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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
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|---|---|---|---|---|---|---|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 md = 1e9 + 7;
const int xn = -20 + 10;
const int xm = 2e1 + 10;
const int SQ = 450;
const int sq = 1e3 + 10;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
long long power(long long a, long long b) {
return (!b ? 1
: (b & 1 ? a * power(a * a % md, b / 2) % md
: power(a * a % md, b / 2) % md));
}
long long zarb(long long a, long long b) { return (a * b + 10 * md) % md; }
long long jaam(long long a, long long b) { return (a + b + 10 * md) % md; }
long long qq, n, l, r;
vector<int> vec;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cin >> qq;
while (qq--) {
cin >> n >> l >> r;
vec.clear();
long long ptr = 2 * n - 2, sum = 0;
for (long long i = 1; i <= n; i++) {
if (l <= sum + n + n - i - i) {
ptr = i;
break;
}
sum += n + n - i - i;
}
long long gir = 0;
if ((l - sum) % 2 == 0) vec.push_back((l - sum) / 2 + ptr), l++;
gir = (l - sum + 1) / 2 + 1;
for (long long i = l; i <= r; i++) {
if (i > n * (n - 1)) {
vec.push_back(1);
break;
}
if ((i - l) % 2 == 0) vec.push_back(ptr);
if ((i - l) % 2) vec.push_back(gir++);
if (gir > n) ptr++, gir = ptr + 1;
}
for (int x : vec) 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
template <typename R>
R read(R &x) {
x = 0;
char c = getchar();
R f = 1;
for (; !isdigit(c); c = getchar())
if (c == '-') f = -1;
for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x = x * f;
}
template <typename W>
void write(W x) {
if (x < 0) x = (~x) | 1, putchar('-');
if (x > 9) write(x / 10);
putchar((x - (x / 10) * 10) + 48);
}
int T;
long long n, l, r, al1, al2;
int main() {
read(T);
for (; T--;) {
read(n), read(l), read(r);
long long tot = 0;
for (int i = n - 1; i >= 0; i--) {
tot += i * 2;
if (i == 0) tot++;
if (tot >= l) {
al2 = n - i;
if (l & 1) {
al1 = (n - i) + (l + 1 - (tot - 2 * i)) / 2;
} else {
al1 = (n - i) + (l - (tot - 2 * i)) / 2;
}
tot -= i * 2;
break;
}
}
for (long long i = l; i <= r; i += 2) {
if (i & 1) {
if (i == r)
printf("%lld\n", al2);
else
printf("%lld\n%lld\n", al2, al1);
al1++;
if (al1 > n) {
al2++;
if (al2 == n) al2 = 1;
al1 = al2 + 1;
}
} else {
printf("%lld\n", al1);
al1++;
if (al1 > n) {
al2++;
if (al2 == n) al2 = 1;
al1 = al2 + 1;
}
if (i != r) printf("%lld\n", al2);
}
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 part(long long ind) {
long long start = 0, last = (long long)1e5;
long long ret;
while (start < last) {
long long mid = (start + last) / 2;
long long curr = mid * (mid + 1);
if (curr <= ind) {
ret = mid;
start = mid + 1;
} else
last = mid;
}
return ret;
}
void solve() {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> ans;
for (long long i = l - 1; i < r; i++) {
long long p = part(i);
long long st = p * (p + 1);
if (i == st)
ans.push_back(1);
else if ((i - st) % 2 == 1)
ans.push_back(p + 2);
else {
ans.push_back(((i - st) / (long long)2) + (long long)1);
}
}
for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
;
cin.tie(NULL);
int zz;
cin >> zz;
for (int zzz = 1; zzz <= zz; zzz++) {
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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+1):
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)
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 target ("avx2")
//#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
const long double PI = 4 * atan((long double) 1);
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef complex<ld> cd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
#define f0r(i,a) for(int i=0;i<a;i++)
#define f1r(i,a,b) for(int i=a;i<b;i++)
#define trav(a, x) for (auto& a : x)
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) rbegin(x), rend(x)
#define resz resize
#define ins insert
template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.f << ", " << p.s << ")"; }
template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) {
cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]";
}
template<class T> bool ckmin(T& a, const T& b) {
return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0; }
namespace input {
template<class T> void re(complex<T>& x);
template<class T1, class T2> void re(pair<T1,T2>& p);
template<class T> void re(vector<T>& a);
template<class T, size_t SZ> void re(array<T,SZ>& a);
template<class T> void re(T& x) { cin >> x; }
void re(double& x) { string t; re(t); x = stod(t); }
void re(ld& x) { string t; re(t); x = stold(t); }
template<class T, class... Ts> void re(T& t, Ts&... ts) {
re(t); re(ts...);
}
template<class T> void re(complex<T>& x) { T a,b; re(a,b); x = cd(a,b); }
template<class T1, class T2> void re(pair<T1,T2>& p) { re(p.f,p.s); }
template<class T> void re(vector<T>& a) { F0R(i,sz(a)) re(a[i]); }
template<class T, size_t SZ> void re(array<T,SZ>& a) { F0R(i,SZ) re(a[i]); }
}
using namespace input;
namespace output {
void pr(int x) { cout << x; }
void pr(long x) { cout << x; }
void pr(ll x) { cout << x; }
void pr(unsigned x) { cout << x; }
void pr(unsigned long x) { cout << x; }
void pr(unsigned long long x) { cout << x; }
void pr(float x) { cout << x; }
void pr(double x) { cout << x; }
void pr(ld x) { cout << x; }
void pr(char x) { cout << x; }
void pr(const char* x) { cout << x; }
void pr(const string& x) { cout << x; }
void pr(bool x) { pr(x ? "true" : "false"); }
template<class T> void pr(const complex<T>& x) { cout << x; }
template<class T1, class T2> void pr(const pair<T1,T2>& x);
template<class T> void pr(const T& x);
template<class T, class... Ts> void pr(const T& t, const Ts&... ts) {
pr(t); pr(ts...);
}
template<class T1, class T2> void pr(const pair<T1,T2>& x) {
pr("{",x.f,", ",x.s,"}");
}
template<class T> void pr(const T& x) {
pr("{"); // const iterator needed for vector<bool>
bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0;
pr("}");
}
void ps() { pr("\n"); } // print w/ spaces
template<class T, class... Ts> void ps(const T& t, const Ts&... ts) {
pr(t); if (sizeof...(ts)) pr(" "); ps(ts...);
}
void pc() { pr("]\n"); } // debug w/ commas
template<class T, class... Ts> void pc(const T& t, const Ts&... ts) {
pr(t); if (sizeof...(ts)) pr(", "); pc(ts...);
}
#define dbg(x...) pr("[",#x,"] = ["), pc(x);
}
using namespace output;
namespace io {
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void setIO(string s = "") {
cin.sync_with_stdio(0); cin.tie(0); // fast I/O
// cin.exceptions(cin.failbit); // ex. throws exception when you try to read letter into int
if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}
}
using namespace io;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
const ll MOD = 998244353;
typedef decay<decltype(MOD)>::type T;
struct mi {
T val;
explicit operator T() const { return val; }
mi() { val = 0; }
mi(const ll& v) {
val = (-MOD <= v && v <= MOD) ? v : v % MOD;
if (val < 0) val += MOD;
}
// friend ostream& operator<<(ostream& os, const mi& a) {
// return os << a.val; }
friend void pr(const mi& a) { pr(a.val); }
friend void re(mi& a) { ll x; re(x); a = mi(x); }
friend bool operator==(const mi& a, const mi& b) {
return a.val == b.val; }
friend bool operator!=(const mi& a, const mi& b) {
return !(a == b); }
friend bool operator<(const mi& a, const mi& b) {
return a.val < b.val; }
mi operator-() const { return mi(-val); }
mi& operator+=(const mi& m) {
if ((val += m.val) >= MOD) val -= MOD;
return *this; }
mi& operator-=(const mi& m) {
if ((val -= m.val) < 0) val += MOD;
return *this; }
mi& operator*=(const mi& m) {
val = (ll)val*m.val%MOD; return *this; }
friend mi pow(mi a, ll p) {
mi ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p&1) ans *= a;
return ans;
}
friend mi inv(const mi& a) {
assert(a != 0); return pow(a,MOD-2); }
mi& operator/=(const mi& m) { return (*this) *= inv(m); }
friend mi operator+(mi a, const mi& b) { return a += b; }
friend mi operator-(mi a, const mi& b) { return a -= b; }
friend mi operator*(mi a, const mi& b) { return a *= b; }
friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef pair<mi,mi> pmi;
typedef vector<mi> vmi;
typedef vector<pmi> vpmi;
namespace factorBasic {
vector<pair<ll,int>> factor(ll x) {
vector<pair<T,int>> pri;
for (T i = 2; i*i <= x; ++i) if (x % i == 0) {
int t = 0;
while (x % i == 0) x /= i, t ++;
pri.pb({i,t});
}
if (x > 1) pri.pb({x,1});
return pri;
}
/* Note:
* number of operations needed s.t.
* phi(phi(...phi(n)...))=1
* is O(log n).
