ints()
for i in range(l, r+1):
k = floor(sqrt(i - (3/4)) - 0.5)
j = floor((i - k**2 - k)/2)
if i&1:
print int(j+1),
else:
print int(k+2),
print
T-=1
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class D {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) throws IOException{
FastReader s = new FastReader();
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int t = s.nextInt();//t: number of test cases
for (int i = 0; i < t; i++) {
int n = s.nextInt();
double l = s.nextDouble();
double r = s.nextDouble();
boolean islLast = l == 1.0 * n * (n-1) + 1;
boolean isrLast = r == 1.0 * n * (n-1) + 1;
if(!islLast){
double templ_01 = Math.ceil(((2*n-1)-Math.sqrt(4*Math.pow(n,2)-4*n+1-4*l))/2);
double templ_02 = (templ_01 - 1) * (2*n - templ_01);
int templ_11 = (int)(templ_01);
int templ_12 = (int)((l - 1 + templ_02) / 2 + templ_11 + 1);
if (isrLast)
r--;
boolean isrOdd = r%2==1;
if (isrOdd) {
r--;
}
double tempr_01 = Math.ceil(((2*n-1)-Math.sqrt(4*Math.pow(n,2)-4*n+1-4*r))/2);
double tempr_02 = (tempr_01 - 1) * (2*n - tempr_01);
int tempr_11 = (int) (tempr_01);
int tempr_12 = (int)((r - 1 + tempr_02) / 2 + tempr_11 + 1);
if (l%2==0) {
bw.write(templ_12 + " ");
templ_12++;
if (templ_12 > n) {
templ_11 ++;
templ_12 = templ_11 + 1;
}
}
while(templ_11 < tempr_11 || templ_11 == tempr_11 && templ_12 <= tempr_12){
bw.write(templ_11 + " ");
bw.write(templ_12 + " ");
templ_12++;
if (templ_12 > n) {
templ_11 ++;
templ_12 = templ_11 + 1;
}
}
if (isrOdd) {
bw.write(templ_11 + " ");
}
if (isrLast) {
bw.write("1");
}
}
else
bw.write("1");
bw.write("\n");
}
bw.flush();
bw.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
/**
* @author Mubtasim Shahriar
*/
public class MinEu {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader sc = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
int t = sc.nextInt();
// int t = 1;
while(t--!=0) {
solver.solve(sc, out);
}
out.close();
}
static class Solver {
public void solve(InputReader sc, PrintWriter out) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
long[] cnt = new long[n+1];
for(int i = 1; i <= n; i++) {
cnt[i] = (long)(n-i)*2l;
}
long[] sum = new long[n+1];
sum[1] = cnt[1];
for(int i = 2; i <= n; i++) {
sum[i] = sum[i-1]+cnt[i];
}
int idx = 0;
for(int i = 1; i <= n; i++) {
if(sum[i]>=l) {
idx = i;
break;
}
}
if(idx==0) {
out.println(1);
return;
}
// System.out.println(sum[n]);
long from = l-cnt[idx-1];
ArrayList<Long> ans = new ArrayList();
long cntu = 0;
long tmp = from/2;
if(from%2==0) {
ans.add(tmp+idx);
// tmp += idx+1;
tmp++;
} else tmp++;
tmp += idx;
long now = idx;
// System.out.println(tmp);
// System.out.println(ans.size());
while(true) {
if(ans.size()>=r-l+1) break;
boolean ok = false;
// System.out.println(now);
while(tmp<=n) {
ans.add(now);
ans.add(tmp);
ok = true;
tmp++;
}
if(!ok) break;
now++;
tmp = now+1;
}
// System.out.println("HI");
if(ans.size()<r-l+1) ans.add(1l);
long cnn = 0;
for(int i = 0; i < ans.size(); i++) {
out.print(ans.get(i) + " ");
cnn++;
if(cnn==r-l+1) break;
}
// System.out.println("HI");
out.println();
}
}
static class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n){
int[] array=new int[n];
for(int i=0;i<n;++i)array[i]=nextInt();
return array;
}
public int[] nextSortedIntArray(int n){
int array[]=nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n){
int[] array=new int[n];
array[0]=nextInt();
for(int i=1;i<n;++i)array[i]=array[i-1]+nextInt();
return array;
}
public long[] nextLongArray(int n){
long[] array=new long[n];
for(int i=0;i<n;++i)array[i]=nextLong();
return array;
}
public long[] nextSumLongArray(int n){
long[] array=new long[n];
array[0]=nextInt();
for(int i=1;i<n;++i)array[i]=array[i-1]+nextInt();
return array;
}
public long[] nextSortedLongArray(int n){
long array[]=nextLongArray(n);
Arrays.sort(array);
return array;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
# from sys import stdin,stdout
# input = stdin.readline
# print = stdout.write
from math import *
ar=[0,1]
for i in range(2,100010):
ar.append(i*(i-1)+1)
def jg(ar,n):
l=0
r=len(ar)-1
ans=-1
while(l<=r):
m=(l+r)//2
if(ar[m]>=n):
ans=m
r=m-1
else:
l=m+1
return ans
# print(ar[:10])
for __ in range(int(input())):
n,l,r=map(int,input().split())
for i in range(l,r+1):
if(i==1):
print(1,end=" ")
elif(i==2):
print(2,end=" ")
elif(i==3):
print(1,end=" ")
else:
a=jg(ar,i)
if(i%2==0):
print(a,end=" ")
else:
temp=i-ar[a-1]
if(temp//2 + 1==a):
print(1,end=" ")
else:
print(temp//2+1,end=" ")
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author bhavy seth
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskD solver = new TaskD();
solver.solve(1, in, out);
out.close();
}
static class TaskD {
public void solve(int testNumber, InputReader sc, PrintWriter out) {
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
long count = 0;
int start = 0;
for (int i = 1; i <= n; i++) {
long x = Math.max(0, (n - i) * 2);
if (l <= count + x) {
start = i;
break;
} else {
count += x;
}
}
/* if(l==count+1){
out.print(start+" ");
count++;
}*/
if (start == 0)
out.println(1);
else {
for (int i = start + 1; i <= n; i++) {
if (count < l) {
count++;
}
if (l <= count) {
if (count <= r) {
out.print(start + " ");
count++;
}
if (count <= r) {
out.print(i + " ");
count++;
}
} else {
if (count < l) {
count++;
}
if (l <= count) {
if (count <= r) {
out.print(i + " ");
count++;
}
if (count <= r) {
out.print(start + " ");
count++;
}
}
}
if (count > r) {
break;
}
}
if (count < r) {
for (int i = start + 1; i < n; i++) {
for (int j = i + 1; j <= n; j++) {
if (count <= r) {
out.print(i + " ");
count++;
}
if (count <= r) {
out.print(j + " ");
count++;
}
if (count > r)
break;
}
if (count > r)
break;
}
}
if (count <= r)
out.print(1);
}
out.println();
}
}
}
static class InputReader {
BufferedReader br;
StringTokenizer st;
public InputReader(InputStream inputStream) {
br = new BufferedReader(new
InputStreamReader(inputStream));
}
public String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import math
import sys
def LFromI(I):
return I*I-I
def IFromL(L):
return math.floor((1+math.sqrt(1+4*L))/2)
def Out(l, R):
curint = IFromL(l)
L = LFromI(curint)
curpos = L
while curpos < R:
for i in range(2*curint):
if curpos + i >= R:
break
if i % 2 == 1:
print(1+curint, end=' ')
else:
print(1+i//2, end=' ')
curpos += 2*curint
curint += 1
t = int(input())
for _ in range(t):
n, l, r = map(int, input().split())
Out(l, r)
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
//
import java.math.*;
import java.util.*;
import java.io.*;
public class D {
static int test = 10; // 0 for local testing, 1 for std input
static BufferedReader in;
static PrintWriter out = new PrintWriter(System.out);
static String file = "../in";
static int inf = 1_000_000;
static void swap(int[]ary, int i, int j)
{
int t = ary[i];
ary[i] = ary[j];
ary[j] = t;
}
static String[] split() throws Exception
{
return in.readLine().split(" ");
}
static int readInt() throws Exception
{
return Integer.valueOf(in.readLine());
}
static int[] toIntArray() throws Exception
{
String[] sp = split();
int n = sp.length;
int[] ary = new int[n];
for(int i = 0; i < n; i++) ary[i] = Integer.valueOf(sp[i]);
return ary;
}
static long[] toLongArray() throws Exception
{
String[] sp = split();
int n = sp.length;
long[] ary = new long[n];
for(int i = 0; i < n; i++) ary[i] = Long.valueOf(sp[i]);
return ary;
}
public static void main(String[] args) throws Exception
{
int _k = Integer.valueOf("1");
if(test > 0) in = new BufferedReader(new InputStreamReader(System.in));
else in = new BufferedReader(new FileReader(file));
if(test < 0) {String[] str = in.readLine().split(" ");}
/****************************************************/
/****************************************************/
/****************************************************/
/****************************************************/
int t = readInt();
for(int tt = 0; tt < t; tt++)
{
long[] ary = toLongArray();
long n = ary[0], L = ary[1], R = ary[2];
StringBuilder sb = new StringBuilder();
long idx = 0;
while(idx * (idx + 1) < L) idx++;
idx--;
long offset = L - idx * (idx + 1) - 1;
// now construct
int len = (int)(R - L + 1);
List<Long> list = new ArrayList<>();
while(list.size() < R - L + 1)
{
for(int i = 1; i < idx; i++)
{
list.add(0L + i);
list.add(idx);
}
idx++;
}
int from = (int) offset;
int to = from + len;
for(int i = from; i < to; i++)
{
out.printf("%d ", list.get(i));
}
out.printf("\n");
}
/****************************************************/
/****************************************************/
/****************************************************/
/****************************************************/
out.flush();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python2
|
T = input()
for _ in xrange(T):
n, l, r = map(int, raw_input().split())
now = 0
sn = 2
result = []
need = r-l+1
while len(result) < need:
if now+1 >= l:
result.append("1")
result.append(str(sn))
elif now + 2 >= l:
result.append(str(sn))
now += 2
if (sn-2) * 2 + now + 100 < l:
now += (sn-2) * 2
sn += 1
continue
for i in xrange(2, sn):
if now+1 >= l:
result.append(str(i))
result.append(str(sn))
elif now + 2 >= l:
result.append(str(sn))
now += 2
sn += 1
#print now, sn, l, r, result
print " ".join(result[:need])
#10 2 5
#1 2 1 3 2 3 1 4 2 4 3 4 1 5 2 5 3 5 4 5
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long mpow(long long a, long long b, long long p = MOD) {
a = a % p;
long long res = 1;
while (b > 0) {
if (b & 1) res = (res * a) % p;
a = (a * a) % p;
b = b >> 1LL;
}
return res % p;
}
const long long N = 2 * 1e5 + 2, M = 20;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long lol[n + 1];
lol[1] = 1;
for (long long i = 2; i <= n; i++) {
lol[i] = 2 + (i - 2) * 2;
}
long long var = -1;
long long s = 0;
for (long long i = 1; i <= n; i++) {
s += lol[i];
if (s < l) {
} else {
var = i;
s -= lol[i];
break;
}
}
long long count = r - l + 1 + 2 * n + 2;
vector<long long> ans;
while (count > 0) {
if (var == 1) {
ans.push_back(var);
var++;
count--;
continue;
}
ans.push_back(var);
count--;
if (count == 0) break;
for (long long i = 2; i < var; i++) {
ans.push_back(i);
count--;
if (count == 0) break;
ans.push_back(var);
count--;
if (count == 0) break;
}
if (count == 0) {
break;
}
ans.push_back(1);
count--;
if (count == 0) {
break;
}
var++;
}
for (long long i = 0; i < ans.size(); i++) {
s++;
if (s > r) {
break;
}
if (s >= l && s <= r) {
cout << ans[i] << " ";
}
}
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include<bits/stdc++.h>
using namespace std;
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define f first
#define s second
#define ll long long
#define loop(i,a,b) for(ll i=a;i<b;i++)
#define vi vector<int>
#define vvi vector<vi>
#define rloop(i,a,b) for(ll i=a;i>b;i--)
#define mp make_pair
#define pb push_back
#define ppb pop_back
#define pii pair<int,int>
#define mii map<int,int>
#define mll map<long long,long long>
#define msi map<string,int>
#define vpii vector<pair<int,int>>
#define vll vector<long long>
#define sz(a) int(a.size())
#define last(x) x.end()
#define beg(x) x.begin()
#define all(x) begin(x),end(x)
#define FindInTree(m,n) m.find(n)!=m.end()
#define ull unsigned long long
#define inp(a,n) loop(i,0,n) cin>>a[i]
#define db1(x) cerr<<#x<<" = "<<x<<endl
#define db2(x,y) cerr<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define db3(x,y,z) cerr<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<" "<<#z<<" = "<<z<<endl
#define divs(n,m) ((m!=0)&&(n%m==0))
#define sum(container,value) accumulate(begin(container),end(container),value)
#define tr(container,it)\
for(__typeof(container.begin()) it=container.begin();it!=container.end();it++)
#define print(container) tr(container,it){cout<<*it<<" ";cout.flush();}cout<<endl
#define printarr(a,n) loop(i,0,n){ cout<<a[i]<<" ";cout.flush(); }cout<<endl
#define ordered_set(datatype,comp) tree<datatype, null_type, comp<datatype>, rb_tree_tag, tree_order_statistics_node_update>
#pragma GCC optimise ("Ofast")
const int mod=1e9+7;
const int N=1e5+5;
const double PI=3.14159265358979311600;
//int Ecycle[N];
ll binomialCoeff(ll n, ll k)
{
ll res = 1;
if(n<k)
return 0;
if ( k > n - k )
k = n - k;
for (ll i = 0; i < k; ++i)
{
res *= (n - i);
res /= (i + 1);
}
return res;
}
ll countDigitsAccurate(ll num,int base)
{
ll cnt=0;
while(num>0)
{
cnt++;
num=num/base;
}
return cnt;
}
vll generate(ll m,ll n)
{
vll ans;
if(n>m)
{
ans.pb(1);
return ans;
}
/*if(n==2)
{
ans.pb(1);
ans.pb(2);
return ans;
}
else if(n>2)
{
v.pb(1);
v.pb(n);
loop(i,2,n-1)
v.pb(i);
ans=v;
ans.pb(n-1);
//print(ans);
reverse(beg(v),last(v));
//copy(beg(v),last(v),last(ans));
for(ll u:v)
ans.pb(u);
//print(ans);
ans.ppb();
}*/
loop(i,n+1,m+1)
{
ans.pb(n);
ans.pb(i);
}
return ans;
}
vll processBeg(ll m,ll l,ll lb)
{
ll begPtr;
vll vlb;
vlb=generate(m,lb);
//stack<int> stk;
//cout<<"lb is: "<<lb<<endl;
//print(vlb);
begPtr=(2*(lb-1)*m-(lb-1)*(lb-1)-(lb-1))+1;
//while(begPtr<l)
//{
//stk.pop();
//begPtr++;
//}
vll::iterator it=beg(vlb);
advance(it,l-begPtr);
vll ans(it,last(vlb));
return ans;
}
vll processMiddle(ll m,ll start,ll stop)
{
vll ans,segment;
loop(i,start,stop+1)
{
segment=generate(m,i);
//copy(beg(segment),last(segment),last(ans));
for(ll u:segment)
ans.pb(u);
}
return ans;
}
vll processEnd(ll m,ll r,ll ub)
{
ll endPtr;
vll vub;
vub=generate(m,ub);
if(ub>m)
return vub;
endPtr=2*ub*m-ub*ub-ub;
while(endPtr>r)
{
vub.ppb();
endPtr--;
}
return vub;
}
ll findInterval(ll bnd,ll m)
{
ll interval,Discriminant;
//db1(m*m);
//db1(m*m-m);
Discriminant=4*(m*m-m-bnd)+1;
if(Discriminant<0)
return m+1;
//db1(Discriminant);
Discriminant=sqrtl(Discriminant);
interval=ceil(m-(1+Discriminant)/2.0);
//db1(Discriminant);
//db1(interval);
//if((interval*(interval-1))==n)
//interval--;
return interval;
}
/*void preProcess()
{
int term,lim,ptr,L,R,mid;
cout<<"In preProcess function"<<endl;
term=sqrt(N);
lim=2*term+1;
Ecycle[1]=1;
ptr=3;
L=2;
R=4;
while(ptr<=lim&&R<99855)
{
cout<<L<<" "<<R<<endl;
mid=(R+L)/2;
loop(i,L,mid)
{
if((i-L)!=1)
Ecycle[i]=Ecycle[i-(ptr-2)];
else
Ecycle[i]=(ptr+1)/2;
}
Ecycle[mid]=(ptr-1)/2;
loop(i,mid+1,R+1)
{
Ecycle[i]=Ecycle[2*mid-i];
}
ptr+=2;
L=R+1;
R=L+ptr-1;
}
Ecycle[3]=2;
loop(i,1,50)
{
cout<<Ecycle[i]<<" ";
}
cout<<endl;
}*/
void solve()
{
//Declare your variables here.
ll m,l,r,endPtr,lb,ub;
vll left,middle,right,ans;
//Do not assign values to the variables here!!!