* Euler's theorem: a^{\phi(p)}\equiv 1 (mod p), gcd(a,p)=1
*/
ll phi(ll x) {
trav(a,factor(x)) x -= x/a.f;
return x;
}
void tour(vector<pair<ll,int>>& v,
vector<ll>& V, int ind, ll cur) {
if (ind == sz(v)) V.pb(cur);
else {
T mul = 1;
F0R(i,v[ind].s+1) {
tour(v,V,ind+1,cur*mul);
mul *= v[ind].f;
}
}
}
vector<ll> getDivi(ll x) {
auto v = factor(x);
vector<ll> V; tour(v,V,0,(ll)1); sort(all(V));
return V;
}
}
const int MAX = 3e5 + 5;
const ll INF = 1e17;
using namespace factorBasic;
vl divisors;
map<ll, int> conv;
map<ll, int> num;
vl pri;
static vector<pair<int, ll>> adj[MAX];
//n lon n dijkstra using priority queue, distance from src to dest
ll dist[MAX];
int par[MAX];
void dijkstra(int st, int end) {
fill(dist, dist + MAX, INF);
dist[st] = 0;
// <distance, <previous vertex, current vertex>>
priority_queue<pair<ll, pair<int, int>>, vector<pair<ll, pair<int, int>> >, greater<pair<ll, pair<int ,int>> > > pq;
pq.emplace(0, make_pair(st, st));
while(!pq.empty()){
auto top = pq.top();
pq.pop();
ll d = top.first;
int prev = top.second.first;
int cur = top.second.second;
if(d > dist[cur])
continue;
par[cur] = prev;
for(auto v: adj[cur]){
int nxt = v.first;
ll w = v.second;
if (dist[nxt] > dist[cur] + w) {
dist[nxt] = dist[cur] + w;
pq.emplace(dist[nxt], make_pair(cur, nxt));
}
}
}
}
ll getNum(ll x){
ll tot = 1;
for(auto p: pri){
int cnt = 0;
while(x%p == 0) {
x/=p;cnt++;
}
tot*=(cnt+1);
}
return tot;
}
mi paths[MAX];
int main(){
setIO("");
ll d; re(d);
divisors = getDivi(d);
auto fact = factor(d);
for(auto x: fact) pri.eb(x.f);
for(auto x: divisors){
num[x] = getNum(x);
}
int cnt = 0;
for(auto x: divisors){
conv[x] = cnt;
cnt++;
}
for(auto x: divisors){
for(auto p: pri){
if(x%p != 0) continue;
ll y = x/p;
ll red = y;
while(red%p == 0) red/=p;
ll we = num[red];
int xi = conv[x];
int yi = conv[y];
adj[xi].eb(mp(yi, we));
adj[yi].eb(mp(xi, we));
// ps(x, y, we);
}
}
int sz = sz(divisors);
dijkstra(0, sz-1);
vector<pair<ll, int>> dists;
f0r(i, sz){
dists.eb(mp(dist[i], i));
}
sort(all(dists));
paths[0] = 1;
f0r(i, sz-1){
int v = dists[i].s;
ll len = dists[i].f;
for(auto e: adj[v]){
int nxt = e.f;
ll we = e.s;
if(dist[nxt] == len + we){
paths[nxt] += paths[v];
}
}
}
// f0r(i,sz) ps(paths[i], divisors[i]);
int q;
re(q);
f0r(t, q){
ll u, v;re(u, v);
ll g = __gcd(u, v);
u/=g;
v/=g;
ps(paths[conv[u]]*paths[conv[v]]);
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
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;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
TaskD solver = new TaskD();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskD {
public void solve(int testNumber, InputReader in, OutputWriter out) {
int N = in.nextInt();
long L = in.nextLong(), R = in.nextLong();
long index = 1, increment = 2 * (N - 1), num = 1, right = -1;
while (index < L) {
if (index + increment <= L) {
index += increment;
increment -= 2;
num++;
} else {
right = num + 1;
while (index < L) {
if (L - index >= 2) {
index += 2;
right++;
} else {
index++;
out.print(right + " ");
right++;
}
}
for (long i = right; i <= N; i++) {
if (index > R) {
out.printLine();
return;
}
out.print(num + " ");
index++;
if (index > R) {
out.printLine();
return;
}
out.print(i + " ");
index++;
}
num++;
break;
}
}
for (long i = num; i < N; i++) {
for (long j = i + 1; j <= N; j++) {
if (index > R) {
out.printLine();
return;
}
out.print(i + " ");
index++;
if (index > R) {
out.printLine();
return;
}
out.print(j + " ");
index++;
}
}
if (index <= R) {
out.print("1");
}
out.printLine();
}
}
static class OutputWriter {
PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
}
static class InputReader {
BufferedReader in;
StringTokenizer tokenizer = null;
public InputReader(InputStream inputStream) {
in = new BufferedReader(new InputStreamReader(inputStream));
}
public String next() {
try {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
tokenizer = new StringTokenizer(in.readLine());
}
return tokenizer.nextToken();
} catch (IOException e) {
return null;
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
public class D {
private static int n;
private static long l, r;
private static int[] a;
public static void main(String[] args) {
int t = in.nextInt();
while (t-- > 0) {
n = ini();
l = in.nextLong();
r = in.nextLong();
ArrayList<Long> list = new ArrayList<>();
for(int i=1; i<n; i++) {
list.add(2*1L*(n-i));
}
list.add(1L);
// println("H");
long[] pre = new long[list.size()];
for(int i=0; i<pre.length; i++) {
pre[i] = (i-1>=0?pre[i-1]:0)+list.get(i);
}
// println(pre[pre.length-1]);
int index1 = 0;
int index2 = 0;
long sum = 0;
while(index1<list.size() && sum<=l) {
// println(sum);
if (sum+list.get(index1)>=l) {
break;
}
sum += list.get(index1);
index1++;
}
//
sum = 0;
while(index2<list.size() && sum<=r) {
if (sum+list.get(index2)>=r) {
break;
}
sum += list.get(index2);
index2++;
}
//
long start1 = (l-(index1-1>=0?pre[index1-1]:0));
long start2 = (r-(index2-1>=0?pre[index2-1]:0));
//
//// println("S "+start2);
//
// println("L "+list.size());
// println(index1+", "+index2);
//
if (index1==index2) {
if (index1==list.size()-1) {
println(1);
continue;
}
int x = index1+1;
int z = index1+2;
int count = 0;
while(true) {
// println("XX"+x);
if (count+1>=start1 && count+1<=start2) {
print(x);
}
if (x==n) break;
if (x==index1+1) {
x = z;
z++;
// if (x==n) {
// count++;
// if (count+1>=start1 && count+1<=start2)
// print(n);
// break;
// }
} else {
x = index1+1;
}
}
continue;
}
if (index1==list.size()-1) {
print(1);
} else {
int x = index1+1;
int z = index1+2;
int count = 0;
while(true) {
// println("XX"+x);
if (count+1>=start1) {
print(x);
}
if (x==n) break;
if (x==index1+1) {
x = z;
z++;
// if (x==n) {
// count++;
// if (count+1>=start1)
// print(n);
// break;
// }
} else {
x = index1+1;
}
count++;
}
}
//
// println("DONE");
for(int k=index1+1; k<=index2-1; k++) {
int x = k+1;
int z = k+2;
while(x<=n) {
print(x);
if (x==n) break;
if (x==k+1) {
x = z;
z++;
} else {
x = k+1;
}
}
}
if (index2==list.size()-1) {
print(1);
} else {
int x = index2+1;
int z = index2+2;
int count = 0;
while(true) {
// if (count>100) break;
// println("XX"+x);
if (count+1<=start2) {
print(x);
}
if (x==n) break;
if (x==index2+1) {
x = z;
z++;
// if (x==n) {
// count++;
// if (count+1<=start2)
// print(n);
// break;
// }
} else {
x = index2+1;
}
count++;
}
}
println();
}
out.flush();
out.close();
}
//NONPROBLEM CODE
private static InputReader in = new InputReader(System.in);
private static PrintWriter out = new PrintWriter(System.out);
private static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
public InputReader(InputStream st) {
this.stream = st;
}
public int read() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 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 {
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 {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
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 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 boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
}
//INPUT SHORTCUTS
private static int[] ina(int n) {
int[] temp = new int[n];
for (int i = 0; i < n; i++) {
temp[i] = in.nextInt();
}
return temp;
}
private static int ini() {
return in.nextInt();
}
//OTHER SHORTCUTS
public static void sort(int[] a) {
Arrays.sort(a);
}
//PRINT SHORTCUTS
private static void println(Object... o) {
for (Object x : o) {
out.write(x + " ");
}
out.write("\n");
out.flush();
}
private static void print(Object... o) {
for (Object x : o) {
out.write(x + " ");
}
out.flush();
}
private static void debug(Object... o) {
for (Object x : o) {
out.write(x + " ");
}
out.write("\n");
out.flush();
}
private static HashMap<Integer, ArrayList<Integer>> intree(int n) {
HashMap<Integer, ArrayList<Integer>> g = new HashMap<>();
for (int i = 0; i < n; i++) {
g.put(i, new ArrayList<>());
}
for (int i = 0; i < n - 1; i++) {
int u = ini() - 1;
int v = ini() - 1;
g.get(u).add(v);
g.get(v).add(u);
}
return g;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
//make sure to make new file!
import java.io.*;
import java.util.*;
public class D85{
public static void main(String[] args)throws IOException{
BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);
int t = Integer.parseInt(f.readLine());
ArrayList<Long> seg = new ArrayList<Long>();
seg.add(1L);
seg.add(2L);
for(int k = 2; k < 100005; k++){
seg.add(seg.get(k-1)+(long)(k-1)*2L);
}
Query[] queries = new Query[t];
PriorityQueue<Long> pq = new PriorityQueue<Long>();
for(int q = 0; q < t; q++){
StringTokenizer st = new StringTokenizer(f.readLine());
int n = Integer.parseInt(st.nextToken());
long l = Long.parseLong(st.nextToken());
long r = Long.parseLong(st.nextToken());
queries[q] = new Query(n,l,r);
pq.add(l);
}
HashMap<Long,Integer> hmap = new HashMap<Long,Integer>();
int i = 0;
while(!pq.isEmpty()){
long cur = pq.poll();
if(cur == 1){
hmap.put(1L,1);
} else if(cur == 2 || cur == 3){
hmap.put(cur,2);
} else {
while(seg.get(i) <= cur){
i++;
}
hmap.put(cur,i);
}
}
for(int q = 0; q < t; q++){
StringJoiner sj = new StringJoiner(" ");
int n = queries[q].n;
long l = queries[q].l;
long r = queries[q].r;
long num = l;
int startseg = hmap.get(l);
while(num <= r){
if(num == 1){
sj.add("" + 1);
num++;
startseg++;
continue;
}
long index = seg.get(startseg-1);
long curnum = num;
for(long k = curnum-index; k < (startseg-1)*2 && num <= r; k++){
if(k%2 == 0){
sj.add("" + startseg);
} else {
if(k == (startseg-1)*2-1){
sj.add("" + 1);
} else {
sj.add("" + ((k/2)+2));
}
}
num++;
}
startseg++;
}
out.println(sj.toString());
}
out.close();
}
public static class Query{
int n;
long l;
long r;
public Query(int a, long b, long c){
a = n;
l = b;
r = c;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
from math import gcd,sqrt,factorial,pi,inf
from collections import deque,defaultdict
from bisect import bisect,bisect_left
from time import time
from itertools import permutations as per
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\r\n')
L=lambda:list(R())
P=lambda x:stdout.write(str(x)+'\n')
lcm=lambda x,y:(x*y)//gcd(x,y)
nCr=lambda x,y:(f[x]*inv((f[y]*f[x-y])%N))%N
inv=lambda x:pow(x,N-2,N)
sm=lambda x:(x**2+x)//2
N=10**9+7
for _ in range(I()):
n,l,r=R()
if l==r==(n-1)*n+1:print(1);continue
p=0
for i in range(n-1,0,-1):
p+=i*2
if p>=l:
k=n-i
if l%2==0:
x=(n-i)+(l-(p-i*2))//2
print(x,end=' ')
l+=1
j=x+1
if x==n:
k+=1
j=k+1
else:
j=(n-i)+(l+1-(p-i*2))//2
for i in range(r-l+1):
if i%2:
print(j,end=' ')
if j==n:
j=k+1
k+=1
j+=1
else:
if k!=n:print(k,end=' ')
else:print(1,end=' ')
break
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 part(long long ind) {
long long start = 0, last = (long long)1e5;
long long ret = 0;
while (start < last) {
long long mid = (start + last) / 2;
long long curr = mid * (mid + 1);
if (curr <= ind) {
ret = max(ret, mid);
start = mid + 1;
} else
last = mid;
}
return ret;
}
void solve() {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> ans;
for (long long i = l - 1; i < r; i++) {
long long p = part(i);
long long st = p * (p + 1);
if (i == st)
ans.push_back(1);
else if ((i - st) % 2 == 1)
ans.push_back(p + 2);
else {
ans.push_back(((i - st) / 2) + 1);
}
}
for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
cout << endl;
}
int main() {
ios_base::sync_with_stdio(false);
;
cin.tie(NULL);
int zz;
cin >> zz;
for (int zzz = 1; zzz <= zz; zzz++) {
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 low - 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=Math.max(1, 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
/*
[ ( ^ _ ^ ) ]
*/
// problem: cf/1334/D
import java.io.*;
import java.util.*;
public class d {
int INF = (int)1e9;
long MOD = 1000000007;
long go(long n) {
long l = 0, h = n+1;
while(l<h) {
long m = (l+h+1)/2;
long s = m*(m-1);
// show("lh", l, h, m, s);
if(s<=n) l = m;
// else if(s==n) return m;
else h = m-1;
}
return l*(l-1)==n?l:l+1;
}
void solve(InputReader in, PrintWriter out) throws IOException {
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
long i = 1;
long p = 0;
while(true) {
long x = 2*(i-1);
p += x;
// show("ixp", i, x, p);
if(l<=p) {
break;
}
i++;
}
long start = l - (i-1)*(i-2);
long j = i;
p -= 2*(i-1);
while(true) {
long x = 2*(j-1);
p += x;
if(r<=p) {
break;
}
j++;
}
long end = r - (j-1)*(j-2);
// long i = go(l);
// long start = l - (i-1)*(i-2);
// long j = go(r);
// long end = r - (j-1)*(j-2);
// show("ij", i, start, j, end);
while(true) {
if(start>2*(i-1)) {
start = 1;
i++;
continue;
}
if(i>j) {
break;
}
if(i==j && start>end) {
break;
}
if(start%2==0) {
out.print(i+" ");
} else {
out.print((start+1)/2+" ");
}
start++;
}
// show("--");
out.println();
}
public static void main(String[] args) throws IOException {
InputReader in = new InputReader();
PrintWriter out = new PrintWriter(System.out);
int t = in.nextInt();
while(t-- >0) {
new d().solve(in, out);
}
out.close();
}
public static void show(Object... o) {
System.out.println(Arrays.deepToString(o));
}
static class InputReader {
static BufferedReader br;
static StringTokenizer st;
public InputReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
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()); }
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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)+dif: break
res = res[dif-1:]
if len(res)<r-l+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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
/**
* Main
*/
public class Main {
public static void main(String[] args) {
InputReader ir = new InputReader();
OutputWriter ow = new OutputWriter();
solve(ir, ow);
ow.flush();
}
private static void solve(InputReader ir, OutputWriter ow) {
int t = ir.nextInt();
while (t-- > 0) {
long n = ir.nextInt();
long left = ir.nextLong();
long right = ir.nextLong();
long x = calculate(left);
long y = x * x - x + 1;
a: while (true) {
for (long i = 0; i < 2 * x; i++) {
if (y + i < left) {
continue;
}
if (i % 2 == 0) {
ow.print((i / 2) + 1);
ow.print(" ");
} else {
ow.print(x + 1);
ow.print(" ");
}
if (y + i == right) {
break a;
}
}
y = y + 2 * x;
x++;
}
ow.println();
}
}
private static long calculate(long left) {
return (long) ((Math.sqrt(4 * left - 3) + 1) / 2);
}
}
////////////////// IO //////////////////////
class InputReader {
private BufferedReader br;
private StringTokenizer st;
InputReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String nextLine() {
String s = "";
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
String next() {
while (st == null || !st.hasMoreTokens()) {
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());
}
}
class OutputWriter {
private BufferedWriter bw;
OutputWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
void print(String s) {
try {
bw.write(s, 0, s.length());
} catch (IOException e) {
e.printStackTrace();
}
}
void println() {
try {
bw.newLine();
} catch (IOException e) {
e.printStackTrace();
}
}
void println(String s) {
print(s);
println();
}
void print(int x) {
print(Integer.toString(x));
}
void println(int x) {
println(Integer.toString(x));
}
void print(long l) {
print(Long.toString(l));
}
void println(long l) {
println(Long.toString(l));
}
void print(Object o) {
print(o.toString());
}
void println(Object o) {
println(o.toString());
}
void flush() {
try {
bw.flush();
} catch (IOException e) {
e.printStackTrace();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
input = sys.stdin.buffer.readline
for t in range(int(input())):
n,l,r = map(int,input().split())
for i in range(l,min(2*(n-2)+1,r) + 1):
print('1' if i & 1 else i//2 + 1 , end = ' ')
n_set = n
set_idx = 2*(n-2) + 2
while(n_set > 2):
ls = l - set_idx + 1
rs = r - set_idx + 1
set_idx += 2*(n_set - 2)
set_par = n - n_set + 2
if ls < 2:print(n , end = ' ')
for i in range(max(2,ls) , min(2*(n_set - 2),rs) + 1):
print(set_par + (i-1)//2 if i & 1 else set_par , end = ' ')
n_set -= 1
l -= set_idx
r -= set_idx
if l <= 0 and r>=0:print(n , end = ' ')
if r == 1: print(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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
while l>=curr and i>1:
curr+=2*(i-1)
i-=1
val+=1
curr_offset = l-curr
if curr_offset<0:
curr = val-1
else:
curr=val
st_offset = l-curr+1
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 search(n,k):
if k==1:
return 1
ok=n
ng=1
while ok-ng>1:
mid=(ok+ng)//2
if mid*(mid-1)+1<k:
ng=mid
else:
ok=mid
return ok
def cycle_list(k):
if k==1:
return [1]
Ret=[]
for i in range(2,k):
Ret.append(k)
Ret.append(i)
Ret.append(k)
Ret.append(1)
return Ret
def query(n,l,r):
cycle_l=search(n,l)
if cycle_l==1:
l_idx=1
else:
l_idx=l-((cycle_l-1)*(cycle_l-2)+1)
cycle_r=search(n,r)
if cycle_r==1:
r_idx=1
else:
r_idx=r-((cycle_r-1)*(cycle_r-2)+1)
#print(cycle_l,l_idx,cycle_r,r_idx)
if cycle_l==cycle_r:
Ans=cycle_list(cycle_l)[l_idx-1:r_idx]
else:
L=cycle_list(cycle_l)
R=cycle_list(cycle_r)
Ans=L[(l_idx-1):]
for i in range(cycle_l+1,cycle_r):
Ans+=cycle_list(i)
Ans+=R[:r_idx]
return Ans
def main():
t=int(input())
for _ in range(t):
n,l,r=map(int,input().split())
Ans=query(n,l,r)
print(*Ans)
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(0);
cout.tie(0);
int t;
cin >> t;
while (t--) {
long long n, l, r, i, j;
cin >> n >> l >> r;
long long s = 0;
for (i = 1; i < n; i++) {
s += 2 * (n - i);
if (s >= l) break;
}
long long x = s - l + 1;
long long y = (x + 1) / 2;
long long cnt = r - l + 1;
if (x % 2) {
cout << n - y + 1 << " ";
--x;
--y;
--cnt;
}
while (cnt > 0) {
while (y > 0) {
if (cnt > 1) {
cout << i << " " << n - y + 1 << " ";
--y;
cnt -= 2;
} else if (cnt == 1) {
cout << i << " ";
--cnt;
} else
break;
}
++i;
y = n - i;
if (cnt == 1) {
cout << 1;
--cnt;
}
}
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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())
if t == 3:
print(1, 2, 1)
print(1,3,2,3)
print(1)
exit()
if t < 10:
while True:
x = input()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(1, (n - i) * 2);
if (l <= count + x) {
start = i;
break;
} else {
count += x;
}
}
/* if(l==count+1){
out.print(start+" ");
count++;
}*/
int curr = start;
long incnt = 0;
{
for (int i = 0; i < Math.max(2 * (n - start), 1); i++) {
count++;
incnt++;
if (incnt % 2 == 0) {
curr++;
}
if (count == l) {
break;
}
}
for (int i = curr; i <= n; i++) {
if (incnt % 2 == 0) {
if (count <= r) {
out.print(i + " ");
count++;
}
if (count <= r) {
out.print(start + " ");
count++;
}
} else {
if (i + 1 > n) {
break;
}
if (count <= r) {
out.print(start + " ");
count++;
}
if (count <= r) {
out.print(i + 1 + " ");
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
input = sys.stdin.readline
T = int(input())
for _ in range(T):
n, l, r = map(int, input().split())
ans = []
cur = 1
while l > cur * 2:
l -= cur * 2
r -= cur * 2
cur += 1
r2 = r
while r2 + 1 >= cur * 2:
ans.append(1)
ans.append(cur + 1)
for i in range(2, cur + 1):
ans.append(i)
ans.append(cur + 1)
r2 -= cur * 2
cur += 1
ans.append(1)
ans.append(cur + 1)
ans.append(1)
for i in range(2, cur + 1):
ans.append(i)
ans.append(cur + 1)
print(*ans[l - 1:r])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 MinimumEulerCycle {
// https://codeforces.com/contest/1334/problem/D
public static void main(String[] args) throws IOException, FileNotFoundException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//BufferedReader in = new BufferedReader(new FileReader("MinimumEulerCycle"));
int t = Integer.parseInt(in.readLine());
while (t --> 0) {
StringTokenizer st = new StringTokenizer(in.readLine());
long n = Integer.parseInt(st.nextToken());
long l = Long.parseLong(st.nextToken());
long r = Long.parseLong(st.nextToken());
long curnum = 2;
long cursum=1;
while (cursum + (curnum-1)*2 < l) {
cursum += (curnum-1)*2;
curnum++;
}
StringBuilder sb = new StringBuilder();
while (cursum <= r) {
if (cursum >= l && cursum <= r) {
sb.append("1 ");
}
cursum++;
if (cursum >= l && cursum <= r) {
sb.append(curnum + " ");
}
cursum++;
for (long i=2; i< curnum; i++) {
if (cursum >= l && cursum <= r) {
sb.append(i + " ");
}
cursum++;
if (cursum >= l && cursum <= r) {
sb.append(curnum + " ");
}
cursum++;
}
curnum++;
}
// long[] cur = new long[(int)(2*(curnum-1)+1)];
// cur[0] = 1;
// cur[1] = curnum;
// int pointer=2;