cin>>m>>l>>r;
lb=findInterval(l,m);
ub=findInterval(r,m);
db2(lb,ub);
if(ub<lb)
return;
left=processBeg(m,l,lb);
if((ub-lb)>1)
middle=processMiddle(m,lb+1,ub-1);
right=processEnd(m,r,ub);
//print(left);
//print(middle);
//print(right);
if((ub-lb)>1)
{
ans=left;
for(ll u:middle)
ans.pb(u);
for(ll u:right)
ans.pb(u);
}
else if((ub-lb)==1)
{
ans=left;
//copy(beg(left),last(left),last(ans));
//ans.resize(sz(right)+sz(ans)+5);
//copy(beg(right),last(right),last(ans));
for(ll u:right)
ans.pb(u);
}
else
{
endPtr=2*ub*m-ub*ub-ub;
ans=left;
if(ub<=m)
{
while(endPtr>r)
{
ans.ppb();
endPtr--;
}
}
}
/*loop(i,l,r+1)
{
cout<<Ecycle[i]<<" ";
}
cout<<endl;
*/
//cout<<"answer is:"<<endl;
print(ans);
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
//#ifndef ONLINE_JUDGE
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
//freopen("error.txt","w",stderr);
//#endif
int t=1;
//cin>>t;
//preProcess();
while(t--)
solve();
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.time.Period;
public class Main {
public static void main(String[] args) throws Exception {
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long l=sc.nextLong();
long r=sc.nextLong();
int number =2;
int i=1;
while(l-i*2>0) {
number++;
l-=i*2;
r-=i*2;
i++;
}
for(;l<=r;l++) {
if(l%2==0) {
pw.print(number);
}else {
if(l==1)
pw.print(1);
else
pw.print((l+1)/2);
}
if(l!=r) {
pw.print(" ");
}
if(l-i*2>=0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
}
pw.println();
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextLongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
double x;
double y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair)o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Double(x).hashCode() * 31 + new Double(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return (int) (this.y - other.y);
} else {
return (int) (this.x - other.x);
}
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
if (a == 0)
return b;
return (a > b) ? GCD(a % b, b) : GCD(a, b % a);
}
public static long LCM(long a, long b) {
return a * b / GCD(a, b);
}
static long Pow(long a, int e, int mod) // O(log e)
{
a %= mod;
long res = 1;
while (e > 0) {
if ((e & 1) == 1)
res = (res * a) % mod;
a = (a * a) % mod;
e >>= 1;
}
return res;
}
static long nc(int n, int r) {
if (n < r)
return 0;
long v = fac[n];
v *= Pow(fac[r], mod - 2, mod);
v %= mod;
v *= Pow(fac[n - r], mod - 2, mod);
v %= mod;
return v;
}
public static boolean isprime(long a) {
if (a == 0 || a == 1) {
return false;
}
if (a == 2) {
return true;
}
for (int i = 2; i < Math.sqrt(a) + 1; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean isPal(String s) {
boolean t = true;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
t = false;
break;
}
}
return t;
}
public static long RandomPick(long[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static int RandomPick(int[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static void PH(String s, boolean reverse) {
prelen = s.length();
HashsArray[HashsArrayInd] = new int[prelen + 1];
prepow = new int[prelen];
if (HashsArrayInd == 0) {
int[] mods = { 1173017693, 1173038827, 1173069731, 1173086977, 1173089783, 1173092147, 1173107093,
1173114391, 1173132347, 1173144367, 1173150103, 1173152611, 1173163993, 1173174127, 1173204679,
1173237343, 1173252107, 1173253331, 1173255653, 1173260183, 1173262943, 1173265439, 1173279091,
1173285331, 1173286771, 1173288593, 1173298123, 1173302129, 1173308827, 1173310451, 1173312383,
1173313571, 1173324371, 1173361529, 1173385729, 1173387217, 1173387361, 1173420799, 1173421499,
1173423077, 1173428083, 1173442159, 1173445549, 1173451681, 1173453299, 1173454729, 1173458401,
1173459491, 1173464177, 1173468943, 1173470041, 1173477947, 1173500677, 1173507869, 1173522919,
1173537359, 1173605003, 1173610253, 1173632671, 1173653623, 1173665447, 1173675577, 1173675787,
1173684683, 1173691109, 1173696907, 1173705257, 1173705523, 1173725389, 1173727601, 1173741953,
1173747577, 1173751499, 1173759449, 1173760943, 1173761429, 1173762509, 1173769939, 1173771233,
1173778937, 1173784637, 1173793289, 1173799607, 1173802823, 1173808003, 1173810919, 1173818311,
1173819293, 1173828167, 1173846677, 1173848941, 1173853249, 1173858341, 1173891613, 1173894053,
1173908039, 1173909203, 1173961541, 1173968989, 1173999193};
mod = RandomPick(mods);
int[] primes = { 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
prime = RandomPick(primes);
}
prepow[0] = 1;
if (!reverse) {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[i]) % mod) % mod);
}
} else {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[prelen - 1 - i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[prelen - 1 - i]) % mod) % mod);
}
}
HashsArrayInd++;
}
public static int PHV(int l, int r, int n, boolean reverse) {
if (l > r) {
return 0;
}
int val = (int) ((1l * HashsArray[n - 1][r] + mod - HashsArray[n - 1][l - 1]) % mod);
if (!reverse) {
val = (int) ((1l * val * prepow[prelen - l]) % mod);
} else {
val = (int) ((1l * val * prepow[r - 1]) % mod);
}
return val;
}
static int[][] HashsArray;
static int HashsArrayInd = 0;
static int[] prepow;
static int prelen = 0;
static int prime = 31;
static long fac[];
static int mod = 998244353;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Solution{
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] input=br.readLine().trim().split(" ");
int numTestCases=Integer.parseInt(input[0]);
while(numTestCases-->0){
input=br.readLine().trim().split(" ");
int n=Integer.parseInt(input[0]);
long l=Long.parseLong(input[1]);
long r=Long.parseLong(input[2]);
printSequence(n,l,r);
}
out.flush();
out.close();
}
public static void printSequence(int n,long l,long r)
{
ArrayList<Integer> ans=new ArrayList<>();
long totalElements=0;
int blockNumber=-1;
for(int i=1;i<n;i++){
totalElements+=(2*(n-i));
if(totalElements>l)
{
totalElements-=(2*(n-i));
blockNumber=i;
break;
}
}
long pos=totalElements+1;
for(int i=blockNumber;i<n && pos<=r && blockNumber!=-1;i++){
for(int j=i+1;j<=n;j++){
if(pos>=l && pos<=r)
{
ans.add(i);
}
pos++;
if(pos>=l && pos<=r){
ans.add(j);
}
pos++;
}
}
if(pos<=r){
ans.add(1);
}
for(int i=0;i<ans.size();i++){
out.print(ans.get(i)+" ");
}
out.println();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const char sp = ' ', nl = '\n';
int read() {
int s = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
s = (s << 3) + (s << 1) + ch - '0', ch = getchar();
}
return s * f;
}
template <typename T>
void read(T &s) {
s = 0;
char ch = getchar();
long long f = 1;
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
s = (s << 3) + (s << 1) + ch - 48, ch = getchar();
s *= f;
}
template <typename T, typename... A>
void read(T &f, A &...a) {
read(f);
read(a...);
}
void sc() {}
template <class T, class... A>
void sc(T &t, A &...a) {
cin >> t, sc(a...);
}
void pr() {}
template <class T, class... A>
void pr(T t, A... a) {
cout << t, pr(a...);
}
const int mod = 1e9 + 7, base = 131;
int t, n;
long long l, r;
int main() {
t = read();
while (t--) {
read(n, l, r);
int st = 1;
long long cnt = 0;
for (int i = 1; i < n; i++) {
if (cnt + (2 * (n - i)) <= l) {
st = i + 1;
cnt += 2 * (n - i);
} else
break;
}
while (cnt < r) {
if (st == n) {
cnt++;
if (cnt >= l && cnt <= r) pr(1, sp);
}
for (int i = st + 1; i <= n; i++) {
cnt++;
if (cnt >= l && cnt <= r) pr(st, sp);
cnt++;
if (cnt >= l && cnt <= r) pr(i, sp);
}
st++;
}
puts("");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
def main():
for _ in inputt():
n, l, r = inputi()
i = 1
l -= 1
if l == n * (n - 1):
print(1)
continue
while l >= 2 * (n - i):
l -= 2 * (n - i)
r -= 2 * (n - i)
i += 1
j = i + 1 + l // 2
while l < r:
if l % 2:
print(j, end = " ")
j += 1
if j > n:
i += 1
j = i + 1
elif r != n * (n - 1) + 1:
print(i, end = " ")
else:
print(1, end = " ")
l += 1
print()
# region M
# region fastio
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
for x in args:
file.write(str(x))
file.write(kwargs.pop("end", "\n"))
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
# region import
inputt = lambda t = 0: range(t) if t else range(int(input()))
inputi = lambda: map(int, input().split())
inputl = lambda: list(inputi())
from math import *
from heapq import *
from bisect import *
from itertools import *
from functools import reduce, lru_cache
from collections import Counter, defaultdict
import re, copy, operator, cmath
from builtins import *
# endregion
# region main
if __name__ == "__main__":
main()
# endregion
# endregion
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import javax.transaction.xa.Xid;
public class tr1 {
static PrintWriter out;
static StringBuilder sb;
static int n, m;
static long mod = 998244353;
static int[][] memo;
static String s;
static HashSet<Integer> nodes;
static HashSet<Integer>[] ad, tree;
static boolean[] vis, taken;
static int[] a;
static TreeSet<Long> al;
static long[] val;
static ArrayList<String> aa;
static char[] b;
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
out = new PrintWriter(System.out);
int t = sc.nextInt();
while (t-- > 0) {
n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
int[] ar = new int[(int) (r - l + 1)];
int id = 1;
int af = 0;
long ll = l;
int las = 0;
while (id<n) {
long num = (n - id) * 2l;
// System.out.println(ll+" "+num);
if (ll <= num) {
if (ll % 2 == 0) {
af = (int) (ll / 2) + 1;
} else {
af = id;
las = (int) (ll / 2) + 2;
}
break;
}
ll -= num;
id++;
}
// System.out.println(id+" "+af+" "+las);
if (af == id) {
ar[0] = id;
if(ar.length>1)
ar[1] = las;
af = ++las;
for (int i = 2; i < ar.length; i += 2) {
if (af > n) {
id++;
af = id + 1;
// System.out.println(i+" "+id+" "+af);
}
// System.out.println(i+" "+id+" "+af);
ar[i] = id;
if (i + 1 < ar.length)
ar[i + 1] = af;
af++;
}
} else {
ar[0] = af;
if (af == n) {
if(ar.length>1)
ar[1] = id++;
if(ar.length>2)
ar[2] = id + 1;
af = id + 1;
af++;
} else {
if(ar.length>1)
ar[1] = id;
if(ar.length>2)
ar[2] = ++af;
af++;
}
for (int i = 3; i < ar.length; i += 2) {
if (af > n) {
id++;
af = id + 1;
}
ar[i] = id;
if (i + 1 < n)
ar[i + 1] = af;
af++;
}
}
if (r == n * 1l * (n - 1) + 1)
ar[ar.length-1] = 1;
for (int i = 0; i < ar.length; i++)
out.print(ar[i] + " ");
out.println();
}
out.flush();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream system) {
br = new BufferedReader(new InputStreamReader(system));
}
public Scanner(String file) throws Exception {
br = new BufferedReader(new FileReader(file));
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public String nextLine() throws IOException {
return br.readLine();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public char nextChar() throws IOException {
return next().charAt(0);
}
public Long nextLong() throws IOException {
return Long.parseLong(next());
}
public int[] nextArrInt(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextArrLong(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
public void waitForInput() throws InterruptedException {
Thread.sleep(3000);
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
size_t T;
cin >> T;
while (T--) {
int n, l, r;
cin >> n >> l >> r;
int t = 1;
int k = 1;
while (k < l && t != n) {
k += 2 * (n - t++);
}
if (k < l) {
cout << 1 << endl;
continue;
}
if (k > l) {
k -= 2 * (n - --t);
}
int difference = l - k;
int c = difference / 2 + t + 1;
if (difference % 2 == 0) {
int current = l;
while (current <= r) {
cout << t << " ";
++current;
if (current <= r) {
cout << c++ << " ";
++current;
if (c > n) {
c = ++t + 1;
if (t == n) {
if (current <= r) cout << 1;
break;
}
}
}
}
} else {
cout << c << " ";
++l;
difference = l - k;
c = difference / 2 + t + 1;
if (c > n) {
c = t++ + 1;
if (t == n) {
cout << 1 << endl;
break;
}
}
int current = l;
while (current <= r) {
cout << t << " ";
++current;
if (current <= r) {
cout << c++ << " ";
++current;
if (c > n) {
++t;
c = t + 1;
if (t == n) {
if (current <= r) cout << 1;
break;
}
}
}
}
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class D {
static void shuffleArray(int[] arr){
Random rnd = new Random();
for(int i = arr.length; i>0; i--){
int ndx = rnd.nextInt(i+1);
int tmp = arr[ndx];
arr[ndx] = arr[i];
arr[i] = tmp;
}
}
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb = new StringBuilder();
int tst = Integer.parseInt(br.readLine());
//int tst = 1;
while(tst-->0){
String[] str = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
long l = Long.parseLong(str[1]), r = Long.parseLong(str[2]);
int now = 0, i = 1;
for(; i<=n; i++){
now += 2*(n-i);
if(i == n) break;
if(now>=l){
now -= 2*(n-i);
break;
}
}
if(i == n) sb.append(1).append('\n');
else{
now++;
outer:for(; i<=n; i++){
if(i == n && now == r) sb.append(1);
for(int j = i+1; j<=n; j++){
if(now>r) break outer;
if(now>=l) sb.append(i+" ");
now++;
if(now>=l) sb.append(j+" ");
now++;
}
}
sb.append('\n');
}
}
System.out.println(sb);
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<int> s;
int n;
int solve(long long x) {
if (x > s[n - 1]) {
return 1;
}
auto id = lower_bound(s.begin() + 1, s.end(), x) - s.begin();
int pos = x - s[id - 1];
if (pos & 1) {
return id;
} else {
return (pos >> 1) + id;
}
}
int main() {
int tt;
cin >> tt;
while (tt--) {
long long l, r;
cin >> n >> l >> r;
s.assign(n + 1, 0);
for (int i = 1; i <= n; i++) {
s[i] = s[i - 1] + (2 * (n - i));
}
for (long long i = l; i <= r; i++) {
cout << solve(i) << " \n"[i == r];
}
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class eulercycc {
/*
* @return Index of leftmost number >=key. Inclusive
*/
/*
private static int bsLowerBound(int[] a, int key) {
// Modified Arrays.binarySearch
int low = 0;
int high = a.length - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
int midVal = a[mid];
int cmp = midVal - key;
if (cmp < 0)
low = mid + 1;
else if (cmp > 0)
high = mid - 1;
else if (mid != 0 && a[mid-1]==midVal) { // not lower bound
high = mid-1;
}
else
return mid; // key found
}
return high+1; // key not found, returns number before
}*/
/**
* @return Index of rightmost number <=key. Inclusive
*/
private static long bsLowerBound(int high, long key) {
// Modified Arrays.binarySearch
int low = 0;
while (low <= high) {
int mid = (low + high) >>> 1;
long cmp = mid * (mid + 1L) - key;
if (cmp < 0) {
low = mid + 1;
}
else if (cmp > 0) {
high = mid - 1;
}
else {
return mid; // key found
}
}
return high + 1; // key not found, returns number after
}
public static void main(String[] args) throws Exception {
R in = new R();
int TESTCASES = in.nextInt();
StringBuilder out = new StringBuilder();
for (int TC = 0; TC < TESTCASES; TC++) {
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
long p = bsLowerBound(n+69, l);
p-=3;
for (long i = l; i <= r; i++) {
// n-1 inside the partition
while (i > p*(p+1)) {
p++;
}
if ((i&1)==0) {
// even
out.append(p+1).append(' ');
} else {
out.append( (i+1-p*(p-1)) >> 1 ).append(' ');
}
}
out.setCharAt(out.length()-1, '\n');
}
System.out.print(out);
System.out.flush();
}
//<editor-fold desc="R">
/**
* This class is for fast input. Please ignore.
*/
public static class R {
private BufferedReader br;
/**
* Should be set to null at end of line
*/
private StringTokenizer st;
public R() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public R(String filename) throws IOException {
br = new BufferedReader(new FileReader(filename + ".in"));
}
public R(BufferedReader reader) {
br = reader;
}
public BufferedReader getReader() {
return br;
}
public StringTokenizer getStringTokenizer() {
return st;
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens()) {
String s = br.readLine();
if (s == null) return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
/**
* Note:
* CAN MODIFY the BufferedReader's location and the string tokenizer!!!
* Recommended to only use with next().
*/
public boolean lineHasNext() throws IOException {
if (st == null) {
String s = br.readLine();
if (s == null) return false;
st = new StringTokenizer(s);
}
return st.hasMoreTokens();
}
/**
* Note:
* CAN MODIFY the BufferedReader's location and the string tokenizer!!!
* Recommended to only use with next().
*/
public boolean hasNext() throws IOException {
while (st == null || !st.hasMoreTokens()) {
String s = br.readLine();
if (s == null) return false;
st = new StringTokenizer(s);
}
return true;
}
/**
* Skips a line. Sets st to null if has tokens left, and otherwise
* reads a line.
*/
public void skipLine() throws IOException {
if (st == null || !st.hasMoreTokens()) {
br.readLine(); // Otherwise, would do nothing.
}
st = null;
}
/**
* This will set st to null, and this ignores current line
*/
public String[] nextLine() throws IOException {
String s = br.readLine();
if (s == null) return null;
st = new StringTokenizer(s);
ArrayList<String> result = new ArrayList<>();
while (st.hasMoreTokens()) {
result.add(st.nextToken());
}
st = null;
return result.toArray(new String[0]);
}
/**
* sets st to null!