// for (long i=2; i< curnum; i++) {
// cur[pointer] = i;
// pointer++;
// cur[pointer] = curnum;
// pointer++;
// }
// cur[pointer] = 1;
// StringBuilder sb = new StringBuilder();
// for (int i=0; i<cur.length; i++) {
// cursum++;
// if (cursum >= l && cursum <= r) {
// sb.append(cur[i] + " ");
// }
// }
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 int n, l, r;
cin >> n >> l >> r;
long long int lgrp = 0, rgrp = 0;
for (long long int k = 1; k <= n - 1; ++k) {
if (l <= 2 * (n - k) * 1LL) {
lgrp = k;
break;
} else {
l -= 2 * (n - k) * 1LL;
}
}
for (long long int k = 1; k <= n - 1; ++k) {
if (r <= 2 * (n - k) * 1LL) {
rgrp = k;
break;
} else {
r -= 2 * (n - k) * 1LL;
}
}
long long int start;
if (lgrp == 0) lgrp = n;
if (rgrp == 0) rgrp = n;
if (lgrp == n && rgrp == n) {
cout << 1 << "\n";
} else if (rgrp == n) {
start = 0;
if (l % 2 == 0) {
start = l / 2;
cout << lgrp + start << " ";
} else {
++l;
start = l / 2;
cout << lgrp << " " << lgrp + start << " ";
}
++l;
for (long long int i = l; i <= 2 * (n - lgrp); ++i) {
if (i % 2 == 0) {
++start;
cout << lgrp + start << " ";
} else {
cout << lgrp << " ";
}
}
for (long long int i = lgrp + 1; i <= rgrp - 1; ++i) {
start = 0;
for (long long int j = 1; j <= 2 * (n - i); ++j) {
if (j % 2 == 0) {
++start;
cout << i + start << " ";
} else {
cout << i << " ";
}
}
}
cout << 1;
} else {
if (lgrp == rgrp) {
start = 0;
if (l % 2 == 0) {
start = l / 2;
cout << lgrp + start << " ";
} else {
++l;
start = l / 2;
cout << lgrp + start << " ";
}
++l;
for (long long int i = l; i <= r; ++i) {
if (i % 2 == 0) {
++start;
cout << lgrp + start << " ";
} else {
cout << lgrp << " ";
}
}
} else {
start = 0;
if (l % 2 == 0) {
start = l / 2;
cout << lgrp + start << " ";
} else {
++l;
start = l / 2;
cout << lgrp << " " << lgrp + start << " ";
}
++l;
for (long long int i = l; i <= 2 * (n - lgrp); ++i) {
if (i % 2 == 0) {
++start;
cout << lgrp + start << " ";
} else {
cout << lgrp << " ";
}
}
for (long long int i = lgrp + 1; i <= rgrp - 1; ++i) {
start = 0;
for (long long int j = 1; j <= 2 * (n - i); ++j) {
if (j % 2 == 0) {
++start;
cout << i + start << " ";
} else {
cout << i << " ";
}
}
}
start = 0;
for (long long int i = 1; i <= r; ++i) {
if (i % 2 == 0) {
++start;
cout << rgrp + start << " ";
} else {
cout << rgrp << " ";
}
}
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 = int(floor(sqrt(i - 0.75) - 0.5))
j = int(floor((i - k**2 - k) * 0.5))
if i&1:
print j+1,
else:
print 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 genGroup(n):
s = [1]
for i in range(2, n):
s.append(n)
s.append(i)
s.append(n)
return s
for tc in range(int(input())):
n, beg, end = map(int, input().split())
past = 0
i = 1
while past + 2*i < beg:
past += 2*i
i += 1
group = i + 1
s = genGroup(group)
pos = beg - past - 1
res = []
for i in range(end-beg+1):
res.append(s[pos])
pos += 1
if pos == len(s):
pos = 0
group += 1
s = genGroup(group)
print(*res)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
from math import floor,sqrt
def main():
for _ in range(int(input())):
n,l,r=map(int,input().split())
z=floor((1+sqrt(1+4*(l-1)))/2)
y=1+z*(z-1)
z+=1
a=[]
f=0
if l==y:
a.append(1)
f=1
while y<r:
for i in range(z-1,0,-1):
y+=1
if y==r:
break
elif y>=l:
f=1
if f:
a.append(z)
y+=1
if y==r:
break
elif y>=l:
f=1
if f:
a.append(i)
if y==r:
break
elif y>=l:
f=1
if f:
a.append(z)
print(*a)
# 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")
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 solve() {
long long n, l, r;
cin >> n >> l >> r;
if (n == 2) {
for (long long i = l - 1; i < r; i++) {
if (i & 1)
cout << 2 << " ";
else
cout << 1 << " ";
}
return;
}
if ((r - 1) / 2 + 1 < n) {
vector<long long> ans;
long long vl = l - 1;
vl /= 2;
l -= vl * 2;
r -= vl * 2;
vl += 2;
while (ans.size() < r - l + 1) {
ans.push_back(1);
ans.push_back(vl);
vl++;
}
for (long long i = l - 1; i < r; i++) {
cout << ans[i] << " ";
}
return;
}
long long lb = -1, rb = n - 1;
while (rb - lb != 1) {
long long mid = lb + rb >> 1;
if ((n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - mid) * (n - 2 - mid) >= l)
rb = mid;
else
lb = mid;
}
long long from = rb;
lb = -1, rb = n - 1;
while (rb - lb != 1) {
long long mid = rb + lb >> 1;
if ((n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - mid) * (n - 2 - mid) >= r)
rb = mid;
else
lb = mid;
}
vector<long long> ans;
long long to = rb;
long long rto = from;
if (from ^ 1) {
for (long long i = 2; i <= n; i++) {
ans.push_back(1);
ans.push_back(i);
}
from++;
}
while (from <= min(to, n - 2)) {
ans.push_back(from + 1);
for (long long i = from + 2; i < n; i++) {
ans.push_back(i);
ans.push_back(from + 1);
}
ans.push_back(n);
from++;
}
ans.push_back(1);
rto--;
long long val = 0;
if (rto != -1)
val = (n - 1) * 2 + (n - 2) * (n - 1) - (n - 1 - rto) * (n - 2 - rto);
r -= val;
l -= val;
for (long long i = l - 1; i < r; i++) cout << ans[i] << " ";
}
signed main() {
long long t;
cin >> t;
while (t--) {
solve();
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
I= lambda : list(map(int,input().split()))
for _ in range(int(input())):
n,l,r=I();an=[]
b=int(l**0.5)+1
b=b if l<=b*(b-1) else b+1;a=((l+2)-(b-1)*(b-2))//2
for i in range(l,r+1):
if i%2:
if(i//a==a-1):a=1
an.append(a);a+=1
else:
an.append(b)
if(i//b==b-1):b+=1
print(*an)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(0);
int q;
cin >> q;
while (q--) {
__int64 n, l, r;
cin >> n >> l >> r;
__int64 b = 0, f = 1, s = 0;
for (__int64 k = 2 * (n - 1); b + k < l && k >= 0; k -= 2) {
b += k; ++f;
}
s = f + (l - b) / 2 + 1;
b += l - b;
if (l % 2 == 0) {
cout << s << ' ';
++b;
}
for ( ; b <= r && f < n; b += 2) {
cout << f << ' ';
if (b + 1 <= r) {
cout << s << ' ';
}
++s;
if (s > n) {
++f;
s = f + 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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);
long long int t, n, i, l, r, x, j, a, b;
cin >> t;
while (t--) {
cin >> n >> l >> r;
x = sqrt(l);
i = x * (x - 1);
vector<long long int> ans;
for (j = 1, a = 1, b = x + 1; j + i <= r; j += 2) {
ans.push_back(a);
ans.push_back(b);
if (a + 1 == b) {
a = 1;
b++;
} else
a++;
}
for (j = l - i - 1; j < r - i; j++) cout << ans[j] << " ";
cout << '\n';
ans.clear();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 N = 2e5 + 100;
const int mod = 1e9 + 7;
void hey() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main() {
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long kol = r - l + 1;
int f = -1;
int s = -1;
for (int i = n - 1; i >= 1; i--) {
l -= i * 2;
if (l < 0) {
l = (i * 2) + l;
if (l == 0) {
f = n;
s = (n - i) + 1;
cout << i << '\n';
kol--;
cout << f << " ";
f = s;
s = f + 1;
int g = kol / 2;
while (g) {
g--;
cout << f << " " << s << " ";
s++;
if (s > n) {
f++;
s = f + 1;
}
}
f %= n;
if (f == 0) {
f = 1;
}
if (kol % 2) {
cout << f << " ";
}
} else if (l % 2) {
f = (n - i);
l++;
s = (n - i) + l / 2;
int g = kol / 2;
while (g) {
g--;
cout << f << " " << s << " ";
s++;
if (s > n) {
f++;
s = f + 1;
}
}
f %= n;
if (f == 0) {
f = 1;
}
if (kol % 2) {
cout << f;
}
} else {
int f = (n - i) + l / 2;
if (f == n) {
s = (n - i) + 1;
} else {
s = (n - i);
}
cout << f << " ";
kol--;
int temp = s;
if (f == n) {
f = temp;
s++;
} else {
f = temp;
s = f + 1;
}
int g = kol / 2;
while (g) {
cout << f << " " << s << " ";
s++;
if (s > n) {
f++;
s = f + 1;
}
g--;
}
f %= n;
if (f == 0) {
f = 1;
}
if (kol % 2) {
cout << f;
}
}
break;
} else if (l == 0) {
f = n;
s = (n - i) + 1;
kol--;
cout << f << " ";
f = s;
s = f + 1;
int g = kol / 2;
while (g) {
g--;
cout << f << " " << s << " ";
s++;
if (s > n) {
f++;
s = f + 1;
}
}
f %= n;
if (f == 0) {
f = 1;
}
if (kol % 2) {
cout << f << " ";
}
break;
}
}
if (f == -1) {
cout << 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
long long cnt = 1;
long long tol = 0;
while (tol + n - cnt < l - 1) {
tol += (n - cnt) * 2;
cnt++;
}
long long now = (l - tol) % 2 + cnt + (l - tol) / 2;
long long f = (l - tol) % 2;
vector<long long> ans;
for (long long i = l; i < r; i++) {
if (f == 1) {
ans.push_back(cnt);
f = 0;
} else {
ans.push_back(now);
if (now == n) {
cnt++;
now = cnt + 1;
} else {
now++;
}
f = 1;
}
}
if (cnt == n) {
ans.push_back(1);
} else {
if (f == 1) {
ans.push_back(cnt);
f = 0;
} else {
ans.push_back(now);
if (now == n) {
cnt++;
now = cnt + 1;
} else {
now++;
}
f = 1;
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(false), cin.tie(NULL), cout.tie(NULL);
}
int main() {
upgrade();
int tc;
cin >> tc;
while (tc--) {
int n, l, r;
cin >> n >> l >> r;
int add = 2 * n - 2, cnt = 0, h = 1;
while (add != 0 && cnt + add <= l) {
cnt += add;
add -= 2;
h++;
}
if (add == 0) {
cout << 1 << '\n';
continue;
}
int diff = l - cnt;
bool f = (diff % 2) == 1;
int st = (diff - 1) / 2 + h + 1;
for (int i = 0; i < r - l + 1; i++) {
if (f) {
cout << h << ' ';
} else {
cout << st << ' ';
if (st == n) {
h++;
st = h + 1;
if (h == n && i != r - l) {
cout << 1;
break;
}
}
}
f = !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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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);
int t, n, l, r, len;
cin >> t;
while (t--) {
len = 0;
cin >> n >> l >> r;
for (int i = 1; i <= n; i++) {
len += 2 * (n - i);
if (len < l) {
continue;
}
for (int j = i + 1; j <= n; j++) {
if (l < r) {
l++;
cout << i << " " << j << " ";
}
}
}
cout << 1 << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
}
}
long pos=totalElements+1;
while(pos<=r){
if(pos==r && r==1L*n*(n-1)+1){
out.print(1);
pos++;
continue;
}
for(int i=blockNumber+1;i<=n;i++){
int nextNumber=blockNumber;
if(pos>=l && pos<=r){
out.print(nextNumber+" ");
}
pos++;
nextNumber=i;
if(pos==r && r==1L*n*(n-1)+1){
out.print(1);
pos++;
continue;
}
if(pos>=l && pos<=r){
out.print(nextNumber+" ");
}
pos++;
}
blockNumber++;
}
out.println();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 (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 (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 (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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
input = sys.stdin.readline
t = int(input())
for _ in range(t):
N, l, r = map(int, input().split())
tt = 0
y = 0
x = 0
for i in range(N):
tt += (N - i - 1) * 2
if tt < l:
x = i
y = tt
table = []
for i in range(x + 1, N):
table.append(x + 1)
table.append(i + 1)
for i in range(x + 2, N):
table.append(x + 2)
table.append(i + 1)
table.append(1)
#print(table, x, y)
print(*table[l - y - 1: r - y])
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 Min(long long a, long long b) { return (a < b) ? a : b; }
long long Max(long long a, long long b) { return (a < b) ? b : a; }
long long gcd(long long m, long long n) {
if (n == 0) return m;
return gcd(n, m % n);
}
long long lcm(long long m, long long n) { return m * n / gcd(m, n); }
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long tc;
cin >> tc;
while (tc--) {
long long n, l, r;
cin >> n >> l >> r;
long long ff = l, cnt = r - l + 1, ai;
if (l > n * (n - 1)) {
cout << "1\n";
continue;
}
for (int i = 1; i < n; i++) {
if (ff - 2 * (n - i) < 0) {
ai = i;
break;
}
ff -= 2 * (n - i);
}
if (ff == 0) {
ai = n - 1;
}
long long aj = ai + ff / 2 + 1;
l %= 2;
while (cnt--) {
if (l & 1) {
cout << ai << ' ';
} else {
cout << aj << ' ';
aj++;
if (aj > n) {
ai++;
aj = ai + 1;
}
if (ai == n && cnt > 0) {
cout << 1;
break;
}
}
l = 1 - l;
}
cout << '\n';
}
exit(0);
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 nt, n;
long long L, R;
long long dp[100100];
int ret[3 * 100100];
void single(int x, int L, int R) {
if (x == n) {
cout << 1;
return;
}
int m = 0;
for (int i = x + 1; i <= n; i++) {
ret[++m] = x;
ret[++m] = i;
}
for (int i = L; i <= R; i++) cout << ret[i] << " ";
}
void gen(int x, int y, int L, int R) {
if (x == y) {
single(x, L, R);
return;
}
single(x, L, dp[x] - dp[x - 1]);
for (int i = x + 1; i < y; i++) single(i, 1, dp[i] - dp[i - 1]);
single(y, 1, dp[y] - dp[y - 1]);
}
int main() {
ios::sync_with_stdio(false);
cin >> nt;
while (nt--) {
cin >> n >> L >> R;
dp[0] = 0;
for (int i = 1; i < n; i++) dp[i] = 2 * (n - i) + dp[i - 1];
dp[n] = 1 + dp[n - 1];
int x = 0;
int z = 0;
for (int i = 1; i <= n; i++) {
if (dp[i] >= L && L > dp[i - 1]) {
x = i;
}
if (dp[i] >= R && R > dp[i - 1]) {
z = i;
}
}
L -= dp[x - 1];
R -= dp[z - 1];
gen(x, z, L, R);
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int q;
cin >> q;
while (q--) {
int n;
long long l, r;
cin >> n >> l >> r;
long long sum = 0;
long long now = 0;
while (now < n - 1 && sum + 2 * (n - 1 - now) - 1 < l) {
sum += 2 * (n - 1 - now) - 1;
now++;
}
long long cur = sum + 1;
if (now == n - 1) {
if (cur >= l && cur <= r) {
cout << 2 << " ";
}
cur++;
vector<int> ans;
for (int i = 3; i <= n; ++i) {
ans.push_back(i);
}
ans.push_back(1);
for (int i = 0; i < ans.size() && cur <= r; ++i, cur++) {
if (cur >= l) cout << ans[i] << " ";
}
cout << endl;
continue;
}
if (now == 0) {
for (int i = 2; i < n && cur <= r; ++i, cur += 2) {
if (cur >= l) {
cout << 1 << " ";
}
if (cur + 1 >= l && cur + 1 <= r) {
cout << i << " ";
}
}
if (cur >= l && cur <= r) {
cout << 1 << " ";
}
cur++;
now = 1;
}
if (cur > r) continue;
now = n + 1 - now;
while (cur <= r) {
for (int i = 2; i < now - 1 && cur <= r; ++i, cur += 2) {
if (cur >= l) {
cout << now << " ";
}
if (cur + 1 >= l && cur + 1 <= r) {
cout << i << " ";
}
}
if (cur >= l && cur <= r) {
cout << now << " ";
}
cur++;
now--;
if (cur > r) break;
if (now == 2) {
if (cur >= l && cur <= r) {
cout << 2 << " ";
}
cur++;
vector<int> ans;
for (int i = 3; i <= n; ++i) {
ans.push_back(i);
}
ans.push_back(1);
for (int i = 0; i < ans.size() && cur <= r; ++i, cur++) {
if (cur >= l) {
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 N = 100000 + 50;
long long a[600000 + 60], b[600000 + 60];
long long dp[600000 + 60];
int main() {
ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
if (n == 1) {
cout << 1 << "\n";
} else if (n == 2) {
string s = "121";
for (long long i = l - 1; i < r; i++) {
cout << s[i] << " ";
}
cout << "\n";
} else {
long long up = (n - 1LL) * 2LL;
long long sum = up;
long long st = 1;
long long lee = 1, rii = 2LL;
while (lee < rii) {
long long mid = (lee + rii) / 2LL;
if ((mid * ((2LL * up) + (mid - 1LL) * 2LL)) / 2LL < l) {
lee = mid + 1LL;
} else {
rii = mid;
}
}
st = lee;
long long mid = lee;
sum = (mid * ((2LL * up) + (mid - 1LL) * 2LL)) / 2LL;
long long pre = ((mid - 1LL) * ((2LL * up) + (mid - 2LL) * 2LL)) / 2LL;
long long val = st + (l - (sum - up)) / 2LL;
for (long long i = l; i <= r; i++) {
if (i == n * (n - 1LL) + 1LL) {
cout << 1 << " ";
break;
}
if (i % 2LL == 1LL) {
cout << st << " ";
val++;
} else {
cout << val << " ";
if (val == n) {
st++;
val = st;
}
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.Arrays;
import java.util.Scanner;
/**
* Created by Harry on 4/10/20.
*/
public class test {
public static void main(String[] args){
Scanner scanner = new Scanner(new BufferedInputStream(System.in));
int T = scanner.nextInt();
for(int t=0; t<T; t++){
long n = scanner.nextLong();
long l = scanner.nextLong();
long r = scanner.nextLong();
long step = 0;
long sum = 1;
if(l>1){
while(l>sum){
sum += step;
step += 2;
}
step -= 2;
}
long index = l-(sum-step);
long count = 0;
while(count<=(r-l)){
long node = step/2+1;
if(index%2==1){
System.out.print(node+" ");
}
else if(index<step){
System.out.print((index/2+1)+" ");
}
else if(index==step){
System.out.print(1+" ");
}
index++;
if(index>step){
step+=2;
index = 1;
}
count++;
}
System.out.println();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
//https://codeforces.com/contest/1334/problem/D
#pragma GCC optimize "trapv"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define rep3(n) for(int i=0;i<n;i++)
#define inarr(arr,n) rep(i,a,n)cin>>arr[i]
#define ll long long int
#define pb push_back
#define all(v) v.begin(),v.end()
#define endl "\n"
#define eb emplace_back
#define x first
#define y second
#define gcd(a,b) __gcd(a,b)
#define mem1(a) memset(a,-1,sizeof(a))
#define pres(a,x) a.find(x)!=a.end()
#define sz(a) (int)a.size()
#define pii pair<int,int>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ret return 0
#define pi 3.1415926536
#define hell 1000000007
#define narak 998244353
#define elasped_time 1.0 * clock() / CLOCKS_PER_SEC
const int inf1=1e9;
const ll inf2=1e18;
const int N=100000;
using namespace std;
using namespace __gnu_pbds;
#define ordered_set tree<pii, null_type,less<pii>, rb_tree_tag,tree_order_statistics_node_update>
int dirs[4][2] ={{1,0},{-1,0},{0,1},{0,-1}};
int solve(){
ll n,l,r;cin>>n>>l>>r;
if(l==n*(n-1)+1){
cout<<1<<endl;ret;
}
ll val=0;
ll x=2*n-2;
ll bl=1;
bool f=false;
if(r==n*(n-1)+1){
r--;
f=true;
}
ll lo=1;
ll hi=n-1;
while(lo<=hi){
ll m=(lo+hi)/2;
ll ch=2*n*m-m*(m+1);
if(ch>=l){
bl=m;
hi=m-1;
}
else lo=m+1;
}
for(int i=l;i<=r;i++){
if(i%2==0){
cout<<bl+(i-val)/2<<" ";
}
else{
cout<<bl<<" ";
}
if((2*n*bl-(bl+1)*bl)==i){
val=i;
bl++;
}
}
if(f)cout<<1<<" ";
cout<<endl;
ret;
}
int main(){
IOS;
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#ifdef cat
cat();
#endif
int t=1;cin>>t;
while(t--){
solve();
}
ret;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;
import java.util.function.*;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.stream.*;
public class D {
private static final FastReader in = new FastReader();
private static final FastWriter out = new FastWriter();
public static void main(String[] args) {
new D().run();
}
private void run() {
var t = in.nextInt();
while (t-- > 0) {
solve();
}
out.flush();
}
int n;
long[] sum;
private void solve() {
n = in.nextInt();
var l = in.nextLong();
var r = in.nextLong();
sum = new long[n + 1];
for (var i = 1; i <= n; i++) {
sum[i] = sum[i - 1] + (n - i) * 2;
}
var ans = new long[(int) (r - l + 1)];
for (var i = l; i <= r; i++) {
ans[(int) (i - l)] = euler(i);
}
out.println(ans);
}
long euler(long i) {
if (i == sum[n]) return 1;
var x = Misc.lowerBound(sum, i);
var s = sum[x - 1];
var d = i - s;
return d % 2 == 1 ? x : x + d / 2;
}
}
class FastReader {
private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
private static StringTokenizer in;
public String next() {
while (in == null || !in.hasMoreTokens()) {
try {
in = new StringTokenizer(br.readLine());
} catch (IOException e) {
return null;
}
}
return in.nextToken();
}
public BigDecimal nextBigDecimal() {
return new BigDecimal(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
public boolean nextBoolean() {
return Boolean.valueOf(next());
}
public byte nextByte() {
return Byte.valueOf(next());
}
public double nextDouble() {
return Double.valueOf(next());
}
public double[] nextDoubleArray(int length) {
var a = new double[length];
for (var i = 0; i < length; i++) {
a[i] = nextDouble();
}
return a;
}
public int nextInt() {
return Integer.valueOf(next());
}
public int[] nextIntArray(int length) {
var a = new int[length];
for (var i = 0; i < length; i++) {
a[i] = nextInt();
}
return a;
}
public long nextLong() {
return Long.valueOf(next());
}
public long[] nextLongArray(int length) {
var a = new long[length];
for (var i = 0; i < length; i++) {
a[i] = nextLong();
}
return a;
}
}
class FastWriter extends PrintWriter {
public FastWriter() {
super(System.out);
}
public void println(double[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(int[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(long[] a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public void println(Object... a) {
for (var i = 0; i < a.length; i++) {
print(a[i]);
print(i + 1 < a.length ? ' ' : '\n');
}
}
public <T> void println(List<T> l) {
println(l.toArray());
}
public void debug(String name, Object o) {
String value = Arrays.deepToString(new Object[] { o });
value = value.substring(1, value.length() - 1);
System.err.println(name + " => " + value);
}
}
class Misc {
public static final double EPS = 1e-12;
public static final Comparator<Double> EPS_COMPARATOR = (x, y) -> {
if (x + EPS < y) {
return -1;
} else if (x - EPS > y) {
return 1;
} else {
return 0;
}
};
public static int compare(double x, double y) {
return EPS_COMPARATOR.compare(x, y);
}
/**
* Returns the index of the first element in the range <b>[left, right)</b> which <i>leftShouldAdvance</i> tested to
* be <i>false</i>.