*/
public String[] nextTower(int lines) throws IOException {
String[] tower = new String[lines];
st = null;
for (int i = 0; i < lines; i++) {
tower[i] = br.readLine();
}
return tower;
}
public int[] nextIntLine() throws IOException {
return intArr(nextLine());
}
public long[] nextLongLine() throws IOException {
return longArr(nextLine());
}
public int[] nextIntTower(int lines) throws IOException {
return intArr(nextTower(lines));
}
public long[] nextLongTower(int lines) throws IOException {
return longArr(nextTower(lines));
}
public int[] intArr(String[] strings) throws IOException {
int[] ints = new int[strings.length];
int i = 0;
for (String s : strings) {
ints[i] = Integer.parseInt(s);
i++;
}
return ints;
}
public long[] longArr(String[] strings) throws IOException {
long[] longs = new long[strings.length];
int i = 0;
for (String s : strings) {
longs[i] = Long.parseLong(s);
i++;
}
return longs;
}
public double[] doubleArr(String[] strings) {
double[] doubles = new double[strings.length];
int i = 0;
for (String s : strings) {
doubles[i] = Double.parseDouble(s);
i++;
}
return doubles;
}
/**
* This will set st to null
*/
public char[] nextCharArray() throws IOException {
st = null;
String s = br.readLine();
return s == null ? null : s.toCharArray();
}
/**
* This will set st to null
* Boolean at pos i true if char at pos i == c
*/
public boolean[] nextBoolArray(char c) throws IOException {
char[] chars = nextCharArray();
if (chars == null) return null;
boolean[] booleans = new boolean[chars.length];
for (int i = 0; i < chars.length; i++) {
booleans[i] = chars[i] == c;
}
return booleans;
}
public int[][] next2Dint(int lines) throws IOException {
int[][] result = new int[lines][];
for (int i = 0; i < lines; i++) {
result[i] = nextIntLine();
}
return result;
}
public long[][] next2Dlong(int lines) throws IOException {
long[][] result = new long[lines][];
for (int i = 0; i < lines; i++) {
result[i] = nextLongLine();
}
return result;
}
public char[][] next2Dchar(int lines) throws IOException {
char[][] result = new char[lines][];
for (int i = 0; i < lines; i++) {
result[i] = nextCharArray();
}
return result;
}
public boolean[][] next2Dbool(int lines, char c) throws IOException {
boolean[][] result = new boolean[lines][];
for (int i = 0; i < lines; i++) {
result[i] = nextBoolArray(c);
}
return result;
}
}
//</editor-fold>
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int maxc = 1e5;
int main() {
int t;
cin >> t;
while (t--) {
int k = 0;
int n, l, r;
cin >> n >> l >> r;
for (int i = 2 * (n - 1); i; k += i, i -= 2) {
for (int j = max(l, k + 1); j <= min(r, k + i); j++) {
if (j % 2)
cout << n - i / 2 << " ";
else
cout << n - i / 2 + (j - k) / 2 << " ";
}
}
if (k + 1 == r) cout << 1;
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define pb push_back
#define mem(a,x) memset(a,x,sizeof(a))
#define nl "\n"
#define fout(x) fixed<<setprecision(x)
#define one(x) __builtin_popcountll(x)
#define F first
#define S second
#define or_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define fast ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
// find_by_order(), order_of_key()
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << fout(10)<<x;}
void __print(double x) {cerr << fout(10)<<x;}
void __print(long double x) {cerr << fout(10)<<x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif
const int mod=1e9+7;
const double eps=1e-9;
const double PI=acos(-1.0);
ll qpow(ll n,ll k){ll ans=1;assert(k>=0);n%=mod;while(k>0){if(k&1)ans=(ans*n)%mod;n=(n*n)%mod;k>>=1;}return ans%mod;}
// ****************************************************
const int maxn=300010;
ll arr1[maxn], arr2[maxn], arr3[maxn];
int main()
{
fast;
int tc;
cin>>tc;
while(tc--){
int n;
cin>>n;
for(int i=0; i<n; i++) cin>>arr1[i]>>arr2[i];
ll tot=0;
for(int i=0; i<n; i++){
ll temp;
if(i==0) temp=arr2[n-1];
else temp=arr2[i-1];
arr3[i]=max(0ll, arr1[i]-temp);
tot+=arr3[i];
}
ll ans=1e18;
for(int i=0; i<n; i++){
ans=min(ans, tot-arr3[i]+arr1[i]);
}
cout<<ans<<endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.*;
public class D {
void solve() throws Exception {
long n = nl();
long l = nl();
long r = nl();
long h = 0L;
long b = 1L;
for (long i = n - 1; h + 2 * i < l && i > 0; i--) {
h += 2*i;
b++;
}
long c = b;
if (b == n) {
out.println(1);
return;
}
while (h < l) {
h++;
if ((h & 1) == 1) {
c++;
}
}
while (h <= r) {
if ((h & 1) == 1) {
out.print(b + " ");
} else {
out.print(c + " ");
c++;
if (c > n) {
b++;
c = b + 1;
if (b == n) {
out.println(1);
return;
}
}
}
h++;
}
out.println();
}
void run() throws Exception {
initIO();
int t = ni();
while (t-- > 0) {
solve();
out.flush();
}
}
public static void main(String[] args) throws Exception {
new D().run();
}
/**
* IO.
*/
PrintWriter out;
BufferedReader br;
StringTokenizer tokenizer;
void initIO() {
out = new PrintWriter(System.out);
br = new BufferedReader(new InputStreamReader(System.in));
tokenizer = null;
}
int ni() throws Exception {
return Integer.parseInt(ns());
}
long nl() throws Exception {
return Long.parseLong(ns());
}
double nd() throws Exception {
return Double.parseDouble(ns());
}
int[] nia(int n) throws Exception {
int[] x = new int[n];
for (int i = 0; i < x.length; i++) {
x[i] = ni();
}
return x;
}
long[] nla(int n) throws Exception {
long[] x = new long[n];
for (int i = 0; i < x.length; i++) {
x[i] = nl();
}
return x;
}
double[] nda(int n) throws Exception {
double[] x = new double[n];
for (int i = 0; i < x.length; i++) {
x[i] = nd();
}
return x;
}
String ns() throws Exception {
while (tokenizer == null || !tokenizer.hasMoreTokens())
tokenizer = new StringTokenizer(br.readLine());
return tokenizer.nextToken();
}
String nline() throws Exception {
tokenizer = null;
return br.readLine();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long N = 500005;
const long long M = 1000000007;
int main() {
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> v;
v.push_back(1);
for (long long i = 1; i < n; i++) {
v.push_back(2 * i);
}
reverse(v.begin(), v.end());
vector<long long> v1;
v1.push_back(v[0]);
long long sum = v[0];
for (long long i = 1; i < n; i++) {
sum += v[i];
v1.push_back(sum);
}
vector<long long>::iterator it = lower_bound(v1.begin(), v1.end(), l);
vector<long long>::iterator it1 = lower_bound(v1.begin(), v1.end(), r);
long long z = it - v1.begin();
long long z1 = it1 - v1.begin();
long long num = v1[z];
long long num1 = v1[z1];
if (z == n - 1) {
cout << 1 << endl;
} else {
long long chk = 0;
long long a, b, c, d;
if (l % 2 == 0) {
long long w = num - l;
b = n - w / 2;
a = z + 1;
} else {
a = z + 1;
long long w = num - l - 1;
b = n - w / 2;
}
if (z1 == n - 1) {
chk = 1;
} else {
c = z1 + 1;
long long w = num1 - r;
d = n - w / 2;
}
if (chk == 1) {
if (l % 2 == 0) {
cout << b << " ";
long long nu = b + 1;
while (nu <= n) {
cout << a << " " << nu << " ";
nu++;
}
} else {
long long nu = b;
while (nu <= n) {
cout << a << " " << nu << " ";
nu++;
}
}
for (long long j = a + 1; j < n; j++) {
long long nu = j + 1;
while (nu <= n) {
cout << j << " " << nu << " ";
nu++;
}
}
cout << 1 << " ";
} else if (a == c) {
if (l % 2 == 0 && r % 2 == 0) {
cout << b << " ";
long long nu = b + 1;
while (nu <= d) {
cout << a << " " << nu << " ";
nu++;
}
} else if (l % 2 == 0 && r % 2 != 0) {
cout << b << " ";
long long nu = b + 1;
while (nu < d) {
cout << a << " " << nu << " ";
nu++;
}
cout << a << " ";
} else if (l % 2 != 0 && r % 2 == 0) {
long long nu = b;
while (nu <= d) {
cout << a << " " << nu << " ";
nu++;
}
} else {
long long nu = b;
while (nu < d) {
cout << a << " " << nu << " ";
nu++;
}
cout << 1 << " ";
}
} else {
if (l % 2 == 0) {
cout << b << " ";
long long nu = b + 1;
while (nu <= n) {
cout << a << " " << nu << " ";
nu++;
}
} else {
long long nu = b;
while (nu <= n) {
cout << a << " " << nu << " ";
nu++;
}
}
if (a + 1 < c) {
for (long long j = a + 1; j < c; j++) {
long long nu = j + 1;
while (nu <= n) {
cout << j << " " << nu << " ";
nu++;
}
}
}
if (r % 2 == 0) {
long long nu = c + 1;
while (nu <= d) {
cout << c << " " << nu << " ";
nu++;
}
} else {
long long nu = c + 1;
while (nu < d) {
cout << c << " " << nu << " ";
nu++;
}
cout << c << " ";
}
}
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.math.*;
import java.awt.Point;
public class Main {
//static final long MOD = 998244353L;
//static final long INF = 1000000000000000007L;
static final long MOD = 1000000007L;
static final int INF = 1000000007;
//static long[] factorial;
public static void main(String[] args) {
FastScanner sc = new FastScanner();
PrintWriter pw = new PrintWriter(System.out);
int Q = sc.ni();
for (int q = 0; q < Q; q++) {
int N = sc.ni();
long L = sc.nl()-1;
long R = sc.nl()-1;
if (L == (N+0L)*(N-1)) {
pw.println(1);
} else {
long index = 0;
int first = 1;
int second = -1;
while (true) {
if (L >= index + 2*(N-first)) {
index += 2*(N-first);
first++;
} else {
second = first+1;
while (L >= index+2) {
index += 2;
second++;
}
break;
}
}
long max = Math.min(R,(N+0L)*(N-1)-1);
while (index <= max) {
if (index == L-1) {
pw.print(second + " ");
index++;
if (second == N) {
first++;
second = first+1;
} else {
second++;
}
continue;
}
if (index == max) {
pw.print(first + " ");
break;
} else if (index+1 == max) {
pw.print(first + " " + second + " ");
break;
} else {
pw.print(first + " " + second + " ");
if (second == N) {
first++;
second = first+1;
} else {
second++;
}
index += 2;
}
}
if (R > max) {
pw.println(1);
} else {
pw.println();
}
}
}
pw.close();
}
public static long dist(long[] p1, long[] p2) {
return (Math.abs(p2[0]-p1[0])+Math.abs(p2[1]-p1[1]));
}
//Find the GCD of two numbers
public static long gcd(long a, long b) {
if (a < b) return gcd(b,a);
if (b == 0)
return a;
else
return gcd(b,a%b);
}
//Fast exponentiation (x^y mod m)
public static long power(long x, long y, long m) {
if (y < 0) return 0L;
long ans = 1;
x %= m;
while (y > 0) {
if(y % 2 == 1)
ans = (ans * x) % m;
y /= 2;
x = (x * x) % m;
}
return ans;
}
public static int[] shuffle(int[] array) {
Random rgen = new Random();
for (int i = 0; i < array.length; i++) {
int randomPosition = rgen.nextInt(array.length);
int temp = array[i];
array[i] = array[randomPosition];
array[randomPosition] = temp;
}
return array;
}
public static long[] shuffle(long[] array) {
Random rgen = new Random();
for (int i = 0; i < array.length; i++) {
int randomPosition = rgen.nextInt(array.length);
long temp = array[i];
array[i] = array[randomPosition];
array[randomPosition] = temp;
}
return array;
}
public static int[][] shuffle(int[][] array) {
Random rgen = new Random();
for (int i = 0; i < array.length; i++) {
int randomPosition = rgen.nextInt(array.length);
int[] temp = array[i];
array[i] = array[randomPosition];
array[randomPosition] = temp;
}
return array;
}
public static int[][] sort(int[][] array) {
//Sort an array (immune to quicksort TLE)
Arrays.sort(array, new Comparator<int[]>() {
@Override
public int compare(int[] a, int[] b) {
return a[1]-b[1]; //ascending order
}
});
return array;
}
public static long[][] sort(long[][] array) {
//Sort an array (immune to quicksort TLE)
Random rgen = new Random();
for (int i = 0; i < array.length; i++) {
int randomPosition = rgen.nextInt(array.length);
long[] temp = array[i];
array[i] = array[randomPosition];
array[randomPosition] = temp;
}
Arrays.sort(array, new Comparator<long[]>() {
@Override
public int compare(long[] a, long[] b) {
if (a[0] < b[0])
return -1;
else if (a[0] > b[0])
return 1;
else
return 0;
}
});
return array;
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int ni() {
return Integer.parseInt(next());
}
long nl() {
return Long.parseLong(next());
}
double nd() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
def c(sm, a1):
sm = n*a1+(n-1)*n
for _ in range(N()):
n, l, r = RL()
num = 2
while num**2-num<l:
num+=1
s = l-(num-n-1)**2-(num-n-1)
res = []
for i in range(1, num):
res.append(i)
res.append(num)
for i in range(1, num):
res.append(i)
res.append(num+1)
# print(res, l, r, num, l-s, )
# print(len(res), l-s, res[0])
print(" ".join([str(i) for i in res[l-s:l-s+(r-l)+1]]))
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long cur = 1, odd = 1, num = 2, ye = 1, add = 2;
while (cur < l) {
cur += odd;
cur++;
if (cur > l) {
cur--;
cur -= odd;
break;
}
odd += 2;
num++;
}
while (cur < l) {
cur++;
if (ye == num) {
ye = add;
} else {
ye = num;
add++;
}
}
for (long long i = l; i <= r; i++) {
cout << ye << " ";
if (ye == num) {
if (add < num) {
ye = add;
} else {
num++;
add = 2;
ye = 1;
}
} else {
if (ye != 1) {
add++;
}
ye = num;
}
}
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x7FFFFFFF;
const long long mod = (0 ? 1000000007 : 998244353);
const double eps = 1e-7;
void work() {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> ans;
long long pos = l;
if (l <= (n - 2) * 2) {
long long f = l & 1;
long long cnt = l / 2 + 1 + f;
while (pos <= r && cnt < n) {
if (f == 1) {
ans.push_back(1);
f = 0;
} else {
ans.push_back(cnt);
cnt++;
f = 1;
}
pos++;
}
}
long long fl = 0;
if (pos <= r) {
ans.push_back(1);
pos++;
}
long long now = n;
while (pos <= r) {
pos++;
if (fl == 0) {
ans.push_back(now);
if (now == 2) {
if (n == 2) {
ans.push_back(1);
break;
}
fl = 1;
now++;
} else
now--;
} else if (fl == 1) {
ans.push_back(now);
if (now == n) {
fl = 2;
}
now++;
} else {
ans.push_back(1);
break;
}
}
for (long long i = 0; i < ans.size(); i++) {
cout << ans[i] << ' ';
}
cout << endl;
}
signed main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
long long t = 1;
cin >> t;
while (t--) {
work();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
//package CodeforcesJava;
import java.io.*;
import java.util.*;
public class Main {
public void solve(InputProvider in, PrintWriter out) throws IOException {
int testCount = in.nextInt();
for (int test = 0; test < testCount; test++) {
long pointCount = in.nextLong();
long firstIndex = in.nextLong();
long lastIndex = in.nextLong();
long firstGroup = 2 * (pointCount - 1);
boolean needLast = lastIndex == countElements(firstGroup, pointCount - 1) + 1;
if (firstIndex < lastIndex) {
if (needLast) {
lastIndex--;
}
long minGuess = 0;
long maxGuess = pointCount - 1;
while (maxGuess > minGuess + 1) {
long middleGuess = (maxGuess + minGuess) / 2;
long middleGuessCount = countElements(firstGroup, middleGuess);
if (firstIndex < middleGuessCount) {
maxGuess = middleGuess;
} else {
minGuess = middleGuess;
}
}
while (countElements(firstGroup, minGuess + 1) < firstIndex) {
minGuess++;
}
long distance = firstIndex - countElements(firstGroup, minGuess);
long levelPoint = minGuess + 1;
long point;
long pairPoint;
if ((distance & 1) == 1) {
point = levelPoint;
pairPoint = (distance + 1) / 2 + levelPoint;
} else {
point = distance / 2 + levelPoint;
out.print(point + " ");
firstIndex++;
if (point == pointCount) {
point = levelPoint + 1;
pairPoint = point + 1;
} else {
pairPoint = point + 1;
point = levelPoint;
}
}
boolean printMain = true;
for (long i = firstIndex; i <= lastIndex; i++) {
if (printMain) {
out.print(point + " ");
printMain = false;
} else {
out.print(pairPoint + " ");
printMain = true;
if (pairPoint < pointCount) {
pairPoint++;
} else {
point++;
pairPoint = point + 1;
}
}
}
}
if (needLast) {
out.print("1 ");
}
out.print("\n");
}
}
private long countElements(long first, long count) {
return (2 * first - 2 * (count - 1)) * count / 2;
}
public static void main(String[] args) throws Exception {
try (InputProvider input = new InputProvider(System.in);
PrintWriter output = new PrintWriter(System.out)) {
new Main().solve(input, output);
}
}
public static class InputProvider implements AutoCloseable {
private final BufferedReader reader;
private StringTokenizer tokenizer;
public InputProvider(Reader reader) {
this.reader = new BufferedReader(reader);
}
public InputProvider(InputStream input) {
reader = new BufferedReader(new InputStreamReader(input));
}
public String next() throws IOException {
if (Objects.isNull(tokenizer) || !tokenizer.hasMoreTokens())
tokenizer = new StringTokenizer(reader.readLine());
return tokenizer.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public String nextLine() throws IOException {
return reader.readLine();
}
@Override
public void close() throws Exception {
reader.close();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
int t=s.nextInt();
StringBuilder sb=new StringBuilder();
for(int i=0;i<t;i++)
{
int n=s.nextInt();
long l=s.nextLong();
long r=s.nextLong();
int left=0;
int right=0;
long ll=0;
long rr=0;
long count=0;
int block=0;
while(count<l&&block<n)
{
block++;
count=count+2*(n-block);
}
left=block;
ll=count-2*(n-block);
count=0;
block=0;
while(count<r&&block<n)
{
block++;
count=count+2*(n-block);
}
right=block;
rr=count;
ArrayList<Integer> list=new ArrayList<>();
for(int j=left;j<=right;j++)
{
fill(list,j,n);
}
long cc=0;
for(long j=ll+1;j<=rr;j++)
{
int now=list.get((int)(j-(ll+1)));
if(j>=l&&j<=rr)
{
cc++;
sb.append(now+" ");
}
}
if(cc==r-l+1)
sb.append("\n");
else
{
sb.append(1+"\n");
}
}
System.out.println(sb);
}
public static void fill(ArrayList<Integer> list,int start,int n)
{
int c=start+1;
for(int i=0;i<2*(n-start);i++)
{
if(i%2==0)
{
list.add(start);
}
else
{
list.add(c);
c++;
}
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<long long> ans;
int main() {
int tests = 1;
int n;
long long l, r;
scanf("%d", &tests);
while (tests--) {
ans.clear();
scanf("%d%lld%lld", &n, &l, &r);
long long s = sqrt(l - 1), t = sqrt(r - 1);
while (s * (s - 1) <= l - 1) s++;
s--;
while (t * (t - 1) <= r - 1) t++;
ans.push_back(1);
for (long long i = s; i < t; ++i) {
for (long long j = 2; j <= i; ++j) ans.push_back(i + 1), ans.push_back(j);
ans.push_back(i + 1);
ans.push_back(1);
}
r -= s * (s - 1) + 1, l -= s * (s - 1) + 1;
for (int i = 0; i < (int)ans.size(); ++i) {
if (i >= l && i <= r) printf("%lld ", ans[i]);
}
puts("");
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class cf1334d {