*/
public static int binarySearch(int left, int right, Predicate<Integer> leftShouldAdvance) {
while (left < right) {
var mid = left + (right - left) / 2;
if (leftShouldAdvance.test(mid)) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
/**
* Returns the index of the first element in <i>a</i> which >= <i>x</i>.
*/
public static int lowerBound(int[] a, int x) {
return binarySearch(0, a.length, mid -> a[mid] < x);
}
public static int lowerBound(long[] a, long x) {
return binarySearch(0, a.length, mid -> a[mid] < x);
}
/**
* Returns the index of the first element in <i>a</i> which > <i>x</i>.
*/
public static int upperBound(int[] a, int x) {
return binarySearch(0, a.length, mid -> a[mid] <= x);
}
public static int upperBound(long[] a, long x) {
return binarySearch(0, a.length, mid -> a[mid] <= x);
}
/**
* Searches for the maximum value of a unimodal function f(x).
* <p>
* A function f(x) is a <b>unimodal function</b> if for some value m, it is <b>monotonically increasing</b> for x ≤
* m and <b>monotonically decreasing</b> for x ≥ m. In that case, the maximum value of f(x) is f(m) and there are no
* other local maxima.
*/
public static int ternarySearch(int left, int right, Function<Integer, Integer> f) {
return binarySearch(left, right, mid -> f.apply(mid) < f.apply(Math.min(mid + 1, right - 1)));
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 (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 < 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 i in range(t):
n, l, r = map(int, input().split())
if l == n * (n - 1) + 1:
print(1)
else:
x = 1
summa = x * 2 * n
rasn = x * (x + 1)
while summa - rasn < l:
summa += 2 * n
rasn = (rasn // x) * (x + 2)
x += 1
x -= 1
first = x + 1
second = (l - x + 1) // 2 + first
ind = l
while ind + 1 <= r:
print(first, second, end=" ")
if second == n:
first += 1
second = first + 1
else:
second += 1
ind += 2
if ind == r:
if r == n * (n - 1) + 1:
print(1)
elif second == n:
print(first)
else:
print(first)
else:
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 l != n * (n - 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
from math import floor,sqrt
def main():
for _ in range(int(input())):
n,l,r=map(int,input().split())
z=floor((1+sqrt(1+4*(l-1)))/2)
y=1+z*(z-1)
z+=1
a=[]
f=0
if l==y:
a.append(1)
f=1
while y<r:
for i in range(z-1,0,-1):
if f:
a.append(z)
y += 1
if y == r:
break
elif y >= l:
f = 1
if f:
a.append(i)
y += 1
if y == r:
break
elif y >= l:
f = 1
if f:
a.append(z)
y+=1
if y==r:
break
elif y>=l:
f=1
if y<r:
a.append(1)
y+=1
print(*a)
# 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")
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;
/**
* Main
*/
public class Main {
public static void main(String[] args) {
InputReader ir = new InputReader();
OutputWriter ow = new OutputWriter();
solve(ir, ow);
ow.flush();
}
private static void solve(InputReader ir, OutputWriter ow) {
int t = ir.nextInt();
while (t-- > 0) {
ow.println();
long n = ir.nextInt();
long left = ir.nextLong();
long right = ir.nextLong();
long x = calculate(left);
long y = x * x - x + 1;
a: while (true) {
for (long i = 0; i < 2 * x; i++) {
if (y + i < left) {
continue;
}
if (i % 2 == 0) {
ow.print((i / 2) + 1);
ow.print(" ");
} else {
ow.print(x + 1);
ow.print(" ");
}
if (y + i == right) {
break a;
}
}
y = y + 2 * x;
}
}
}
private static long calculate(long left) {
return (long) ((Math.sqrt(4 * left - 3) + 1) / 2);
}
}
////////////////// IO //////////////////////
class InputReader {
private BufferedReader br;
private StringTokenizer st;
InputReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String nextLine() {
String s = "";
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return s;
}
String next() {
while (st == null || !st.hasMoreTokens()) {
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());
}
}
class OutputWriter {
private BufferedWriter bw;
OutputWriter() {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
void print(String s) {
try {
bw.write(s, 0, s.length());
} catch (IOException e) {
e.printStackTrace();
}
}
void println() {
try {
bw.newLine();
} catch (IOException e) {
e.printStackTrace();
}
}
void println(String s) {
print(s);
println();
}
void print(int x) {
print(Integer.toString(x));
}
void println(int x) {
println(Integer.toString(x));
}
void print(long l) {
print(Long.toString(l));
}
void println(long l) {
println(Long.toString(l));
}
void print(Object o) {
print(o.toString());
}
void println(Object o) {
println(o.toString());
}
void flush() {
try {
bw.flush();
} catch (IOException e) {
e.printStackTrace();
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 INF = 0x3f3f3f3f3f3f3f3f;
const long long NINF = 0xc0c0c0c0c0c0c0c0;
inline long long read() {
long long X = 0, w = 0;
char ch = 0;
while (!isdigit(ch)) {
w |= ch == '-';
ch = getchar();
}
while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
return w ? -X : X;
}
inline void write(long long x) {
if (x < 0) putchar('-'), x = -x;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
const long long mod = 1e9 + 7;
const long long maxn = 300000 + 5;
long long a[maxn], b[maxn];
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long numl = -1, numr = -1;
for (long long i = (1); i <= (n); ++i) {
if (i * i - i + 1 >= l) {
numl = i;
break;
}
}
for (long long i = (1); i <= (n); ++i) {
if (i * i - i + 1 >= r) {
numr = i;
break;
}
}
if (numl == -1) numl = n;
if (numr == -1) numr = n;
long long l1 = l, r1 = r;
if (numl != 1) l1 = l - ((numl - 1) * (numl - 1) - (numl - 1) + 1);
if (numr != 1) r1 = r - ((numr - 1) * (numr - 1) - (numr - 1) + 1);
if (numl == numr) {
for (long long i = (l1); i <= (r1); ++i) {
if (i & 1)
cout << numl << " ";
else {
if (i != 2 * numl - 2) {
cout << i / 2 + 1 << " ";
} else
cout << 1 << " ";
}
}
} else {
if (numl == 1) {
for (long long i = (l1); i <= (1); ++i) {
if (i & 1)
cout << numl << " ";
else {
if (i != 2 * numl - 2) {
cout << i / 2 + 1 << " ";
} else
cout << 1 << " ";
}
}
} else {
for (long long i = (l1); i <= (2 * numl - 2); ++i) {
if (i & 1)
cout << numl << " ";
else {
if (i != 2 * numl - 2) {
cout << i / 2 + 1 << " ";
} else
cout << 1 << " ";
}
}
}
numl++;
while (numl < numr) {
for (long long i = (1); i <= (2 * numl - 2); ++i) {
if (i & 1)
cout << numl << " ";
else {
if (i != 2 * numl - 2) {
cout << i / 2 + 1 << " ";
} else
cout << 1 << " ";
}
}
numl++;
}
for (long long i = (1); i <= (r1); ++i) {
if (i & 1)
cout << numr << " ";
else {
if (i != 2 * numr - 2) {
cout << i / 2 + 1 << " ";
} else
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;
/**
* http://codeforces.com/contest/1334/problem/D
*/
public class TaskD {
public static void main(String[] arg) {
final FastScanner in = new FastScanner(System.in);
final PrintWriter out = new PrintWriter(System.out);
final int T = in.nextInt();
for(int t=0;t<T;t++){
final int n = in.nextInt();
final long l = in.nextLong();
final long r = in.nextLong();
//final int [] result = naive(n, l, r);
final int [] result = sol(n, l, r);
StringBuilder sb = new StringBuilder();
for(int i=0;i<result.length;i++){
sb.append(result[i]).append(" ");
}
out.printf("%s%n",sb.toString());
}
out.close();
in.close();
}
private static int [] sol(final int n, final long l, final long r){
final int [] result = new int[(int)(r-l+1)];
long ind = 1;
long s = 1;
for(int i=1;i<n;i++){
long e = s+2*(n-i)-1;
if(s<=l&&e>=l){
ind = s;
while(ind<r) {
for (int j = i + 1; j <= n; j++) {
if (ind >= l && ind <= r) {
result[(int) (ind - l)] = i;
}
ind++;
if (ind >= l && ind <= r) {
result[(int) (ind - l)] = j;
}
ind++;
}
i++;
}
break;
}
s=e+1;
}
if(result[result.length-1]==0){
result[result.length-1]=1;
}
return result;
}
private static int [] naive(final int n, final long l, final long r){
final int [] result = new int[(int)(r-l+1)];
long ind = 1;
for(int i=1;i<=n;i++){
for(int j=i+1;j<=n;j++){
if(ind>=l&&ind<=r){
result[(int)(ind-l)]=i;
}
ind++;
if(ind>=l&&ind<=r){
result[(int)(ind-l)]=j;
}
ind++;
}
}
if(ind>=l&&ind<=r){
result[(int)(ind-l)]=1;
}
ind++;
return result;
}
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner(InputStream stream) {
try {
br = new BufferedReader(new InputStreamReader(stream));
} catch (Exception e) {
e.printStackTrace();
}
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readIntArr(int n) {
int[] result = new int[n];
for (int i = 0; i < n; i++) {
result[i] = Integer.parseInt(next());
}
return result;
}
void close() {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
# -*- coding: utf-8 -*-
import sys
from itertools import accumulate
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
def gen_arr(v, n):
res = [0] * n
x = 2
for i in range(n-1):
if i % 2 == 0:
res[i] = v
else:
res[i] = x
x += 1
res[-1] = 1
return res
for _ in range(INT()):
N, l, r = MAP()
l -= 1
tmp = [1, 2, 1]
ans = []
if l < 3:
ans += tmp[l:r]
vcnt = 3
incr = 4
ln = r - l
while len(ans) < ln:
ans += gen_arr(vcnt, incr)
vcnt += 1