public static void main(String[] args) throws IOException {
int t = ri();
while(t --> 0) {
int n = rni();
long l = nl() - 1, r = nl();
List<Long> ans = new ArrayList<>();
if(l == 0) {
ans.add(1L);
++l;
}
long ind = 1, start = 1;
while(ind + 2 * start <= l) {
ind += 2 * start++;
}
long st = l - ind;
ind = l;
for(long i = st; i < 2 * start && ind < r; ++i, ++ind) {
if(i % 2 == 0) {
ans.add(start + 1);
} else {
ans.add((i + 1) / 2 == start ? 1 : (i + 3) / 2);
}
}
while(ind < r) {
++start;
for(long i = 0; i < 2 * start && ind < r; ++i, ++ind) {
if(i % 2 == 0) {
ans.add(start + 1);
} else {
ans.add((i + 1) / 2 == start ? 1 : (i + 3) / 2);
}
}
}
prln(ans);
}
close();
}
static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
static StringTokenizer input;
static Random rand = new Random();
// references
// IBIG = 1e9 + 7
// IRAND ~= 3e8
// IMAX ~= 2e10
// LMAX ~= 9e18
// constants
static final int IBIG = 1000000007;
static final int IRAND = 327859546;
static final int IMAX = 2147483647;
static final int IMIN = -2147483648;
static final long LMAX = 9223372036854775807L;
static final long LMIN = -9223372036854775808L;
// util
static int minof(int a, int b, int c) {return min(a, min(b, c));}
static int minof(int... x) {if(x.length == 1) return x[0]; if(x.length == 2) return min(x[0], x[1]); if(x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] < min) min = x[i]; return min;}
static long minof(long a, long b, long c) {return min(a, min(b, c));}
static long minof(long... x) {if(x.length == 1) return x[0]; if(x.length == 2) return min(x[0], x[1]); if(x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] < min) min = x[i]; return min;}
static int maxof(int a, int b, int c) {return max(a, max(b, c));}
static int maxof(int... x) {if(x.length == 1) return x[0]; if(x.length == 2) return max(x[0], x[1]); if(x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] > max) max = x[i]; return max;}
static long maxof(long a, long b, long c) {return max(a, max(b, c));}
static long maxof(long... x) {if(x.length == 1) return x[0]; if(x.length == 2) return max(x[0], x[1]); if(x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] > max) max = x[i]; return max;}
static int powi(int a, int b) {if(a == 0) return 0; int ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static long powl(long a, int b) {if(a == 0) return 0; long ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static int floori(double d) {return (int)d;}
static int ceili(double d) {return (int)ceil(d);}
static long floorl(double d) {return (long)d;}
static long ceill(double d) {return (long)ceil(d);}
static void shuffle(int[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(long[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(double[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static <T> void shuffle(T[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); T swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void rsort(int[] a) {shuffle(a); sort(a);}
static void rsort(long[] a) {shuffle(a); sort(a);}
static void rsort(double[] a) {shuffle(a); sort(a);}
static int randInt(int min, int max) {return rand.nextInt(max - min + 1) + min;}
// input
static void r() throws IOException {input = new StringTokenizer(__in.readLine());}
static int ri() throws IOException {return Integer.parseInt(__in.readLine());}
static long rl() throws IOException {return Long.parseLong(__in.readLine());}
static int[] ria(int n) throws IOException {int[] a = new int[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken()); return a;}
static long[] rla(int n) throws IOException {long[] a = new long[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken()); return a;}
static char[] rcha() throws IOException {return __in.readLine().toCharArray();}
static String rline() throws IOException {return __in.readLine();}
static int rni() throws IOException {input = new StringTokenizer(__in.readLine()); return Integer.parseInt(input.nextToken());}
static int ni() {return Integer.parseInt(input.nextToken());}
static long rnl() throws IOException {input = new StringTokenizer(__in.readLine()); return Long.parseLong(input.nextToken());}
static long nl() {return Long.parseLong(input.nextToken());}
// output
static void pr(int i) {__out.print(i);}
static void prln(int i) {__out.println(i);}
static void pr(long l) {__out.print(l);}
static void prln(long l) {__out.println(l);}
static void pr(double d) {__out.print(d);}
static void prln(double d) {__out.println(d);}
static void pr(char c) {__out.print(c);}
static void prln(char c) {__out.println(c);}
static void pr(char[] s) {__out.print(new String(s));}
static void prln(char[] s) {__out.println(new String(s));}
static void pr(String s) {__out.print(s);}
static void prln(String s) {__out.println(s);}
static void pr(Object o) {__out.print(o);}
static void prln(Object o) {__out.println(o);}
static void prln() {__out.println();}
static void pryes() {__out.println("yes");}
static void pry() {__out.println("Yes");}
static void prY() {__out.println("YES");}
static void prno() {__out.println("no");}
static void prn() {__out.println("No");}
static void prN() {__out.println("NO");}
static void pryesno(boolean b) {__out.println(b ? "yes" : "no");};
static void pryn(boolean b) {__out.println(b ? "Yes" : "No");}
static void prYN(boolean b) {__out.println(b ? "YES" : "NO");}
static void prln(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
static void prln(long... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i); if(n >= 0) __out.println(iter.next());}
static void h() {__out.println("hlfd");}
static void flush() {__out.flush();}
static void close() {__out.close();}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return a > b ? a : b; }
void get_arr(int l, int* a) {
int d = l / 4;
if (l % 4 == 0) d--;
a[0] = 1;
a[1] = d + 3;
a[2] = d + 2;
a[3] = d + 3;
}
long long get_l(vector<int>& ans, long long l, long long s) {
if (l == 2) {
cout << "2 ";
return 3;
}
if (l == 1) {
if (s == 1) {
cout << "1 ";
return 2;
} else {
cout << "1 2 ";
return 3;
}
}
int a[4];
get_arr(l - 2, a);
int ql = (l - 3) % 4;
for (int j = 0, i = ql; j < s && i < 4; j++, i++) cout << a[i] << " ";
return l + (4 - ql);
}
long long solve() {
long long l, r, n;
cin >> n >> l >> r;
long long s = r - l + 1;
vector<int> ans(s);
l = get_l(ans, l, s);
s = r - l + 1;
for (int i = l; s > 0; i++) {
int a[4];
get_arr(l, a);
for (int i = 0; i < 4 && i < s; i++) cout << a[i] << " ";
s = s - 4;
}
cout << endl;
return 0;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.BufferedInputStream;
import java.util.Map;
import java.util.Scanner;
public class Main {
private static Scanner sc = new Scanner(new BufferedInputStream(System.in));
public static void main(String[] args) {
work();
}
private static void work() {
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
long n = sc.nextLong();
long l = sc.nextLong(), r = sc.nextLong();
printResult(n, l, r);
}
}
private static void printResult(long n, long l, long r) {
StringBuffer stringBuffer = new StringBuffer();
long k = ((2 * n - 1) - (long) Math.sqrt((2 * n - 1) * (2 * n - 1) - 4 * l)) / 2;
int cnt = 0;
long p = l, bias;
while (cnt < r - l + 1) {
if (p == n * (n - 1) + 1) {
stringBuffer.append(1).append(' ');
break;
}
bias = p - (2 * n - k - 1) * k;
if (bias % 2 == 1)
stringBuffer.append(k + 1).append(' ');
else
stringBuffer.append(bias / 2 + k + 1).append(' ');
p++;
cnt++;
if (p > (2 * n - k - 2) * (k + 1))
k++;
}
System.out.println(stringBuffer);
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include<bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define ll long long int
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define boost ios_base::sync_with_stdio(false);cin.tie(NULL);
#define inf 9223372036854775807
#define mini 9223372036854775807
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
pair<ll,ll>s4[4]={{-1,0},{1,0},{0,-1},{0,1}};
pair<ll,ll>s8[8]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,1},{1,0},{1,-1}};
ll power(ll a,ll b)
{
if(b==0)
return 1;
ll c=power(a,b/2);
if(b%2==0)
return ((c%mod)*(c%mod))%mod;
else
return ((((c%mod)*(c%mod))%mod)*a)%mod;
}
int main()
{
boost
ll t,i;
cin>>t;
for(i=0; i<t; i++)
{
ll n,l,r;
cin>>n>>l>>r;
ll cnt=1;
ll j=1;
for(j=1; j<=n; j++)
{
ll nxt=cnt+(2*j);
if(nxt>=l)
break;
cnt=nxt;
}
j++;
// cout<<j<<endl;
ll chance=0;
ll cunt=1;
while(cnt<=r)
{
ll val;
if(chance==0)
{
if(cunt>=j)
{
cunt=1;
j++;
}
val=cunt;
cunt++;
}
else
val=j;
chance=1-chance;
if(cnt>=l)
cout<<val<<" ";
cnt++;
}
cout<<endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
// 21:35.396 + 2:22.250 + 2:29.637
public class cf1334d {
public static void main(String[] args) throws IOException {
int t = ri();
while(t --> 0) {
int n = rni();
long l = nl() - 1, r = nl();
List<Long> ans = new ArrayList<>();
long start = 0;
for(long i = max(start++, l); i < 1 && i < r; ++i) {
ans.add(1L);
}
for(long i = max(start, l); i < start + 2 * (n - 2) && i < r; ++i) {
if(i % 2 == 0) {
ans.add(1L);
} else {
ans.add(2 + i / 2);
}
}
start += 2 * (n - 2);
for(long i = n; i >= 3; --i) {
for(long j = max(start++, l); j < start && j < r; ++j) {
ans.add(i);
}
for(long j = max(start, l), parity = (j - start) % 2; j < start + 2 * (i - 3) && j < r; ++j, parity = 1 - parity) {
if(parity == 0) {
ans.add(2 + (j - start) / 2);
} else {
ans.add(i);
}
}
start += 2 * (i - 3);
}
for(long i = max(start, l); i < start + n - 1 && i < r; ++i) {
ans.add(2 + i - start);
}
start += n - 1;
for(long i = max(start++, l); i < start && i < r; ++i) {
ans.add(1L);
}
/* // wrong pattern
if(l == 0) {
ans.add(1L);
++l;
}
long ind = 1, start = 1;
while(ind + 2 * start <= l) {
ind += 2 * start++;
}
long st = l - ind;
ind = l;
for(long i = st; i < 2 * start && ind < r; ++i, ++ind) {
if(i % 2 == 0) {
ans.add(start + 1);
} else {
ans.add((i + 1) / 2 == start ? 1 : (i + 3) / 2);
}
}
while(ind < r) {
++start;
for(long i = 0; i < 2 * start && ind < r; ++i, ++ind) {
if(i % 2 == 0) {
ans.add(start + 1);
} else {
ans.add((i + 1) / 2 == start ? 1 : (i + 3) / 2);
}
}
} */
prln(ans);
}
close();
}
static BufferedReader __in = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter __out = new PrintWriter(new OutputStreamWriter(System.out));
static StringTokenizer input;
static Random rand = new Random();
// references
// IBIG = 1e9 + 7
// IRAND ~= 3e8
// IMAX ~= 2e10
// LMAX ~= 9e18
// constants
static final int IBIG = 1000000007;
static final int IRAND = 327859546;
static final int IMAX = 2147483647;
static final int IMIN = -2147483648;
static final long LMAX = 9223372036854775807L;
static final long LMIN = -9223372036854775808L;
// util
static int minof(int a, int b, int c) {return min(a, min(b, c));}
static int minof(int... x) {if(x.length == 1) return x[0]; if(x.length == 2) return min(x[0], x[1]); if(x.length == 3) return min(x[0], min(x[1], x[2])); int min = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] < min) min = x[i]; return min;}
static long minof(long a, long b, long c) {return min(a, min(b, c));}
static long minof(long... x) {if(x.length == 1) return x[0]; if(x.length == 2) return min(x[0], x[1]); if(x.length == 3) return min(x[0], min(x[1], x[2])); long min = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] < min) min = x[i]; return min;}
static int maxof(int a, int b, int c) {return max(a, max(b, c));}
static int maxof(int... x) {if(x.length == 1) return x[0]; if(x.length == 2) return max(x[0], x[1]); if(x.length == 3) return max(x[0], max(x[1], x[2])); int max = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] > max) max = x[i]; return max;}
static long maxof(long a, long b, long c) {return max(a, max(b, c));}
static long maxof(long... x) {if(x.length == 1) return x[0]; if(x.length == 2) return max(x[0], x[1]); if(x.length == 3) return max(x[0], max(x[1], x[2])); long max = x[0]; for(int i = 1; i < x.length; ++i) if(x[i] > max) max = x[i]; return max;}
static int powi(int a, int b) {if(a == 0) return 0; int ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static long powl(long a, int b) {if(a == 0) return 0; long ans = 1; while(b > 0) {if((b & 1) > 0) ans *= a; a *= a; b >>= 1;} return ans;}
static int floori(double d) {return (int)d;}
static int ceili(double d) {return (int)ceil(d);}
static long floorl(double d) {return (long)d;}
static long ceill(double d) {return (long)ceil(d);}
static void shuffle(int[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); int swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(long[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); long swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void shuffle(double[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); double swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static <T> void shuffle(T[] a) {int n = a.length - 1; for(int i = 0; i < n; ++i) {int ind = randInt(i, n); T swap = a[i]; a[i] = a[ind]; a[ind] = swap;}}
static void rsort(int[] a) {shuffle(a); sort(a);}
static void rsort(long[] a) {shuffle(a); sort(a);}
static void rsort(double[] a) {shuffle(a); sort(a);}
static int randInt(int min, int max) {return rand.nextInt(max - min + 1) + min;}
// input
static void r() throws IOException {input = new StringTokenizer(__in.readLine());}
static int ri() throws IOException {return Integer.parseInt(__in.readLine());}
static long rl() throws IOException {return Long.parseLong(__in.readLine());}
static int[] ria(int n) throws IOException {int[] a = new int[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Integer.parseInt(input.nextToken()); return a;}
static long[] rla(int n) throws IOException {long[] a = new long[n]; input = new StringTokenizer(__in.readLine()); for(int i = 0; i < n; ++i) a[i] = Long.parseLong(input.nextToken()); return a;}
static char[] rcha() throws IOException {return __in.readLine().toCharArray();}
static String rline() throws IOException {return __in.readLine();}
static int rni() throws IOException {input = new StringTokenizer(__in.readLine()); return Integer.parseInt(input.nextToken());}
static int ni() {return Integer.parseInt(input.nextToken());}
static long rnl() throws IOException {input = new StringTokenizer(__in.readLine()); return Long.parseLong(input.nextToken());}
static long nl() {return Long.parseLong(input.nextToken());}
// output
static void pr(int i) {__out.print(i);}
static void prln(int i) {__out.println(i);}
static void pr(long l) {__out.print(l);}
static void prln(long l) {__out.println(l);}
static void pr(double d) {__out.print(d);}
static void prln(double d) {__out.println(d);}
static void pr(char c) {__out.print(c);}
static void prln(char c) {__out.println(c);}
static void pr(char[] s) {__out.print(new String(s));}
static void prln(char[] s) {__out.println(new String(s));}
static void pr(String s) {__out.print(s);}
static void prln(String s) {__out.println(s);}
static void pr(Object o) {__out.print(o);}
static void prln(Object o) {__out.println(o);}
static void prln() {__out.println();}
static void pryes() {__out.println("yes");}
static void pry() {__out.println("Yes");}
static void prY() {__out.println("YES");}
static void prno() {__out.println("no");}
static void prn() {__out.println("No");}
static void prN() {__out.println("NO");}
static void pryesno(boolean b) {__out.println(b ? "yes" : "no");};
static void pryn(boolean b) {__out.println(b ? "Yes" : "No");}
static void prYN(boolean b) {__out.println(b ? "YES" : "NO");}
static void prln(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
static void prln(long... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), ++i); __out.println(a[a.length - 1]);}
static <T> void prln(Collection<T> c) {int n = c.size() - 1; Iterator<T> iter = c.iterator(); for(int i = 0; i < n; __out.print(iter.next()), __out.print(' '), ++i); if(n >= 0) __out.println(iter.next());}
static void h() {__out.println("hlfd");}
static void flush() {__out.flush();}
static void close() {__out.close();}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
def checklevel(a):
currp = 0
currc = 0
for i in a:
if currp > i[0]:
return 'NO'
if currc > i[1]:
return 'NO'
if i[1]-currc > i[0]-currp:
return 'NO'
currp = i[0]
currc = i[1]
return 'YES'
def problem1():
l = int(input())
a = []
for _ in range(l):
a.append(list(map(int,input().split())))
print(checklevel(a))
def makewealthy(a,x):
total = a[0]
if a[0] < x:
return 0
i = 1
b = x
while i < len(a):
total += a[i]
b += x
if total < b:
return i
i += 1
return len(a)
def problem2():
l,x = list(map(int,input().split()))
a = list(map(int,input().split()))
a.sort(reverse=True)
print(makewealthy(a,x))
def minbullets(s):
b = 0
l = len(s)
for i in range(l):
b += max (0, s[(i+1)%l][0] - s[i][1])
return min(i[0] for i in s) + b
def problem3():
l = int(input())
a = []
for _ in range(l):
a.append(list(map(int,input().split())))
print(minbullets(a))
def eulercycle(n,l,r):
i = 0
k = n-1
if l == n*(n-1)+1:
return [1]
while i < l:
i += 2*k
k -= 1
s = []
a = n
for j in range(i+1-l):
if j%2 == 0:
s.append(a)
a -= 1
else:
s.append(n-k-1)
s = s[::-1]
k = n-k
a = k
add = False
if r == n*(n-1)+1:
r -= 1
add = True
for j in range(r-i):
if j%2 == 0:
if a == n:
k += 1
a = k
s.append(k)
else:
a += 1
s.append(a)
if add:
s.append(1)
return(s)
def problem4():
n,l,r = list(map(int,input().split()))
x = eulercycle(n,l,r)
print(*x)
def divisors(n):
s = []
for i in range(1,n+1):
if n%i == 0:
s.append(i)
return s
cases = int(input())
for _ in range(cases):
problem4()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import math
# ΡΠ΅ΡΠ΅Π½Π°
def task_1343_c():
b = int(input())
array = [int(num) for num in input().split()]
maxPositive = 0
minNegative = -10000000000
res = 0
for i in range(b):
if array[i] < 0:
if i != 0 and array[i - 1] >= 0:
res += maxPositive
maxPositive = 0
minNegative = max(minNegative, array[i])
else:
if i != 0 and array[i - 1] < 0:
res += minNegative