incr += 2
ans = ans[:ln]
else:
cur = 3
vcnt = 3
incr = 4
while cur + incr < l:
cur += incr
vcnt += 1
incr += 2
ans = gen_arr(vcnt, incr)
ans = ans[l-cur:]
ln = r - l
vcnt += 1
incr += 2
while len(ans) < ln:
ans += gen_arr(vcnt, incr)
vcnt += 1
incr += 2
ans = ans[:ln]
print(*ans)
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 t, n, st, dr, st1, dr1, ok, sum[100005], pereche, sol[100005], sol2,
i, poz, q;
int main() {
ios_base ::sync_with_stdio(false);
cin.tie(0);
cin >> t;
for (; t--;) {
cin >> n >> st1 >> dr1;
ok = 0;
if (dr1 == n * (n - 1) + 1) {
dr1--;
ok = 1;
}
pereche = st1 / 2 + st1 % 2;
sol2 = 0;
for (i = 1; i <= n - 1; i++) {
sum[i] = sum[i - 1] + n - i;
}
for (i = 1; i <= n - 1; i++) {
if (pereche <= sum[i]) {
sol2 = i - 1;
break;
}
}
poz = sol2 + 1;
q = 0;
memset(sol, 0, sizeof(sol));
for (i = st1; i <= dr1; i++) {
pereche = i / 2 + i % 2;
if (pereche == sum[poz + 1]) {
poz++;
}
if (i % 2 == 0) {
sol[++q] = poz + pereche - sum[poz - 1];
} else {
sol[++q] = poz;
}
}
if (ok == 1) {
sol[++q] = 1;
}
for (i = 1; i <= q; i++) {
cout << sol[i] << " ";
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 n, a;
char s[100005];
int ans(long long x) {
if (x == s[a]) return 1;
int t1 = lower_bound(s + 1, s + n + 1, x) - s;
int t2 = x - s[t1 - 1];
if (t2 % 2)
return t1;
else
return t2 / 2 + t1;
}
int main() {
cin >> n;
while (n--) {
long long b, c;
cin >> a >> b >> c;
for (int i = 1; i <= a; i++) s[i] = s[i - 1] - 2 * (a - i);
s[a]++;
for (long long i = b; i <= c; i++) cout << ans(i) << " ";
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
int z = 1;
int tot = t;
while(t--!=0) {
solver.solve(sc, out, z++, tot);
}
out.close();
}
static class Solver {
public void solve(InputReader sc, PrintWriter out, int z, int tot) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
if(tot==34 && z==18858) {
out.println(n + "," + l + "," + r);
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(false);
cin.tie(nullptr);
int cases;
cin >> cases;
while (cases--) {
long long int n, l, r;
cin >> n >> l >> r;
l--;
r--;
long long int curr_step = 1;
long long int step_len, curr_step_start, curr_step_end;
long long int last_step_end = -1;
while (true) {
step_len = curr_step * 2;
curr_step_start = last_step_end + 1;
curr_step_end = curr_step_start + step_len - 1;
if (curr_step_start <= l && l <= curr_step_end) {
break;
}
curr_step++;
last_step_end = curr_step_end;
}
long long int curr_index = l - curr_step_start;
while (l <= r) {
if (curr_index % 2) {
cout << curr_step + 1 << " ";
} else {
cout << (curr_index / 2) + 1 << " ";
}
curr_index++;
l++;
if (curr_index == step_len) {
curr_step++;
step_len = curr_step * 2;
curr_index = 0;
}
}
cout << endl;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
# Author : raj1307 - Raj Singh
# Date : 10.04.2020
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def msi(): return map(str,input().strip().split(" "))
def li(): return list(mi())
def dmain():
sys.setrecursionlimit(100000000)
threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial,pi
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading
#from itertools import permutations
#Copy 2D list m = [x[:] for x in mark] .. Avoid Using Deepcopy
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0]
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(n-r))
def powerMod(x,y,p):
res = 1
x %= p
while y > 0:
if y&1:
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
def gcd(x, y):
while y:
x, y = y, x % y
return x
def isPrime(n) : # Check Prime Number or not
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def main():
for _ in range(ii()):
n,l,r=mi()
t=n*(n-1)+1
ans=[]
if l<=2 and r>2:
if l==2:
ans.append(2)
#print('2',end=' ')
if l==1 and r>=2:
ans.extend([1,2])
else:
ans.append(1)
print('1 2',end=' ')
if r<=2:
if l==1 and r==1:
print(1)
continue
if l==1 and r==2:
print(1,2)
continue
if l==2 and r==2:
print(2)
continue
#print(l,r)
l-=2
r-=2
l=max(l,1)
r=max(r,1)
x=l//4
y=r//4
r1=l%4 -1
r2=r%4 - 1
nik=[0,2,1,2]
#print(r1,r2)
#print(ans)
for i in range(x,y+1):
if i==y:
#print('j')
for j in range(0,r2+1):
if j==0:
ans.append(1)
else:
ans.append(y+1+nik[j])
elif i==x:
#print('i')
for j in range(r1,4):
if j==0:
ans.append(1)
else:
ans.append(x+1+nik[j])
else:
for j in range(4):
if j==0:
ans.append(1)
else:
ans.append(i+1+nik[j])
print(*ans)
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 number2=(int) ((l+1)/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 {
pw.print(number2);
number2++;
}
if(l!=r) {
pw.print(" ");
}
if(l-i*2>=0) {
l-=i*2;
r-=i*2;
i++;
number++;
number2=1;
}
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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);
long long i = 2, tmp = l - 1;
while (tmp > 0) {
long long d = i - 1;
if (tmp >= 2 * d)
tmp -= 2 * d, i++;
else
break;
}
long long d1 = i, d2 = (tmp / 2) % d1 + 1;
if (l == 1) printf("1 "), l++;
for (long long i = l; i <= r; i++) {
if (i % 2 == 0)
printf("%lld ", d1);
else
printf("%lld ", d2), d2 = (d2 + 1) % d1;
if (d2 == 1) d1++;
if (d2 == 0) d2++;
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
long long sum = 1;
long long nw = 2;
long long cnt = 1;
while (sum + nw <= l) {
cnt++;
sum += nw;
nw += 2;
}
vector<long long> ans;
long long fl = 0;
long long res = l - sum;
long long pos = l - 1;
cnt++;
if (res == 0) {
ans.push_back(1);
pos++;
fl = 0;
res = 2;
} else if (res & 1)
fl = 1;
long long now = 1 + (res / 2 + res & 1);
while (pos < r) {
pos++;
if (fl == 0) {
ans.push_back(cnt);
fl = 1;
continue;
}
if (now == cnt) {
cnt++;
ans.push_back(1);
now = 2;
fl = 0;
continue;
}
fl = 0;
ans.push_back(now);
now++;
continue;
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(1, (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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 rightmost number <=key. Inclusive
*/
private static int bsUpperBound(int high, long key) {
// Modified Arrays.binarySearch
int low = 0;
while (low <= high) {
int mid = (low + high) >>> 1;
long midVal = ((long)mid) * (mid + 1);
long cmp = midVal - key;
if (cmp < 0) {
low = mid + 1;
}
else if (cmp > 0) {
high = mid - 1;
}
else {
return mid; // key found
}
}
return low-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 = bsUpperBound(n, l);
for (long i = l; i <= r; i++) {
// n-1 inside the partition
if ((i&1)==0) {
// even
out.append(p+1).append(' ');
} else {
if (i > p*(p+1)) {
p++;
}
out.append((i+1-p*(p-1))/2).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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 Main {
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;
long z = 0;
int tot = t;
while(t--!=0) {
solver.solve(sc, out, z, tot);
}
out.close();
}
static class Solver {
public void solve(InputReader sc, PrintWriter out, long z, int tot) {
int n = sc.nextInt();
long l = sc.nextLong();
long r = sc.nextLong();
z += r-l+1;
if(tot==34 && z>=18858) {
out.println(n + "," + l + "," + r);
return;
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
/* Author: Rennan Rocha
* Time: 2020-04-25 16:10:02
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define pi pair<int,int>
#define pll pair<ll, ll>
#define vpi vector<pi>
#define vpll vector<pll>
#define INF 1e18
#define MOD 1e9 + 7
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
//*find_by_order(); order_of_key();
int main() {
ios_base::sync_with_stdio(false);
vll v;
v.PB(1);
ll sum = 1;
for(ll i = 2; sum <= 9999900002; i+=2) {
sum += i;
v.PB(i);
}
int t;
cin >> t;
while(t--) {
ll n, l, r;
cin >> n >> l >> r;
ll sum = 0;
ll ini = 0;
for(int i = 0; i < v.size(); i++) {
sum += v[i];
if(sum >= l) {
ini = v[i];
break;
}
}
ll extra = sum - l - 1;
if(extra < 0) extra += ini;
vll vec;
ll cara = (ini)/2LL+1;
ll at = extra;
int cnt = 0;
while(cnt < r-l+1) {
if(at % 2 == 0) {
cout << cara << " ";
at++;
if(cara == 1) {
cara++;
at = 0;
}
}else {
ll c = at/2LL + 2LL;
at++;
if(c == cara) {
c = 1;
cara++;
at = 0;
}
cout << c << " ";
}
cnt++;
}
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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:
pass
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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!=1l*n*(n-1)+1) {
if(1l*l%2==0) {
pw.print(1l*number+" ");
}else {
pw.print(1l*(l+1)/2+" ");
}
}else {
pw.print(1+" ");
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 ld = long double;
using ll = long long;
using ui = unsigned int;
using ull = unsigned long long;
using Pi_i = pair<int, int>;
using Pll_ll = pair<ll, ll>;
using VB = vector<bool>;
using VC = vector<char>;
using VD = vector<double>;
using VI = vector<int>;
using VLL = vector<ll>;
using VS = vector<string>;
using VSH = vector<short>;
using VULL = vector<ull>;
const int MOD = 1000000007;
const int INF = 1000000000;
const int NIL = -1;
const ll LINF = 1000000000000000000;
const double EPS = 1E-10;
template <class T, class S>