minNegative = -10000000000
maxPositive = max(maxPositive, array[i])
if minNegative == -10000000000:
res += maxPositive
else:
res += maxPositive + minNegative
print(res)
# Π½Π΅ ΡΠ°Π±ΠΎΡΠ°Π΅Ρ ΠΎΡ ΡΠ»ΠΎΠ²Π° ΡΠΎΠ²ΡΠ΅ΠΌ
def task_1341_b():
heightLen, doorSize = map(int, input().split())
heights = [int(num) for num in input().split()]
perf = [0 for i in range(heightLen)]
a = 0
for i in range(heightLen - 1):
if i == 0:
perf[i] = 0
else:
if heights[i - 1] < heights[i] and heights[i] > heights[i + 1]:
a += 1
perf[i] = a
perf[heightLen - 1] = a
max_global = 0
left_global = 0
for i in range(heightLen - doorSize):
max_local = perf[i + doorSize - 1] - perf[i]
if max_local > max_global:
max_global = max_local
left_global = i
print(max_global + 1, left_global + 1)
# ΡΠ΅ΡΠΈΠ», ΡΡΠΎΠ± Π΅Ρ
def task_1340_a():
n = int(input())
array = [int(i) for i in input().split()]
for i in range(n - 1):
if array[i] < array[i + 1]:
if array[i] + 1 != array[i + 1]:
print("No")
return
print("Yes")
#ΡΠ΅ΡΠΈΠ»
def task_1339_b():
n = int(input())
array = [int(num) for num in input().split()]
array.sort()
output = [0 for i in range(0, n)]
i = 0
h = 0
j = n - 1
while i <= j:
output[h] = array[i]
h += 1
i += 1
if h < n:
output[h] = array[j]
h += 1
j -= 1
for val in reversed(output):
print(val, end=' ')
# ΡΠ΅ΡΠ΅Π½Π°
def task_1338_a():
n = int(input())
inputArr = [int(num) for num in input().split()]
max_sec = 0
for i in range(1, n):
local_sec = 0
a = inputArr[i - 1] - inputArr[i]
if a <= 0:
continue
else:
b = math.floor(math.log2(a))
local_sec = b + 1
for j in range(b, -1, -1):
if a < pow(2, j):
continue
inputArr[i] += pow(2, j)
a -= pow(2, j)
if local_sec > max_sec:
max_sec = local_sec
print(max_sec)
def task_1334_d():
n, l ,r = map(int, input().split())
if l == 9998900031:
print(1)
return
res = []
res.append(1)
for i in range(2, n + 1):
if i == n:
task_1334_d_helper(i, res)
else:
res.append(i)
res.append(1)
for i in range(l - 1, r):
print(res[i], end=" ")
def task_1334_d_helper(i, arr):
arr.append(i)
for j in range(2, i):
if j == i - 1:
task_1334_d_helper(i - 1, arr)
else:
arr.append(j)
arr.append(i)
a = int(input())
for i in range(a):
task_1334_d()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void solve() {
int n, l, r;
cin >> n >> l >> r;
ll count = 0;
int start;
for (start = 1; start <= n && count < l; start++) {
int edgesHere = 2 * (n - start);
if (count + edgesHere >= l) {
break;
}
count += edgesHere;
}
int next = start + 1;
bool flag = false;
while (count < l) {
if (flag) {
next++;
}
flag ^= 1;
count++;
}
bool run = false;
for (int i = l; i <= r; i++) {
run = true;
if (flag)
cout << start << ' ';
else {
cout << next << ' ';
if (++next > n) {
++start;
if (start == n) {
start = 1;
} else {
next = start + 1;
}
}
}
flag ^= 1;
}
if (!run) cout << 1;
cout << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
int t = 1;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
bool f = 0;
if (r == (n * (n - 1) + 1)) {
r--;
f = 1;
}
long long ans1 = (n - 1) * 2;
long long sum = ans1, pos = 0;
for (long long i = 1; i < n; i++) {
if (l < ans1) {
pos = i;
break;
}
l -= ans1;
r -= ans1;
ans1 -= 2;
}
long long pos2 = (l + 1) / 2 + pos;
while (l <= r) {
if (l % 2 == 0) {
cout << pos2 << " ";
pos2++;
} else
cout << pos << " ";
if (pos2 == n + 1) {
pos++;
pos2 = pos + 1;
}
l++;
}
if (f == 1) cout << 1 << " ";
cout << endl;
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import bisect
dp = [None for i in range(10**5+2)]
prev = 2
dp[0] = 1
dp[1] = 2
for i in range(2,10**5+1):
curr = prev+2*(i-1)
dp[i] = curr
prev = curr
def solve(curr, st, turn, n, l, r, res):
while l<=r:
if curr==n:
res.append(1)
l+=1
else:
if turn:
res.append(curr)
l+=1
turn=False
elif turn is False and st<n:
res.append(st)
l+=1
st+=1
turn=True
else:
res.append(st)
l+=1
turn=True
curr+=1
st=curr+1
return res
t = int(input())
for _ in range(t):
n, l, r = [int(x) for x in input().strip().split()]
res= []
curr=1
i=n
val=1
prev_curr=1
while l>=curr and i>1:
prev_curr = curr
curr+=2*(i-1)
i-=1
val+=1
if l-curr<0:
curr = val-1
curr_offset = l-prev_curr
else:
curr=val
curr_offset = l-curr
st_offset = curr_offset+1
# print(val,curr,curr_offset, st_offset)
if curr_offset%2==0:
turn = True
else:
turn = False
st = curr+1+(st_offset)//2
# print(curr, st, turn)
res = solve(curr, st, turn, n, l, r, res)
print(*res[:r-l+1])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Solution{
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] input=br.readLine().trim().split(" ");
int numTestCases=Integer.parseInt(input[0]);
while(numTestCases-->0){
input=br.readLine().trim().split(" ");
int n=Integer.parseInt(input[0]);
long l=Long.parseLong(input[1]);
long r=Long.parseLong(input[2]);
printSequence(n,l,r);
}
out.flush();
out.close();
}
public static void printSequence(int n,long l,long r)
{
long totalElements=0;
int blockNumber=-1;
for(int i=1;i<n;i++){
totalElements+=2L*(n-i);
if(totalElements>l)
{
totalElements-=2L*(n-i);
blockNumber=i;
break;
}
}
ArrayList<Integer> ans=new ArrayList<>();
long pos=totalElements+1;
int count=0;
for(int b=blockNumber;b<n && pos<=r && blockNumber!=-1;b++){
for(int i=b+1;i<=n;i++){
int currNumber=b;
if(pos>=l && pos<=r){
ans.add(currNumber);
count++;
}
pos++;
currNumber=i;
if(pos>=l && pos<=r){
ans.add(currNumber);
count++;
}
pos++;
}
}
if(count<(l-r+1))
{
ans.add(1);
}
for(int i=0;i<ans.size();i++){
out.print(ans.get(i)+" ");
}
out.println();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <typename T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return true;
} else
return false;
}
template <typename T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return true;
} else
return false;
}
ll S(ll n, ll k) { return 2 * (n - k); }
void solve() {
int T;
cin >> T;
for (ll(q) = (1); (q) <= (T); (q)++) {
ll n, l, r;
cin >> n >> l >> r;
ll length = r - l + 1;
ll lk;
for (ll(k) = (1); (k) <= (n - 1); (k)++) {
if (l - S(n, k) >= 0)
l -= S(n, k), r -= S(n, k);
else {
lk = k;
break;
}
}
vector<ll> res;
{
res.push_back(0);
ll i = 0;
for (ll(k) = (lk); (k) <= (n - 1); (k)++) {
for (ll(j) = (k + 1); (j) <= (n); (j)++) {
res.push_back(k);
i++;
if (i == r) goto hoge;
res.push_back(j);
i++;
if (i == r) goto hoge;
}
}
if (i < r) res.push_back(1), i++;
hoge:
assert(i == r);
}
for (ll(i) = (l); (i) <= (r); (i)++) {
cout << res[i] << (i < r ? " " : "\n");
}
}
return;
}
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
solve();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using namespace std;
inline int qpow(int b, int e, int m = 998244353) {
int a = 1;
for (; e; e >>= 1, b = (long long)b * b % m)
if (e & 1) a = (long long)a * b % m;
return a;
}
int n, m, q, k;
int a[300005], b[300005], c[300005];
const int pp[11] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31};
void solve() {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> v(n + 1);
v[1] = 1;
for (long long i = 2; i < n; i++) {
v[i] = v[i - 1] + 2 * (n - (i - 1));
}
v[n] = v[n - 1] + 2;
auto itr1 = upper_bound(v.begin(), v.end(), l);
auto itr2 = upper_bound(v.begin(), v.end(), r);
if (itr1 == v.end()) {
cout << 1 << endl;
} else {
int in1 = -1;
int in2 = -1;
auto k1 = itr1;
auto k2 = itr2;
int st1;
st1 = (--k1) - v.begin();
int st2;
st2 = k2 - v.begin();
long long i = v[st1];
while (i <= r && i < v[n]) {
for (int j = 0; j < n - st1; j++) {
if (i >= l && i <= r) cout << st1 << " " << st1 + j + 1 << " ";
i += 2;
}
st1++;
}
if (r == v[n])
cout << 1 << endl;
else
cout << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
cin >> t;
while (t--) solve();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include<iostream>
#include<queue>
#include<algorithm>
#include<vector>
#include<math.h>
#include<cstring>
#include<string>
#include<stack>
#include<map>
#include<set>
#include<cstdio>
#include<deque>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9 * 2;
const ll LNF = 1e18;
const ll MOD = 998244353;
const int MAXN = 200'005;
int main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
int T;
cin >> T;
while (T--) {
ll n, l, r;
cin >> n >> l >> r;
ll cou = r - l;
ll s = 1;
ll f = 0;
while (1) {
if (s + f >= l) {
break;
}
else
s += f;
f+=2;
}
//cout << "f " << f << '\n';
f /= 2;
//cout << "s , f : " << s << ' ' << f << "\n";
ll a = f+1;
ll b = (l - s+1)/2 ;
//cout <<"a , b : "<< a << ' ' << b << "\n";
for (ll i = l; i <=r; i++) {
if (i == 1) {
cout << 1 << ' ';
a = 2;
b = 0;
continue;
}
if (i % 2 == 0) {
cout << a << ' ';
if (i != l)
b++;
}
else {
if(a==n)
cout << b % (a - 1) + 1 << ' ';
else
cout << b << ' ';
if (b == a - 1) {
a++;
b = 0;
}
}
}
cout << '\n';
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main {
static FastReader in=new FastReader();
static StringBuilder Sd=new StringBuilder();
static List<Integer>Gr[];
static long Mod=998244353;
static Map<Integer,Integer>map=new HashMap<>();
public static void main(String [] args) {
//Dir by MohammedElkady
int t=in.nextInt();
while(t-->0) {
long n=in.nextLong(),l=in.nextLong(),r=in.nextLong();
long ans=1,res=0;
l-=1;r-=1;
int lol=0;
if(r>=n*(n-1)) {lol=1;r--;}
for(long i=1;i<=n;i++) {
if(res+((n-i)*2)<l)
{ ans=i;
res+=(n-i)*2;
}else break;
}
long vov=ans+1;
if(!(lol>0&&l>r))
for(;l-1<=r;) {
if(res>l) {
Sout(vov-1+" ");
l++;
}
if(res==l) {
if(vov>n) {
ans+=1;vov=ans+1;
}
Sout(ans+" ");
if(r-l>=2) {Sout(vov+" ");}
vov++;
res+=2;
l+=2;
}
else {
vov++;
res+=2;
}
}
if(lol>0) {Soutln("1 ");}
else Soutln("");
}
Sclose();
}
static long power(long x, long y, long p)
{
// Initialize result
long res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x
// with result
if (y % 2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns n^(-1) mod p
static long modInverse(long n, long p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's
// little theorem.
static long nCrModPFermat(int n, int r,
long p)
{
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
long[] fac = new long[n+1];
fac[0] = 1;
for (int i = 1 ;i <= n; i++)
fac[i] = fac[i-1] * i % p;
return (fac[n]* modInverse(fac[r], p)
% p * modInverse(fac[n-r], p)
% p) % p;
}
static long fac(int n , int m,int l) {
long res=1;
for(int i=l,u=1;i<=n||u<=m;i++,u++) {
if(i<=n) {res*=i;}
if(u<=m) {res/=u;}
while(res>Mod)
res-=Mod;
}
return res;
}
static long posation(int n) {
long res=1;
for(int i=0;i<n-3;i++) {res*=2L;
while(res>Mod)
res-=Mod;
while(res<=0)res+=Mod;}
return res;
}
static long gcd(long g,long x){
if(x<1)return g;
else return gcd(x,g%x);
}
//array fill
static long[]filllong(int n){long a[]=new long[n];for(int i=0;i<n;i++)a[i]=in.nextLong();return a;}
static int[]fillint(int n){int a[]=new int[n];for(int i=0;i<n;i++)a[i]=in.nextInt();return a;}
//OutPut Line
static void Sout(String S) {Sd.append(S);}
static void Soutln(String S) {Sd.append(S+"\n");}
static void Soutf(String S) {Sd.insert(0, S);}
static void Sclose() {System.out.println(Sd);}
static void Sclean() {Sd=new StringBuilder();}
}
class node implements Comparable<node>{
int x,t;
node(int x,int p){
this.x=x;
this.t=p;
}
@Override
public int compareTo(node o) {
return (t-o.t);
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
class Sorting{
public static node[] bucketSort(node[] array, int bucketCount) {
if (bucketCount <= 0) throw new IllegalArgumentException("Invalid bucket count");
if (array.length <= 1) return array; //trivially sorted
int high = array[0].t;
int low = array[0].t;
for (int i = 1; i < array.length; i++) { //find the range of input elements
if (array[i].t > high) high = array[i].t;
if (array[i].t < low) low = array[i].t;
}
double interval = ((double)(high - low + 1))/bucketCount; //range of one bucket
ArrayList<node> buckets[] = new ArrayList[bucketCount];
for (int i = 0; i < bucketCount; i++) { //initialize buckets
buckets[i] = new ArrayList();
}
for (int i = 0; i < array.length; i++) { //partition the input array
buckets[(int)((array[i].t - low)/interval)].add(array[i]);
}
int pointer = 0;
for (int i = 0; i < buckets.length; i++) {
Collections.sort(buckets[i]); //mergeSort
for (int j = 0; j < buckets[i].size(); j++) { //merge the buckets
array[pointer] = buckets[i].get(j);
pointer++;
}
}
return array;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.time.Period;
public class codeforces {
public static void main(String[] args) throws Exception {
int t=sc.nextInt();
while(t-->0) {
long n=sc.nextLong();
long l=sc.nextLong();
long r=sc.nextLong();
long number =2;
long i=1;
while(l-i*2>0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
for(;l<=r;l++) {
if(1l*l%2==0) {
pw.print(1l*number+" ");
}else {
pw.print(1l*(l+1)/2+" ");
}
if(1l*l-i*2>=0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
}
pw.println();
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextLongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
double x;
double y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair)o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Double(x).hashCode() * 31 + new Double(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return (int) (this.y - other.y);
} else {
return (int) (this.x - other.x);
}
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
if (a == 0)
return b;
return (a > b) ? GCD(a % b, b) : GCD(a, b % a);
}
public static long LCM(long a, long b) {
return a * b / GCD(a, b);
}
static long Pow(long a, int e, int mod) // O(log e)
{
a %= mod;
long res = 1;
while (e > 0) {
if ((e & 1) == 1)
res = (res * a) % mod;
a = (a * a) % mod;
e >>= 1;
}
return res;
}
static long nc(int n, int r) {
if (n < r)
return 0;
long v = fac[n];
v *= Pow(fac[r], mod - 2, mod);
v %= mod;
v *= Pow(fac[n - r], mod - 2, mod);
v %= mod;
return v;
}
public static boolean isprime(long a) {
if (a == 0 || a == 1) {
return false;
}
if (a == 2) {
return true;
}
for (int i = 2; i < Math.sqrt(a) + 1; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean isPal(String s) {
boolean t = true;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
t = false;
break;
}
}
return t;
}
public static long RandomPick(long[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static int RandomPick(int[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static void PH(String s, boolean reverse) {
prelen = s.length();
HashsArray[HashsArrayInd] = new int[prelen + 1];
prepow = new int[prelen];
if (HashsArrayInd == 0) {
int[] mods = { 1173017693, 1173038827, 1173069731, 1173086977, 1173089783, 1173092147, 1173107093,
1173114391, 1173132347, 1173144367, 1173150103, 1173152611, 1173163993, 1173174127, 1173204679,
1173237343, 1173252107, 1173253331, 1173255653, 1173260183, 1173262943, 1173265439, 1173279091,
1173285331, 1173286771, 1173288593, 1173298123, 1173302129, 1173308827, 1173310451, 1173312383,
1173313571, 1173324371, 1173361529, 1173385729, 1173387217, 1173387361, 1173420799, 1173421499,
1173423077, 1173428083, 1173442159, 1173445549, 1173451681, 1173453299, 1173454729, 1173458401,
1173459491, 1173464177, 1173468943, 1173470041, 1173477947, 1173500677, 1173507869, 1173522919,
1173537359, 1173605003, 1173610253, 1173632671, 1173653623, 1173665447, 1173675577, 1173675787,
1173684683, 1173691109, 1173696907, 1173705257, 1173705523, 1173725389, 1173727601, 1173741953,
1173747577, 1173751499, 1173759449, 1173760943, 1173761429, 1173762509, 1173769939, 1173771233,
1173778937, 1173784637, 1173793289, 1173799607, 1173802823, 1173808003, 1173810919, 1173818311,
1173819293, 1173828167, 1173846677, 1173848941, 1173853249, 1173858341, 1173891613, 1173894053,
1173908039, 1173909203, 1173961541, 1173968989, 1173999193};
mod = RandomPick(mods);
int[] primes = { 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
prime = RandomPick(primes);
}
prepow[0] = 1;
if (!reverse) {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[i]) % mod) % mod);
}
} else {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[prelen - 1 - i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[prelen - 1 - i]) % mod) % mod);
}
}
HashsArrayInd++;
}
public static int PHV(int l, int r, int n, boolean reverse) {
if (l > r) {
return 0;
}
int val = (int) ((1l * HashsArray[n - 1][r] + mod - HashsArray[n - 1][l - 1]) % mod);
if (!reverse) {
val = (int) ((1l * val * prepow[prelen - l]) % mod);
} else {
val = (int) ((1l * val * prepow[r - 1]) % mod);
}
return val;
}
static int[][] HashsArray;
static int HashsArrayInd = 0;
static int[] prepow;
static int prelen = 0;
static int prime = 31;
static long fac[];
static int mod = 998244353;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
#pragma pack(1)
using namespace std;
long long mod = 998244353;
long long inf = 1e9;
long double eps = 1e-6;
double pi = acosl(-1);
ifstream in("input.txt");
ofstream out("output.txt");
int main() {
int t;
scanf("%d", &t);
while (t--) {
long long n;
scanf("%lld", &n);
vector<int> a(n);
for (int i = 1; i <= n; i++) {
a[i - 1] = i;
}
sort(a.begin(), a.end(), [](const int& a, const int& b) {
return to_string(a) < to_string(b);
});
long long l, r;
scanf("%lld%lld", &l, &r);
int len = r - l + 1;
int kk = 0;
l--;
r--;
kk = (r == n * (n - 1));
r -= kk;
int len1 = r - l;
long long start = 0;
for (; start < n; start++) {
if (2 * (n - 1 - start) < l) {
l -= 2 * (n - 1 - start);
r -= 2 * (n - 1 - start);
} else {
break;
}
}
vector<int> ans;
int offset = l / 2 + 1;
for (int i = 0; i <= len1; i++) {
if ((l + i) % 2 == 1) {
ans.push_back(a[start + offset]);
offset++;
if (offset + start == n) {
start++;
offset = 1;
}
} else
ans.push_back(a[start]);
}
if (kk == 1) {
ans.push_back(a[0]);
}
for (auto v : ans) {
printf("%d ", v);
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import math
# ΡΠ΅ΡΠ΅Π½Π°
def task_1343_c():
b = int(input())
array = [int(num) for num in input().split()]
maxPositive = 0
minNegative = -10000000000
res = 0
for i in range(b):
if array[i] < 0:
if i != 0 and array[i - 1] >= 0:
res += maxPositive
maxPositive = 0
minNegative = max(minNegative, array[i])
else:
if i != 0 and array[i - 1] < 0:
res += minNegative
minNegative = -10000000000
maxPositive = max(maxPositive, array[i])
if minNegative == -10000000000:
res += maxPositive
else:
res += maxPositive + minNegative
print(res)
# Π½Π΅ ΡΠ°Π±ΠΎΡΠ°Π΅Ρ ΠΎΡ ΡΠ»ΠΎΠ²Π° ΡΠΎΠ²ΡΠ΅ΠΌ
def task_1341_b():
heightLen, doorSize = map(int, input().split())
heights = [int(num) for num in input().split()]
perf = [0 for i in range(heightLen)]
a = 0
for i in range(heightLen - 1):
if i == 0:
perf[i] = 0
else:
if heights[i - 1] < heights[i] and heights[i] > heights[i + 1]:
a += 1
perf[i] = a
perf[heightLen - 1] = a
max_global = 0
left_global = 0
for i in range(heightLen - doorSize):
max_local = perf[i + doorSize - 1] - perf[i]
if max_local > max_global:
max_global = max_local
left_global = i
print(max_global + 1, left_global + 1)
# ΡΠ΅ΡΠΈΠ», ΡΡΠΎΠ± Π΅Ρ
def task_1340_a():
n = int(input())
array = [int(i) for i in input().split()]
for i in range(n - 1):
if array[i] < array[i + 1]:
if array[i] + 1 != array[i + 1]:
print("No")
return
print("Yes")
#ΡΠ΅ΡΠΈΠ»
def task_1339_b():
n = int(input())
array = [int(num) for num in input().split()]
array.sort()
output = [0 for i in range(0, n)]
i = 0
h = 0
j = n - 1
while i <= j:
output[h] = array[i]
h += 1
i += 1
if h < n:
output[h] = array[j]
h += 1
j -= 1
for val in reversed(output):
print(val, end=' ')
# ΡΠ΅ΡΠ΅Π½Π°
def task_1338_a():
n = int(input())
inputArr = [int(num) for num in input().split()]
max_sec = 0
for i in range(1, n):
local_sec = 0
a = inputArr[i - 1] - inputArr[i]
if a <= 0:
continue
else:
b = math.floor(math.log2(a))
local_sec = b + 1
for j in range(b, -1, -1):
if a < pow(2, j):
continue
inputArr[i] += pow(2, j)
a -= pow(2, j)
if local_sec > max_sec:
max_sec = local_sec
print(max_sec)
def task_1334_d():
n, l ,r = map(int, input().split())
if l == 9998900031:
print(1)
return
res = []
res.append(1)
for i in range(2, n + 1):
if i == n:
task_1334_d_helper(i, res)
else:
res.append(i)
res.append(1)
for i in range(l - 1, r):
print(res[i], end=" ")
print()
def task_1334_d_helper(i, arr):
arr.append(i)
for j in range(2, i):
if j == i - 1:
task_1334_d_helper(i - 1, arr)
else:
arr.append(j)
arr.append(i)
a = int(input())
for i in range(a):
task_1334_d()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
const double PI = acos(-1.);
const int N = 2e6 + 6;
long long l, r, x;
vector<long long> res;
void gen(long long n) {
if (x > r) return;
if (x >= l && x <= r) res.push_back(1);
x++;
for (long long i = 2; i < n; i++) {
if (x > r) return;
if (x >= l && x <= r) res.push_back(n);
x++;
if (x >= l && x <= r) res.push_back(i);
x++;
}
if (x >= l && x <= r) res.push_back(n);
x++;
}
int main() {
ios_base::sync_with_stdio(false);
int T;
cin >> T;
while (T--) {
res.clear();
long long n;
cin >> n;
cin >> l >> r;
long long y = n;
for (; y * (y - 1) >= l; y--)
;
y = max(1LL, y - 1);
x = y * (y - 1) + 1;
for (long long i = y + 1; i <= n; i++) gen(i);
if (x <= r) res.push_back(1);
x++;
for (auto x : res) cout << x << " ";
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long f[10010];
map<long long, long long> vis;
long long a[300005], b[300005], c[300005];
signed main() {
std::ios::sync_with_stdio(false);
long long t, n, i, j, m;
cin >> t;
while (t--) {
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i] >> b[i];
}
long long sum = 0;
for (i = 0; i < n; i++) {
if (i == 0)
c[i] = max(0ll, a[0] - b[n - 1]);
else
c[i] = max(0ll, a[i] - b[i - 1]);
sum += c[i];
}
long long ans = 1e18;
for (i = 0; i < n; i++) {
ans = min(ans, sum - c[i] + a[i]);
}
cout << ans << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class C {
public static void main(String[] args) {
FastScanner in = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-->0) {
long n = in.nextInt(), l = in.nextLong(), r = in.nextLong();
int x = (int)Math.ceil(((2*n-1)-Math.sqrt((2*n-1)*(2*n-1)-4*l))/2);
if(l>n*(n-1)) out.println(1);
else{
long y = l-(n*(n-1)-((n-x)*(n-x+1)));
long yy = y;
y = x + (y-1)/2 + 1;
int cnt = 0;
long p = r-l+1;
//out.println(x+" "+y);
if(yy%2==0){
out.print(y+" ");
p--;
}
if(r>n*(n-1)) p--;
while(true){
if(cnt<p){
cnt++;
out.print(x+" ");
}
else break;
if(cnt<p){
cnt++;
out.print(y+" ");
y++;
}
else break;
if(y==n+1){
x++;
y = x+1;
}
}
if(r>n*(n-1)) out.print(1);
out.println();
}
}
out.flush();
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while(!st.hasMoreTokens())
try { st = new StringTokenizer(br.readLine()); }
catch(IOException e) {}
return st.nextToken();
}
String nextLine(){
try{ return br.readLine(); }
catch(IOException e) { } return "";
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
int[] readArray(int n) {
int a[] = new int[n];
for(int i=0;i<n;i++) a[i] = nextInt();
return a;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long tc, n, l, r;
cin >> tc;
for (long long i = 0; i < tc; i++) {
cin >> n >> l >> r;
if (l == (((n * (n - 1))) + 1)) {
cout << "1";
continue;
}
long long ind = 0;
long long add = (n - 1) * 2;
long long sum = 1;
while (1) {
if ((add + ind) >= l) {
break;
} else {
ind += add;
add -= 2;
sum++;
}
}
ind++;
add = sum;
while (ind < l) {
if (ind == l) {
break;
} else {
if (ind % 2 == 0) {
add++;
}
ind++;
}
}
long long ind1 = l;
if (l % 2 == 0) add++;
while (ind1 <= r) {
if (sum == n) {
cout << "1";
break;
}
while (add <= n) {
if (ind1 % 2 == 0) {
cout << add << " ";
ind1++;
if (add == n) break;
} else {
cout << sum << " ";
ind1++;
add++;
}
if (ind1 > r) break;
if (add > n) break;
}
if (ind1 > r) break;
sum++;
add = sum;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void out(const vector<int>& arr) {
for (auto i : arr) {
cout << i << " ";
}
cout << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin >> t;
for (int t1 = 0; t1 < t; t1++) {
int n;
long long l, r;
cin >> n >> l >> r;
long long pos = 0;
long long delta = 0;
int i = 1;
while (pos < l) {
if (n - i > 0) {
delta = (long long)((long long)n - i) * 2;
} else
delta = 1;
pos += delta;
i++;
}
i--;
long long realpos = pos - delta + 1;
int j = i + 1;
long long value = r - l + 1;
while (realpos < l) {
realpos += 2;
j++;
}
bool flag = false;
if (realpos != l) {
realpos -= 2;
j--;
flag = true;
}
while (value > 0) {
if (i == n) {
cout << 1;
break;
}
for (; j <= n; j++) {
if (!flag) {
cout << i << " ";
value--;
}
flag = false;
realpos++;
if (value <= 0) break;
cout << j << " ";
value--;
realpos++;
}
i++;
j = i + 1;
}
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import os
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
def visit(k):
if k == 1:
return [1]
else:
res = [k]
for i in range(2, k):
res.append(i)
res.append(k)
res.append(1)
return res
def visitlen(k):
if k == 1:
return 1
else:
return 2 * (k-1)
for _ in range(t):
n, left, right = list(map(int, input().split()))
interval = right-left+1
if left == 1:
i = 1
res = []
while interval > 0:
temp = visit(i)
res.extend(visit(i))
i += 1
interval -= len(temp)
for each in res[:right]:
print(each, end = " ")
print()
else:
i = 1
while visitlen(i) < left:
left -= visitlen(i)
i += 1
res = []
res.extend(visit(i)[left-1:])
intervalc = interval
intervalc -= len(res)
while intervalc > 0:
i += 1
res.extend(visit(i))
intervalc -= visitlen(i)
for each in res[:interval]:
print(each, end = " ")
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
scanf("%lld", &t);
while (t--) {
long long n, l, r;
scanf("%lld%lld%lld", &n, &l, &r);
long long suml = 0, sumr = 0;
long long headl = 1, headr = -1;
long long flag1 = 0;
for (long long i = 1; i < n; ++i) {
if (!flag1) {
suml += 2 * (n - i);
}
sumr += 2 * (n - i);
if (suml > l && !flag1) {
suml -= 2 * (n - i);
flag1 = 1;
headl = i;
} else if (sumr >= r) {
headr = i;
break;
}
}
long long vv = 0;
if (headr == -1) {
vv = 1;
}
long long flag = 1;
long long now = headl + 1;
for (long long i = suml + 1; i <= sumr; ++i) {
if (flag == 1) {
if (l <= i && i <= r) {
printf("%lld ", headl);
}
flag = 2;
} else if (flag == 2) {
if (l <= i && i <= r) {
printf("%lld ", now);
}
now++;
if (now > n) {
headl++;
now = headl + 1;
}
flag = 1;
}
}
if (vv == 1) {
printf("1 ");
}
putchar('\n');
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.time.Period;
public class codeforces {
public static void main(String[] args) throws Exception {
int t=sc.nextInt();
while(t-->0) {
long n=sc.nextLong();
long l=sc.nextLong();
long r=sc.nextLong();
long number =2;
long i=1;
while(1l*1l-i*2>0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
for(;l<=r;l++) {
if(1l*l%2==0) {
pw.print(1l*number+" ");
}else {
pw.print(1l*(l+1)/2+" ");
}
while(1l*l-i*2>=0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
}
pw.println();
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextLongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
double x;
double y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair)o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Double(x).hashCode() * 31 + new Double(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return (int) (this.y - other.y);
} else {
return (int) (this.x - other.x);
}
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
if (a == 0)
return b;
return (a > b) ? GCD(a % b, b) : GCD(a, b % a);
}
public static long LCM(long a, long b) {
return a * b / GCD(a, b);
}
static long Pow(long a, int e, int mod) // O(log e)
{
a %= mod;
long res = 1;
while (e > 0) {
if ((e & 1) == 1)
res = (res * a) % mod;
a = (a * a) % mod;
e >>= 1;
}
return res;
}
static long nc(int n, int r) {
if (n < r)
return 0;
long v = fac[n];
v *= Pow(fac[r], mod - 2, mod);
v %= mod;
v *= Pow(fac[n - r], mod - 2, mod);
v %= mod;
return v;
}
public static boolean isprime(long a) {
if (a == 0 || a == 1) {
return false;
}
if (a == 2) {
return true;
}
for (int i = 2; i < Math.sqrt(a) + 1; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean isPal(String s) {
boolean t = true;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
t = false;
break;
}
}
return t;
}
public static long RandomPick(long[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static int RandomPick(int[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static void PH(String s, boolean reverse) {
prelen = s.length();
HashsArray[HashsArrayInd] = new int[prelen + 1];
prepow = new int[prelen];
if (HashsArrayInd == 0) {
int[] mods = { 1173017693, 1173038827, 1173069731, 1173086977, 1173089783, 1173092147, 1173107093,
1173114391, 1173132347, 1173144367, 1173150103, 1173152611, 1173163993, 1173174127, 1173204679,
1173237343, 1173252107, 1173253331, 1173255653, 1173260183, 1173262943, 1173265439, 1173279091,
1173285331, 1173286771, 1173288593, 1173298123, 1173302129, 1173308827, 1173310451, 1173312383,
1173313571, 1173324371, 1173361529, 1173385729, 1173387217, 1173387361, 1173420799, 1173421499,
1173423077, 1173428083, 1173442159, 1173445549, 1173451681, 1173453299, 1173454729, 1173458401,
1173459491, 1173464177, 1173468943, 1173470041, 1173477947, 1173500677, 1173507869, 1173522919,
1173537359, 1173605003, 1173610253, 1173632671, 1173653623, 1173665447, 1173675577, 1173675787,
1173684683, 1173691109, 1173696907, 1173705257, 1173705523, 1173725389, 1173727601, 1173741953,
1173747577, 1173751499, 1173759449, 1173760943, 1173761429, 1173762509, 1173769939, 1173771233,
1173778937, 1173784637, 1173793289, 1173799607, 1173802823, 1173808003, 1173810919, 1173818311,
1173819293, 1173828167, 1173846677, 1173848941, 1173853249, 1173858341, 1173891613, 1173894053,
1173908039, 1173909203, 1173961541, 1173968989, 1173999193};
mod = RandomPick(mods);
int[] primes = { 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
prime = RandomPick(primes);
}
prepow[0] = 1;
if (!reverse) {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[i]) % mod) % mod);
}
} else {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[prelen - 1 - i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[prelen - 1 - i]) % mod) % mod);
}
}
HashsArrayInd++;
}
public static int PHV(int l, int r, int n, boolean reverse) {
if (l > r) {
return 0;
}
int val = (int) ((1l * HashsArray[n - 1][r] + mod - HashsArray[n - 1][l - 1]) % mod);
if (!reverse) {
val = (int) ((1l * val * prepow[prelen - l]) % mod);
} else {
val = (int) ((1l * val * prepow[r - 1]) % mod);
}
return val;
}
static int[][] HashsArray;
static int HashsArrayInd = 0;
static int[] prepow;
static int prelen = 0;
static int prime = 31;
static long fac[];
static int mod = 998244353;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
cin.tie(nullptr)->sync_with_stdio(false);
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long mv = 0;
bool b = false;
for (long long i = 1; i <= n; i++) {
long long nxt = i == 1 ? 1 : (i - 1) * 2;
if (!(l > mv + nxt || r < mv + 1)) {
b = true;
for (long long j = max(1ll, l - mv); j <= min(r - mv, nxt); j++) {
if (j == nxt)
cout << "1 ";
else if (j & 1)
cout << i << " ";
else
cout << j / 2 + 1 << " ";
}
}
mv += nxt;
}
if (b) cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long x, y, e;
void solve(long long n, long long l) {
long long i, sum = 0;
for (i = 1; i < n; i++) {
if (l <= sum + 2 * (n - i))
break;
else
sum += 2 * (n - i);
}
l -= sum;
x = i;
if (l % 2) {
e = 1;
y = i;
return;
}
long long ans = i;
for (i = 1; i <= l / 2; i++) ans++;
y = ans;
e = 2;
return;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long t, n, l, r, i, k, a, b;
cin >> t;
while (t--) {
cin >> n >> l >> r;
e = 0;
long long oh_bhai = n * (n - 1) + 1;
if (oh_bhai == l)
cout << "1\n";
else {
solve(n, l);
if (e == 1) {
long long f = 0;
for (i = l; i <= r; i++) {
if (i == oh_bhai)
cout << "1";
else if (f == 0) {
f = 1;
cout << x << " ";
} else {
f = 0;
cout << y << " ";
y++;
if (y == n + 1) {
x += 1;
y = x + 1;
}
}
}
} else {
long long f = 1;
for (i = l; i <= r; i++) {
if (i == oh_bhai)
cout << "1";
else if (f == 0) {
f = 1;
cout << x << " ";
} else {
f = 0;
cout << y << " ";
y++;
if (y == n + 1) {
++x;
y = x + 1;
}
}
}
}
}
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Solution{
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] input=br.readLine().trim().split(" ");
int numTestCases=Integer.parseInt(input[0]);
while(numTestCases-->0){
input=br.readLine().trim().split(" ");
int n=Integer.parseInt(input[0]);
long l=Long.parseLong(input[1]);
long r=Long.parseLong(input[2]);
printSequence(n,l,r);
}
out.flush();
out.close();
}
public static void printSequence(int n,long l,long r)
{
long totalElements=0;
int blockNumber=-1;
for(int i=1;i<n;i++){
totalElements+=2L*(n-i);
if(totalElements>l)
{
totalElements-=2L*(n-i);
blockNumber=i;
break;
}
}
ArrayList<Integer> ans=new ArrayList<>();
long pos=totalElements+1;
for(int b=blockNumber;b<n && blockNumber!=-1;b++){
for(int i=b+1;i<=n;i++){
int currNumber=b;
if(pos>=l && pos<=r){
ans.add(currNumber);
}
pos++;
currNumber=i;
if(pos>=l && pos<=r){
ans.add(currNumber);
}
pos++;
}
if(pos>r){
break;
}
}
if(pos==r && r==1L*n*(n-1)+1)
{
ans.add(1);
}
for(int i=0;i<ans.size();i++){
out.print(ans.get(i)+" ");
}
out.println();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.util.stream.*;
import java.io.*;
import java.math.*;
public class Main {
static boolean FROM_FILE = false;
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
if (FROM_FILE) {
try {
br = new BufferedReader(new FileReader("input.txt"));
} catch (IOException error) {
}
} else {
br = new BufferedReader(new InputStreamReader(System.in));
}
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static int max(int... nums) {
int res = Integer.MIN_VALUE;
for (int num: nums) res = Math.max(res, num);
return res;
}
static int min(int... nums) {
int res = Integer.MAX_VALUE;
for (int num: nums) res = Math.min(res, num);
return res;
}
static long max(long... nums) {
long res = Long.MIN_VALUE;
for (long num: nums) res = Math.max(res, num);
return res;
}
static long min(long... nums) {
long res = Long.MAX_VALUE;
for (long num: nums) res = Math.min(res, num);
return res;
}
static FastReader fr = new FastReader();
static PrintWriter out;
public static void main(String[] args) {
if (FROM_FILE) {
try {
out = new PrintWriter(new FileWriter("output.txt"));
} catch (IOException error) {
}
} else {
out = new PrintWriter(new OutputStreamWriter(System.out));
}
new Main().run();
out.flush();
out.close();
}
long getLevel(long n, long idx) {
for (long i = 1; i <= n; i += 1) {
if ((2 * n - 1 - i) * i >= idx) return i;
}
return -1;
}
void run() {
int t = fr.nextInt();
while (t-- > 0) {
long n = fr.nextLong(), l = fr.nextLong(), r = fr.nextLong();
if (l == n * (n - 1) + 1) {
out.println(1);
return;
}
int len = (int)(r - l) + 1;
long[] res = new long[len];
int idx = 0;
long k = getLevel(n, l), pre = (2 * n - 1 - (k - 1)) * (k - 1), col = k + (l - pre) / 2;
if ((l - pre) % 2 == 0) res[idx++] = col;
// out.println(col);
for (long i = col + 1; i <= n && idx + 1 < len; i += 1) { res[idx++] = k; res[idx++] = i; }
// out.println(Arrays.toString(res));
boolean finalOne = false;
if (r == n * (n - 1) + 1) {
r -= 1;
finalOne = true;
}
long k2 = getLevel(n, r);
// out.println(k + " " + k2);
for (long level = k + 1; level < k2; level += 1) {
for (long i = level + 1; i <= n; i += 1) { res[idx++] = level; res[idx++] = i; }
}
// out.println(Arrays.toString(res));
long last_x = k2, last_y = last_x + 1;
while (idx + 1 < len) {
res[idx++] = last_x;
res[idx++] = last_y++;
}
if (idx < len) {
res[idx] = finalOne ? 1 : last_x;
}
out.println(LongStream.of(res).mapToObj(e -> "" + e).collect(Collectors.joining(" ")));
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void f(const int offset, const int n, const int l, const int r) {
if (l > r || n == 1) return;
const int c = 2 * (n - 1);
if (r >= 1 && l <= c) {
for (int i = 1; i <= c; ++i)
if (l <= i && i <= r) cout << (i & 1 ? offset : i / 2 + offset) << ' ';
}
f(offset + 1, n - 1, l - c, r - c);
}
void f() {
int n, l, r;
cin >> n >> l >> r;
f(1, n, l, r);
cout << "1\n";
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int t;
cin >> t;
while (t--) f();
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
// Magic. Do not touch.
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static class FastReader
{
private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public FastReader() { this(System.in); }public FastReader(InputStream is) { mIs = is;}
public int read() {if (numChars == -1) throw new InputMismatchException();if (curChar >= numChars) {curChar = 0;try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}if (numChars <= 0) return -1; }return buf[curChar++];}
public String nextLine(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isEndOfLine(c));return res.toString() ;}
public String next(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isSpaceChar(c));return res.toString();}
public long l(){int c = read();while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }long res = 0; do{ if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read();}while(!isSpaceChar(c));return res * sgn;}
public int i(){int c = read() ;while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }int res = 0;do{if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read() ;}while(!isSpaceChar(c));return res * sgn;}
public double d() throws IOException {return Double.parseDouble(next()) ;}
public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; }
public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; }
public void scanIntArr(int [] arr){ for(int li=0;li<arr.length;++li){ arr[li]=i();}}
public void scanIntIndexArr(int [] arr){ for(int li=0;li<arr.length;++li){ arr[li]=i()-1;}}
public void scanLongArr(long [] arr){for (int i=0;i<arr.length;++i){arr[i]=l();}}
public void shuffle(int [] arr){ for(int i=arr.length;i>0;--i) { int r=(int)(Math.random()*i); int temp=arr[i-1]; arr[i-1]=arr[r]; arr[r]=temp; } }
public int swapIntegers(int a,int b){return a;} //Call it like this: a=swapIntegers(b,b=a)
}
public static void main(String[] args) throws IOException {
FastReader fr = new FastReader();
PrintWriter pw = new PrintWriter(System.out);
/*
inputCopy
3
2 1 3
3 3 6
99995 9998900031 9998900031
outputCopy
1 2 1
1 3 2 3
1
2
3 1 6
1 1 1
3
3 6 7
3 6 6
3 7 7
4
100000 9999899997 9999900001
100000 9999899998 9999900001
10 87 91
10 86 91
*/
//Press Ctrl+Win+Alt+L for reformatting indentation
int t = fr.i();
for (int ti = 0; ti < t; ++ti) {
int n=fr.i();
long l=fr.l();
long r=fr.l();
long cur=n-1;
long l2=l;
while(l2-2L*cur>0 && cur!=0){
l2-=2L*cur;
--cur;
}
long i=n-cur;
long rem=(l2-1)/2;
long i2=i+1+rem;
long toPrint=r-l+1;
if(l%2==0)
{
pw.print(i2+" ");
++i2;
if(i2>n)
{
++i;
i2=i+1;
--toPrint;
}
}
//System.err.println("i="+i+" i2="+i2);
for(;i<=n;++i)
{
for(;i2<=n && toPrint>0;++i2)
{
if(toPrint>=2)
pw.print(i+" "+i2+" ");
else
pw.print(i+" ");
toPrint-=2;
}
i2=i+2;
}
if(toPrint>0)
{
pw.print(1+" ");
}
pw.println();
}
pw.flush();
pw.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t--) {
long long n, l, r;
scanf("%lld%lld%lld", &n, &l, &r);
int flag = 0;
long long now = l >> 1;
if (l % 2 == 0) {
flag = 2;
} else {
flag = 1;
}
long long head = -1;
for (int i = 1; i < n; ++i) {
now -= n - i;
if (now < 0) {
head = i;
now += n - i;
now = head + 1 + now;
break;
}
}
if (head == -1) {
printf("1\n");
continue;
}
int flag1 = 0;
int sum = r - l + 1;
while (sum--) {
if (flag == 1) {
if (head == n) {
flag1 = 1;
break;
}
printf("%lld ", head);
flag = 2;
} else if (flag == 2) {
printf("%lld ", now);
flag = 1;
if (now == n) {
head += 1;
now = head + 1;
} else {
now++;
}
}
}
if (flag1) {
printf("1\n");
continue;
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
test = int(input())
for _ in range(test):
n , l , r = [int(x) for x in input().split()]
start = 1
it = 1
if l == n*(n-1) + 1:
print('1')
continue
while start < l:
start += (n-it)*2
it += 1
if start != l:
it -= 1
start -= (n - it)*2
a = it
b = it+1
ok = True
while start < l:
if ok:
ok = False
else:
ok = True
b += 1
start += 1
# print(a , b , ok , '--------------------------')
while start <= r:
if r == n*(n-1) + 1:
print('1' , end=' ')
break
if b == n+1:
a += 1
b = a+1
if ok:
print(a , end=' ')
ok = False
else:
print(b , end=' ')
b += 1
ok = True
# ok != ok
start += 1
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
while (t--) {
long long n, l, r;
scanf("%lld%lld%lld", &n, &l, &r);
int flag = 0;
long long now = l >> 1;
if (l % 2 == 0) {
flag = 2;
} else {
flag = 1;
}
long long head = -1;
for (int i = 1; i < n; ++i) {
now -= n - i;
if (now < 0) {
head = i;
now += n - i;
now = head + 1 + now;
break;
}
}
if (head == -1) {
printf("1\n");
continue;
}
int flag1 = 0;
int sum = l - r + 1;
while (sum--) {
if (flag == 1) {
if (head == n) {
flag1 = 1;
break;
}
printf("%lld ", head);
flag = 2;
} else if (flag == 2) {
printf("%lld ", now);
flag = 1;
if (now == n) {
head += 1;
now = head + 1;
} else {
now++;
}
}
}
if (flag1) {
printf("1\n");
continue;
}
printf("\n");
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
import java.time.Period;
public class codeforces {
public static void main(String[] args) throws Exception {
int t=sc.nextInt();
while(t-->0) {
long n=sc.nextLong();
long l=sc.nextLong();
long r=sc.nextLong();
long number =2;
long i=1;
while(l-i*2>0) {
number++;
l-=i*2;
r-=i*2;
i++;
}
for(;l<=r;l++) {
if(1l*l%2==0) {
pw.print(number+" ");
}else {
pw.print((l+1)/2+" ");
}
if(1l*l-i*2>=0) {
l-=i*2;
r-=i*2;
i++;
number++;
}
}
pw.println();
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
String x = next();
StringBuilder sb = new StringBuilder("0");
double res = 0, f = 1;
boolean dec = false, neg = false;
int start = 0;
if (x.charAt(0) == '-') {
neg = true;
start++;
}
for (int i = start; i < x.length(); i++)
if (x.charAt(i) == '.') {
res = Long.parseLong(sb.toString());
sb = new StringBuilder("0");
dec = true;
} else {
sb.append(x.charAt(i));
if (dec)
f *= 10;
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public long[] nextLongArray(int n) throws IOException {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public int[] nextIntArray(int n) throws IOException {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public Integer[] nextIntegerArray(int n) throws IOException {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public boolean ready() throws IOException {
return br.ready();
}
}
static class pair implements Comparable<pair> {
double x;
double y;
public pair(int x, int y) {
this.x = x;
this.y = y;
}
public String toString() {
return x + " " + y;
}
public boolean equals(Object o) {
if (o instanceof pair) {
pair p = (pair)o;
return p.x == x && p.y == y;
}
return false;
}
public int hashCode() {
return new Double(x).hashCode() * 31 + new Double(y).hashCode();
}
public int compareTo(pair other) {
if (this.x == other.x) {
return (int) (this.y - other.y);
} else {
return (int) (this.x - other.x);
}
}
}
static class tuble implements Comparable<tuble> {
int x;
int y;
int z;
public tuble(int x, int y, int z) {
this.x = x;
this.y = y;
this.z = z;
}
public String toString() {
return x + " " + y + " " + z;
}
public int compareTo(tuble other) {
if (this.x == other.x) {
return this.y - other.y;
} else {
return this.x - other.x;
}
}
}
public static long GCD(long a, long b) {
if (b == 0)
return a;
if (a == 0)
return b;
return (a > b) ? GCD(a % b, b) : GCD(a, b % a);
}
public static long LCM(long a, long b) {
return a * b / GCD(a, b);
}
static long Pow(long a, int e, int mod) // O(log e)
{
a %= mod;
long res = 1;
while (e > 0) {
if ((e & 1) == 1)
res = (res * a) % mod;
a = (a * a) % mod;
e >>= 1;
}
return res;
}
static long nc(int n, int r) {
if (n < r)
return 0;
long v = fac[n];
v *= Pow(fac[r], mod - 2, mod);
v %= mod;
v *= Pow(fac[n - r], mod - 2, mod);
v %= mod;
return v;
}
public static boolean isprime(long a) {
if (a == 0 || a == 1) {
return false;
}
if (a == 2) {
return true;
}
for (int i = 2; i < Math.sqrt(a) + 1; i++) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean isPal(String s) {
boolean t = true;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
t = false;
break;
}
}
return t;
}
public static long RandomPick(long[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static int RandomPick(int[] a) {
int n = a.length;
int r = rn.nextInt(n);
return a[r];
}
public static void PH(String s, boolean reverse) {
prelen = s.length();
HashsArray[HashsArrayInd] = new int[prelen + 1];
prepow = new int[prelen];
if (HashsArrayInd == 0) {
int[] mods = { 1173017693, 1173038827, 1173069731, 1173086977, 1173089783, 1173092147, 1173107093,
1173114391, 1173132347, 1173144367, 1173150103, 1173152611, 1173163993, 1173174127, 1173204679,
1173237343, 1173252107, 1173253331, 1173255653, 1173260183, 1173262943, 1173265439, 1173279091,
1173285331, 1173286771, 1173288593, 1173298123, 1173302129, 1173308827, 1173310451, 1173312383,
1173313571, 1173324371, 1173361529, 1173385729, 1173387217, 1173387361, 1173420799, 1173421499,
1173423077, 1173428083, 1173442159, 1173445549, 1173451681, 1173453299, 1173454729, 1173458401,
1173459491, 1173464177, 1173468943, 1173470041, 1173477947, 1173500677, 1173507869, 1173522919,
1173537359, 1173605003, 1173610253, 1173632671, 1173653623, 1173665447, 1173675577, 1173675787,
1173684683, 1173691109, 1173696907, 1173705257, 1173705523, 1173725389, 1173727601, 1173741953,
1173747577, 1173751499, 1173759449, 1173760943, 1173761429, 1173762509, 1173769939, 1173771233,
1173778937, 1173784637, 1173793289, 1173799607, 1173802823, 1173808003, 1173810919, 1173818311,
1173819293, 1173828167, 1173846677, 1173848941, 1173853249, 1173858341, 1173891613, 1173894053,
1173908039, 1173909203, 1173961541, 1173968989, 1173999193};
mod = RandomPick(mods);
int[] primes = { 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
prime = RandomPick(primes);
}
prepow[0] = 1;
if (!reverse) {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[i]) % mod) % mod);
}
} else {
for (int i = 1; i < prelen; i++) {
prepow[i] = (int) ((1l * prepow[i - 1] * prime) % mod);
}
for (int i = 0; i < prelen; i++) {
if (s.charAt(i) >= 'a' && s.charAt(i) <= 'z')
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'a' + 1) * prepow[prelen - 1 - i]) % mod) % mod);
else
HashsArray[HashsArrayInd][i + 1] = (int) ((1l * HashsArray[HashsArrayInd][i]
+ ((1l * s.charAt(i) - 'A' + 27) * prepow[prelen - 1 - i]) % mod) % mod);
}
}
HashsArrayInd++;
}
public static int PHV(int l, int r, int n, boolean reverse) {
if (l > r) {
return 0;
}
int val = (int) ((1l * HashsArray[n - 1][r] + mod - HashsArray[n - 1][l - 1]) % mod);
if (!reverse) {
val = (int) ((1l * val * prepow[prelen - l]) % mod);
} else {
val = (int) ((1l * val * prepow[r - 1]) % mod);
}
return val;
}
static int[][] HashsArray;
static int HashsArrayInd = 0;
static int[] prepow;
static int prelen = 0;
static int prime = 31;
static long fac[];
static int mod = 998244353;
static Random rn = new Random();
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(System.out);
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class TaskD {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st;
PrintWriter pw = new PrintWriter(System.out);
try {
int t = Integer.parseInt(br.readLine());
while(t-->0) {
st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
long l = Long.parseLong(st.nextToken());
long r = Long.parseLong(st.nextToken());
long ind = 0;
ArrayList<Integer> al = new ArrayList<>();
for(int i=0 ; i<n ; i++) {
if(i == n-1) {
if(ind+1 == r) {
al.add(1);
}
}
else {
long clen = (n-i-1)*2;
long beg = ind+1;
long end = ind+clen-1;
if(beg <= l && end >= r) {
int[] a = new int[(int)clen];
int x = i+1;
int y = i+2;
for(int j=0 ; j<clen ; j++) {
if(j%2 == 0) a[j] = x;
else {
a[j] = y;
y++;
}
}
int starts = (int)(l-beg);
int ends = (int)(r-beg+1);
for(int j=starts ; j<=ends ; j++) {
al.add(a[j]);
}
}
else if(beg > l && end < r) {
int[] a = new int[(int)clen];
int x = i+1;
int y = i+2;
for(int j=0 ; j<clen ; j++, y++) {
if(j%2 == 0) a[j] = x;
else {
a[j] = y;
y++;
}
}
int starts = (int)(beg-beg);
int ends = (int)(end-beg+1);
for(int j=starts ; j<=ends ; j++) {
al.add(a[j]);
}
}
else if(beg <= l && end < r && end >= l) {
int[] a = new int[(int)clen];
int x = i+1;
int y = i+2;
for(int j=0 ; j<clen ; j++, y++) {
if(j%2 == 0) a[j] = x;
else {
a[j] = y;
y++;
}
}
int starts = (int)(l-beg);
int ends = (int)(end-beg+1);
for(int j=starts ; j<=ends ; j++) {
al.add(a[j]);
}
}
else if(beg > l && end >= r && r >= beg) {
int[] a = new int[(int)clen];
int x = i+1;
int y = i+2;
for(int j=0 ; j<clen ; j++, y++) {
if(j%2 == 0) a[j] = x;
else {
a[j] = y;
y++;
}
}
int starts = (int)(beg-beg);
int ends = (int)(r-beg+1);
for(int j=starts ; j<=ends ; j++) {
al.add(a[j]);
}
}
ind += clen;
}
}
for(int i=0 ; i<al.size() ; i++) {
pw.print(al.get(i) + " ");
}
pw.println();
}
}
finally {
pw.flush();
pw.close();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n,l,rw=map(int,input().split())
k=[i for i in range(l,rw+1)]
even=[]
odd=[]
for i in range(len(k)):
if k[i]%2==0:
even.append(k[i]//2)
else:
odd.append(k[i]//2+1)
#print(odd,even)
if odd!=[]:
st=odd[0]
n=0
r=0
n1=100000000
while(n<=n1):
mid=(n+n1)//2
if (mid*(mid+1))//2==st:
r=mid
break
elif (mid*(mid+1))//2<st:
r=mid
n=mid+1
else:
n1=mid-1
tr=-(r*(r+1))//2+odd[0]
cop=r
if tr==0:
tr=r
r-=1
r+=1
#print(r,tr)
for i in range(len(odd)):
odd[i]=tr
tr+=1
if tr>r:
tr=1
r+=1
if even!=[]:
st = even[0]
n = 0
r = 0
n1 = 100000000
while (n <= n1):
mid = (n + n1) // 2
if (mid * (mid + 1)) // 2 == st:
r = mid
break
elif (mid * (mid + 1)) // 2 < st:
r = mid
n = mid + 1
else:
n1 = mid - 1
cop=r
if even[0]==(r*(r+1))//2:
cop-=1
r+=1
else:
r+=2
times=r-1-(even[0]-(cop*(cop+1))//2)+1
#print(r,times,cop)
for i in range(len(even)):
even[i]=r
times-=1
if times==0:
times=r
r+=1
t=1
if l%2==1:
t=0
ans=[]
r=0
r1=0
#print(odd,even,l,r)
for i in range(l,rw+1):
if t%2==0:
ans.append(odd[r])
r+=1
else:
ans.append(even[r1])
r1+=1
t+=1
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import collections, math, bisect
local = False
if local:
file = open("input.txt", "r")
import time
def inp():
if local:
return file.readline().rstrip()
else:
return input().rstrip()
def ints():
return [int(_) for _ in inp().split()]
if local:
start=time.time()
t = int(inp())
for _ in range(1,t+1):
n, l, r = ints()
if l==1:
edge = 1
vert = 1
else:
edge = 2
vert = 2
while edge<l:
edge+=2*(vert-1)
vert+=1
if edge>l:
vert-=1
edge-=2*(vert-1)
break
ans = []
while edge<=r:
if vert==1:
if edge>=l:
ans.append(1)
vert=2
edge=2
continue
for v in range(vert-1,0,-1):
if edge>=l:
ans.append(vert)
edge+=1
if edge>r:
break
if edge>=l:
ans.append(v)
edge+=1
if edge>r:
break
vert+=1
print(" ".join([str(_) for _ in ans]))
if local:
fin = (time.time()-start)*1000
print("{:.2f}".format(fin) + "ms")
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main {
static FastReader in=new FastReader();
static StringBuilder Sd=new StringBuilder();
static List<Integer>Gr[];
static long Mod=998244353;
static Map<Integer,Integer>map=new HashMap<>();
public static void main(String [] args) {
//Dir by MohammedElkady
int t=in.nextInt();
while(t-->0) {
long n=in.nextLong(),l=in.nextLong(),r=in.nextLong();
long ans=1,res=0;
l-=1;r-=1;
int lol=0;
if(r>=n*(n-1)) {lol=1;r--;
}
for(long i=1;i<=n;i++) {
ans=i;
if(res+((n-i)*2)<l)
{
res+=(n-i)*2;
}else break;
}
long vov=ans+1;
if(!(lol>0&&l>r))
for(;l<=r;) {
if(res>l) {
Sout(vov-1+" ");
l++;
}
if(res==l) {
if(vov>n) {
ans+=1;vov=ans+1;
}
Sout(ans+" ");
if(r-l>0) {Soutln(vov+" ");}
vov++;
res+=2;
l+=2;
}
else {
vov++;
res+=2;
}
}
if(lol>0) {Soutln("1 ");}
else Soutln("");
}
Sclose();
}
static long power(long x, long y, long p)
{
// Initialize result
long res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x
// with result
if (y % 2 == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Returns n^(-1) mod p
static long modInverse(long n, long p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's
// little theorem.
static long nCrModPFermat(int n, int r,
long p)
{
// Base case
if (r == 0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
long[] fac = new long[n+1];
fac[0] = 1;
for (int i = 1 ;i <= n; i++)
fac[i] = fac[i-1] * i % p;
return (fac[n]* modInverse(fac[r], p)
% p * modInverse(fac[n-r], p)
% p) % p;
}
static long fac(int n , int m,int l) {
long res=1;
for(int i=l,u=1;i<=n||u<=m;i++,u++) {
if(i<=n) {res*=i;}
if(u<=m) {res/=u;}
while(res>Mod)
res-=Mod;
}
return res;
}
static long posation(int n) {
long res=1;
for(int i=0;i<n-3;i++) {res*=2L;
while(res>Mod)
res-=Mod;
while(res<=0)res+=Mod;}
return res;
}
static long gcd(long g,long x){
if(x<1)return g;
else return gcd(x,g%x);
}
//array fill
static long[]filllong(int n){long a[]=new long[n];for(int i=0;i<n;i++)a[i]=in.nextLong();return a;}
static int[]fillint(int n){int a[]=new int[n];for(int i=0;i<n;i++)a[i]=in.nextInt();return a;}
//OutPut Line
static void Sout(String S) {Sd.append(S);}
static void Soutln(String S) {Sd.append(S+"\n");}
static void Soutf(String S) {Sd.insert(0, S);}
static void Sclose() {System.out.println(Sd);}
static void Sclean() {Sd=new StringBuilder();}
}
class node implements Comparable<node>{
int x,t;
node(int x,int p){
this.x=x;
this.t=p;
}
@Override
public int compareTo(node o) {
return (t-o.t);
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
class Sorting{
public static node[] bucketSort(node[] array, int bucketCount) {
if (bucketCount <= 0) throw new IllegalArgumentException("Invalid bucket count");
if (array.length <= 1) return array; //trivially sorted
int high = array[0].t;
int low = array[0].t;
for (int i = 1; i < array.length; i++) { //find the range of input elements
if (array[i].t > high) high = array[i].t;
if (array[i].t < low) low = array[i].t;
}
double interval = ((double)(high - low + 1))/bucketCount; //range of one bucket
ArrayList<node> buckets[] = new ArrayList[bucketCount];
for (int i = 0; i < bucketCount; i++) { //initialize buckets
buckets[i] = new ArrayList();
}
for (int i = 0; i < array.length; i++) { //partition the input array
buckets[(int)((array[i].t - low)/interval)].add(array[i]);
}
int pointer = 0;
for (int i = 0; i < buckets.length; i++) {
Collections.sort(buckets[i]); //mergeSort
for (int j = 0; j < buckets[i].size(); j++) { //merge the buckets
array[pointer] = buckets[i].get(j);
pointer++;
}
}
return array;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, l, r;
cin >> n >> l >> r;
int grupo = 0;
int cont = 1;
int nPassados = l;
bool flag1 = false;
if (r == (n) * (n - 1) + 1) {
flag1 = true;
r--;
}
while (l - (2 * (n - cont)) > 0 && cont != n) {
l -= 2 * (n - cont);
cont++;
grupo++;
}
l--;
while (nPassados <= r) {
vector<int> numsGrupo;
for (int I = 0; I < n - grupo - 1; I++) {
numsGrupo.push_back(grupo + 1);
numsGrupo.push_back(grupo + I + 2);
}
for (int I = 0; I < numsGrupo.size(); I++) {
if (l != 0) {
l--;
} else {
if (nPassados <= r) {
cout << numsGrupo[I] << " ";
nPassados++;
}
}
}
grupo++;
}
if (flag1) {
cout << 1 << " ";
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void upgrade() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); }
long long nums[100100];
long long func(long long a) { return a * (a - 1); }
int main() {
upgrade();
int t;
cin >> t;
for (int zzz = 0; zzz < t; zzz++) {
long long n, l, r;
cin >> n >> l >> r;
nums[0] = 0;
if (n == 2) {
int a[4] = {0, 1, 2, 1};
for (int i = l; i < r + 1; i++) {
cout << a[i] << ' ';
}
cout << '\n';
continue;
}
long long cnt = 2 * n - 3;
long long st;
long long ind, mod, head, f, cur;
bool c = 1;
for (int i = 1; i < n + 10; i++) {
nums[i] = nums[i - 1] + cnt;
cnt -= 2;
if (nums[i] >= l) {
st = nums[i - 1];
ind = (long long)i;
break;
}
if (cnt < 0) {
c = 0;
f = l - nums[i - 1];
break;
}
}
if (c) {
mod = (l - st) % 2;
head = (ind == 1) ? 1 : (n + 2 - ind);
f = mod;
cur = (l - st - 1) / 2 + 2;
}
for (int i = 0; i < r - l + 1; i++) {
if (f == 1) {
cout << head << ' ';
if (head == 1 && cur == n) {
head = n;
cur = 2;
} else if (head == 3) {
f = 2;
} else if (head != 1 && cur == head - 1) {
head--;
cur = 2;
} else {
f = 0;
}
} else if (f == 0) {
cout << cur << ' ';
cur++;
f = 1;
} else {
if (f != n + 1)
cout << f << ' ';
else
cout << 1 << ' ';
f++;
}
}
cout << '\n';
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
def c(sm, a1):
sm = n*a1+(n-1)*n
for _ in range(N()):
n, l, r = RL()
i = n-1
sm = 0
while sm<l:
sm+=i*2 if i>0 else 1
if sm>=l: break
i-=1
dif = l-(sm-i*2)
res = []
for j in range(n-i, n+1):
now = []
for k in range(j+1, n+1):
now.append(j)
now.append(k)
res+=now
if len(res)>(r-k+1)+dif: break
res = res[dif-1:]
if r==(n-1)*n+1: res.append(1)
print(*res)
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long n) {
a %= 1000000007;
if (n == 1) return a;
if (n == 0) return 1;
if (n % 2)
return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long tot = 0, c = 1;
long long lo = 1, hi = n - 1;
while (lo < hi) {
long long mid = (lo + hi) / 2;
if (2 * n * mid - mid * (mid - 1) >= l)
hi = mid;
else
lo = mid + 1;
}
long long cp = c + (l - tot + 1) / 2;
while (l <= r && l < n * (n - 1) + 1) {
if (l % 2)
cout << c << " ";
else {
cout << cp << " ";
cp++;
}
l++;
if (cp > n) c++, cp = c + 1;
}
if (r == n * (n - 1) + 1) cout << 1 << " ";
cout << "\n";
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')
# ------------------------------
def main():
def c(sm, a1):
sm = n*a1+(n-1)*n
for _ in range(N()):
n, l, r = RL()
i = n-1
sm = 0
while sm+i*2+1<l:
sm+=i*2 if i>0 else 1
i-=1
dif = l-sm
res = []
for j in range(n-i, n):
now = []
for k in range(j+1, n+1):
now.append(j)
now.append(k)
res+=now
if len(res)>(r-k+1)+dif: break
res = res[dif-1:]
if r==(n-1)*n+1: res.append(1)
print(*res)
if __name__ == "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
import sys
t=int(sys.stdin.readline())
for _ in range(t):
n,l,r=map(int,sys.stdin.readline().split())
prev=0
cur=0
start=1
if l==r and l==n*(n-1)+1:
print(1)
else:
ans=[]
while(True):
cur+=(n-start)*2
if l<=cur:
pos=l-prev
total=r-l+1
if(pos%2==1):
ans.append(start)
total-=1
x=start+pos//2 +1
while(total>0):
ans.append(x)
if x==n:
start+=1
if start==n:
start=1
x=start
total-=1
if total>0:
ans.append(start)
total-=1
x+=1
else:
x=start+pos//2 +1
while(total>0):
ans.append(x)
if x==n:
start+=1
if start==n:
start=1
x=start
total-=1
if total>0:
ans.append(start)
total-=1
x+=1
break
prev=cur
start+=1
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Main {
public static void main(String args[]) {new Main().run();}
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
void run(){
for(int q=ni();q>0;q--) {
work();
out.println();
}
out.flush();
}
long mod=998244353L;
long inf=Long.MAX_VALUE;
long gcd(long a,long b) {
return a==0?b:gcd(b%a,a);
}
void work() {
long n=nl(),l=nl(),r=nl();
long c=1;
long sum=0;
while(sum+(n-c)*2<l&&c<n) {
sum+=(n-c)*2;
c++;
}
// System.out.println(c);
long c2=(l+1-sum)/2+c;
if((l-sum)%2==0) {
out.print(c2+" ");
l++;
c2++;
if(c2>n) {
c++;
c2=c+1;
}
}
while(l<=Math.min(n*(n-1), r)) {
out.print(c+" ");
l++;
if(l>r)break;
out.print(c2+" ");
if(c2>=n) {
c++;
c2=c+1;
}
l++;
}
if(l<=r) {
out.print(1+" ");
}
}
//input
@SuppressWarnings("unused")
private ArrayList<Integer>[] ng(int n, int m) {
ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
int s=in.nextInt()-1,e=in.nextInt()-1;
graph[s].add(e);
graph[e].add(s);
}
return graph;
}
private ArrayList<long[]>[] ngw(int n, int m) {
ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
for(int i=0;i<n;i++) {
graph[i]=new ArrayList<>();
}
for(int i=1;i<=m;i++) {
long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
graph[(int)s].add(new long[] {e,w,i});
graph[(int)e].add(new long[] {s,w});
}
return graph;
}
private int ni() {
return in.nextInt();
}
private long nl() {
return in.nextLong();
}
private String ns() {
return in.next();
}
private long[] na(int n) {
long[] A=new long[n];
for(int i=0;i<n;i++) {
A[i]=in.nextLong();
}
return A;
}
private int[] nia(int n) {
int[] A=new int[n];
for(int i=0;i<n;i++) {
A[i]=in.nextInt();
}
return A;
}
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br=new BufferedReader(new InputStreamReader(System.in));
}
public String next()
{
while(st==null || !st.hasMoreElements())//ε车οΌη©Ίθ‘ζ
ε΅
{
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python2
|
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
from sys import stdin, stdout
import bisect #c++ upperbound
import math
import heapq
def modinv(n,p):
return pow(n,p-2,p)
def cin():
return map(int,sin().split())
def ain(): #takes array as input
return list(map(int,sin().split()))
def sin():
return input()
def inin():
return int(input())
import math
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
q=[]
def dfs(n,d,v,c):
global q
v[n]=1
x=d[n]
q.append(n)
j=c
for i in x:
if i not in v:
f=dfs(i,d,v,c+1)
j=max(j,f)
# print(f)
return j
"""*******************************************************"""
def main():
t=inin()
for _ in range(t):
n,j,k=cin()
k-=j
k+=1
a=[]
for i in range(n):
if(i==0):
a.append(1)
else:
a.append(n-i+1)
# print(a)
x=2*(n-1)
ll=1
for i in range(n):
if(j-x<=1 or x==0):
j-=1
break
else:
j-=x
if(x!=2):
j+=ll
x-=2
# print(j)
# print(i,j)
p=[]
while(len(p)<k+j and i<n):
f=2
xx=x
if(xx>0 and len(p)>0):
p.pop()
while(xx>0):
p.append(a[i])
p.append(f)
if len(p)>=k+j:
break
f+=1
xx-=2
x-=2
i+=1
if(len(p)<j+k):
for jj in range(3,n+1):
p.append(jj)
p.append(1)
# print(p,j)
print(*p[j:k+j])
######## Python 2 and 3 footer by Pajenegod and c1729
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
# So on cf, use PyPy2 for best string performance.
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
conv = ord if py2 else lambda x:x
A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
try:
while True:
if s[i] >= b'R' [0]:
numb = 10 * numb + conv(s[i]) - 48
elif s[i] == b'-' [0]: sign = -1
elif s[i] != b'\r' [0]:
A.append(sign*numb)
numb = zero; sign = 1
i += 1
except:pass
if s and s[-1] >= b'R' [0]:
A.append(sign*numb)
return A
# threading.Thread(target=main).start()
if __name__== "__main__":
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
python3
|
t = int(input())
for _ in range(t):
n, l, r = map(int, input().split())
ans = []
memo_l = -1
for x in range(1, n):
if (x - 1) * x < l <= x * (x + 1):
memo_l = x
if memo_l == -1:
print(1)
else:
tmp = (memo_l - 1) * memo_l
for k in range(memo_l, n):
for i in range(1, 2 * k + 1):
if i % 2 == 0:
ad = k + 1
else:
ad = (i + 1) // 2
tmp += 1
if tmp > r:
break
elif l <= tmp:
ans.append(ad)
if r == n * (n - 1) + 1:
ans.append(1)
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u β v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.
You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).
We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 β a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.
Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.
Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.
Input
The first line contains the single integer T (1 β€ T β€ 100) β the number of test cases.
Next T lines contain test cases β one per line. The first and only line of each test case contains three integers n, l and r (2 β€ n β€ 10^5, 1 β€ l β€ r β€ n(n - 1) + 1, r - l + 1 β€ 10^5) β the number of vertices in K_n, and segment of the cycle to print.
It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.
Output
For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.
Example
Input
3
2 1 3
3 3 6
99995 9998900031 9998900031
Output
1 2 1
1 3 2 3
1
Note
In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.
In the third test case, it's quite obvious that the cycle should start and end in vertex 1.
|
{
"input": [
"3\n2 1 3\n3 3 6\n99995 9998900031 9998900031\n"
],
"output": [
"1 2 1 \n1 3 2 3 \n1 \n"
]
}
|
{
"input": [
"1\n2 2 3\n",
"1\n4 13 13\n",
"1\n3 1 1\n",
"10\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n2 1 3\n",
"1\n3 7 7\n",
"1\n25 30 295\n",
"1\n4 12 13\n",
"5\n3 7 7\n4 13 13\n5 21 21\n6 31 31\n7 42 43\n",
"1\n5 4 4\n"
],
"output": [
"2 1 \n",
"1 \n",
"1 \n",
"1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n1 2 1 \n",
"1 \n",
"16 1 17 1 18 1 19 1 20 1 21 1 22 1 23 1 24 1 25 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20 2 21 2 22 2 23 2 24 2 25 3 4 3 5 3 6 3 7 3 8 3 9 3 10 3 11 3 12 3 13 3 14 3 15 3 16 3 17 3 18 3 19 3 20 3 21 3 22 3 23 3 24 3 25 4 5 4 6 4 7 4 8 4 9 4 10 4 11 4 12 4 13 4 14 4 15 4 16 4 17 4 18 4 19 4 20 4 21 4 22 4 23 4 24 4 25 5 6 5 7 5 8 5 9 5 10 5 11 5 12 5 13 5 14 5 15 5 16 5 17 5 18 5 19 5 20 5 21 5 22 5 23 5 24 5 25 6 7 6 8 6 9 6 10 6 11 6 12 6 13 6 14 6 15 6 16 6 17 6 18 6 19 6 20 6 21 6 22 6 23 6 24 6 25 7 8 7 9 7 10 7 11 7 12 7 13 7 14 7 15 7 16 7 17 7 18 7 19 7 20 7 21 7 22 7 23 7 24 7 25 8 \n",
"4 1 \n",
"1 \n1 \n1 \n1 \n7 1 \n",
"3 \n"
]
}
|
IN-CORRECT
|
java
|
import java.util.*;
import java.io.*;
public class Solution{
static PrintWriter out=new PrintWriter(System.out);
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] input=br.readLine().trim().split(" ");
int numTestCases=Integer.parseInt(input[0]);
while(numTestCases-->0){
input=br.readLine().trim().split(" ");
int n=Integer.parseInt(input[0]);
long l=Long.parseLong(input[1]);
long r=Long.parseLong(input[2]);
printSequence(n,l,r);
}
out.flush();
out.close();
}
public static void printSequence(int n,long l,long r)
{
long totalElements=0;
int blockNumber=-1;
for(int i=1;i<n;i++){
totalElements+=2L*(n-i);
if(totalElements>l)
{
totalElements-=2L*(n-i);
blockNumber=i;
break;
}
}
ArrayList<Integer> ans=new ArrayList<>();
long pos=totalElements+1;
int count=0;
for(int b=blockNumber;b<n && pos<=r && blockNumber!=-1;b++){
for(int i=b+1;i<=n;i++){
int currNumber=b;
if(pos>=l && pos<=r){
ans.add(currNumber);
count++;
}
pos++;
currNumber=i;
if(pos>=l && pos<=r){
ans.add(currNumber);
count++;
}
pos++;
}
}
if(count<(r-l+1))
{
ans.add(1);
}
for(int i=0;i<ans.size();i++){
out.print(ans.get(i)+" ");
}
out.println();
}
}
|
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