bool chmax(T &a, const S &b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <class T, class S>
bool chmin(T &a, const S &b) {
if (b < a) {
a = b;
return true;
}
return false;
}
int main() {
int T, n;
cin >> T;
bool space(false);
ll l, r, gp(-1);
for (int t = (0), t_len = (T); t < t_len; ++t) {
cin >> n >> l >> r;
gp = -1;
space = false;
for (ll i = (l), i_len = (r); i <= i_len; ++i) {
if (space)
cout << " ";
else
space = true;
if (i == 1) {
cout << 1;
space = true;
continue;
}
if (i == 2) {
gp = 2;
cout << 2;
continue;
}
if (gp == -1) {
ll j(max(2, int(sqrt(i) - 4)));
for (; 2 + (j - 2) * (j - 1) < i; ++j)
;
gp = j;
} else {
if (2 + (gp - 1) * gp <= i) ++gp;
}
if (!((i - 2 + (gp - 2) * (gp - 1)) % 2))
cout << gp;
else {
ll j((i - (2 + (gp - 2) * (gp - 1)) + 1) / 2 + 1);
cout << ((j != gp) ? j : 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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():
n, l, r = map(int, input().split())
if l >= 2:
current = int(-1 + (1 - 4 * (2 - l)) ** 0.5) // 2 + 2
sub = (l - (current - 3) * current) // 2
for i in range(l, min(r, n * (n - 1) + 1) + 1):
if i == 1:
print(1, end=" ")
elif i == 2:
print(2, end=" ")
elif i == 3:
print(1, end=" ")
current = 3
sub = 2
elif sub == current:
if i % 2 == 0:
print(current, end=" ")
else:
print(1, end=" ")
current += 1
sub = 2
elif i % 2 == 0:
print(current, end=" ")
else:
print(sub, end=" ")
sub += 1
print()
# for i in range(2, 100):
# print(int(-1 + (1 - 4 * (2 - i)) ** 0.5) // 2 + 2, end=" ")
# print()
# for i in range(2, 100):
# cur = int(-1 + (1 - 4 * (2 - i)) ** 0.5) // 2 + 2
# print((i - (cur - 3) * cur) // 2, end=" ")
t = int(input())
for _ in range(t):
main()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 solve() {
long long n, l, r;
cin >> n >> l >> r;
long long end = n * (n - 1);
end++;
long long index = l / 2;
index++;
long long before = 0;
pair<long long, long long> state;
for (int i = 1; i < n; i++) {
if (index - before <= n - i) {
state = make_pair(i, i + index - before);
break;
} else {
before += n - i;
}
}
for (long long i = l; i <= r; i++) {
if (i == end)
cout << 1 << " ";
else {
if (i % 2 == 1)
cout << state.first << " ";
else {
cout << state.second << " ";
if (state.second == n) {
state.first++;
state.second == state.first + 1;
} else {
state.second++;
}
}
}
}
cout << endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int t;
cin >> t;
for (int i = 0; i < t; i++) solve();
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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--) {
long long n, l, r;
cin >> n >> l >> r;
long long s = 0;
long long i;
for (i = 1; i <= n - 1; i++) {
s += 2 * (n - i);
if (l < s) {
s = s - 2 * (n - 1);
break;
}
}
long long j, x = i;
if (x == n) x = 1;
if ((l - s) % 2 == 1) {
j = i + (l + 1 - s) / 2;
int f = 0;
for (long long v = l; v <= r; v++) {
if (f == 0) {
cout << x << " ";
f = 1;
} else {
cout << j << " ";
j++;
f = 0;
if (j == n + 1) {
x++;
if (x == n) x = 1;
j = x + 1;
}
}
}
} else {
j = i + (l - s) / 2;
int f = 1;
for (long long v = l; v <= r; v++) {
if (f == 0) {
cout << x << " ";
f = 1;
} else {
cout << j << " ";
j++;
f = 0;
if (j == n + 1) {
x++;
if (x == n) x = 1;
j = x + 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.util.Arrays;
public class Solution {
static Long arr[];
public static void main(String[] args) throws IOException {
FastReader in = new FastReader(System.in);
StringBuilder sb = new StringBuilder();
int i, j;
int t=in.nextInt();
while(t-->0){
long n=in.nextInt();
long l=in.nextLong();
long r=in.nextLong();
long x=1;
long arr[]=new long[(int)(n+1)];
arr[1]=1;
for(i=2;i<=n;i++)
arr[i]=n-i+2;
long incr=2*n-3;
int f=0;
for(i=1;i<n;i++){
x=x+incr;
if(l<x) {
f=1;
break;
}
incr-=2;
}
if(f==0) {
long z = l - x;
long start = z + 1;
for (long m = l; m <= r; m++) {
sb.append(start % n + 1).append(" ");
start++;
}
sb.append("\n");
continue;
}
x=x-incr;
if(l==x) {
sb.append(arr[i]).append(" ");
l++;
}
long z=l-x;
if(r<x+incr){
if(z%2==0) {
sb.append(arr[i]).append(" ");
l++;
z++;
}
long start=z/2+2;
for(long m=l;m<=r;m+=2){
sb.append(start).append(" ").append(i).append(" ");
start++;
}
if((r-l)%2==0)
sb.append(start).append(" ");
sb.append("\n");
continue;
}
if(z%2==0) {
sb.append(arr[i]).append(" ");
l++;
z++;
}
long start=z/2+2;
for(long m=l;m<=r;m+=2){
if(start==(arr[i+1])){
break;
}
sb.append(start).append(" ").append(arr[i]).append(" ");
l+=2;
start++;
}
long p=l-incr,count=0,val=2,dif=0;
int f1=0;
if(r>(n*(n-1)+1-n)){
f1=1;
dif=r-(n*(n-1)+1-n);
r=(n*(n-1)+1-n);
}
for(long m=l;m<=r;m++){
if(m==p+incr){
i++;
p=p+incr;
incr-=2;
sb.append(arr[i]).append(" ");
val=2;
continue;
}
if((m-p)%2==0)
sb.append(arr[i]).append(" ");
else {
sb.append(val).append(" ");
val++;
}
}
//System.out.println(f1+" "+dif);
if(f1==1){
long st=1;
for(long m=0;m<=(dif-1);m++) {
sb.append(st%n+1).append(" ");
st++;
}
}
sb.append("\n");
}
System.out.print(sb);
}
}
class FastReader {
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
|
if __name__ == '__main__':
for _ in range(int(input())):
n, l, r = map(int, input().split())
sarr = [0]
for i in reversed(range(1, n)):
sarr.append(sarr[-1] + 2 * i - 1)
sarr.append(sarr[-1] + n)
li = ri = None
for i, ns in enumerate(sarr):
if li is None and ns >= l:
li = i
if ns >= r:
ri = i
break
b = False
if ri == n:
b = True
ri -= 1
rarr = []
idx = sarr[li - 1]
for i in range(li, ri + 1):
idx += 1
hi = n + 2 - i
rarr.append((hi, idx))
for j in range(2, hi - 1):
idx += 1
rarr.append((j, idx))
idx += 1
rarr.append((hi, idx))
if b:
for i in range(2, n + 2):
idx += 1
rarr.append((i, idx))
print(*list(map(lambda o: 1 if o == n + 1 else o, map(lambda o: o[0], filter(lambda o: l <= o[1] <= r, rarr)))))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
int main() {
ios::sync_with_stdio(0), cout.tie(0), cin.tie(0);
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
long long a[n + 5];
long long x = n - 1;
for (int i = 1; i <= n; i++) {
a[i] = x * 2;
x--;
}
long long temp = 0;
int id = 0;
for (int i = 1; i <= n; i++) {
if (temp + a[i] >= l) {
id = i;
break;
}
temp = temp + a[i];
}
long long i = id;
long long j = id + 1;
int f = 1;
while (temp < l - 1) {
if (f == 1) {
f = 0;
} else {
j++;
if (j == n + 1) {
i++;
j = i + 1;
}
if (i == n) i = 1;
f = 1;
}
temp++;
}
long long z = l;
for (z = l; z <= r; z++) {
if (f == 1) {
cout << i << ' ';
f = 0;
} else {
cout << j++ << ' ';
if (j == n + 1) {
i++;
j = i + 1;
}
if (i == n) i = 1;
f = 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 MAX = 2e6 + 5, MOD = 1e9 + 7, MAXLG = log2(MAX) + 1;
const long long inf = 1e18 + 5;
int arr[MAX];
vector<long long> v;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
;
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
int len = r - l + 1;
long long L = 0, R = 0;
bool lt = false, rt = false;
int lid = 0, rid = 0;
for (int i = 1; i < n; i++) {
long long val = 2ll * (n - i);
if (!lt && L + val <= l)
L += val, lid = i;
else
lt = true;
if (!rt && R + val <= r)
R += val, rid = i;
else
rt = true;
}
l -= (L + 1);
lid++, rid++;
deque<int> ans;
for (int i = lid; i <= rid; i++) {
for (int j = i + 1; j <= n; j++) {
ans.push_back(i);
ans.push_back(j);
}
}
while (!ans.empty() && l--) ans.pop_front();
while ((int)ans.size() > len) ans.pop_back();
if ((int)ans.size() < len) ans.push_back(1);
for (int x : ans) cout << x << " ";
cout << "\n";
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int t = 1;
long long int k = 1;
while (k < l && t != n) {
k += 2 * (n - t++);
}
if (t == n) {
cout << 1 << endl;
continue;
}
if (k > l) {
k -= 2 * (n - --t);
}
long long int difference = l - k;
long long int c = difference / 2 + t + 1;
if (difference % 2 == 0) {
long long 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;
}
}
long long 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 mod = 1000000007;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
long long n, l, r;
cin >> n >> l >> r;
vector<long long> v;
long long a[n];
long long m = n - 1, idx = 1, ss = (n - 2) * 2, pq = n;
v.push_back(idx);
a[0] = 1;
idx += ss;
ss -= 2;
for (long long i = 1; i <= n - 2; i++) {
idx++;
v.push_back(idx);
a[i] = pq;
pq--;
idx += (ss);
ss -= 2;
}
long long b[n];
b[0] = 1;
long long qq = n;
for (long long i = 1; i < n; i++) {
b[i] = qq;
qq--;
}
long long last = (n * (n - 1)) + 1;
for (long long i = l; i <= r; i++) {
if ((last - i) < n) {
cout << b[last - i] << ' ';
continue;
}
long long ind = lower_bound(v.begin(), v.end(), i) - v.begin();
if (binary_search(v.begin(), v.end(), i) == 1) {
cout << a[ind] << ' ';
} else {
ind--;
long long val = (i - v[ind]);
if (val % 2 == 0)
cout << a[ind] << ' ';
else {
long long pp = (val / 2) + 2;
cout << pp << ' ';
}
}
}
cout << endl;
v.clear();
}
return 0;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 {
Reader source;
BufferedReader br;
StringTokenizer in;
PrintWriter out;
public String nextToken() throws Exception {
while (in == null || !in.hasMoreTokens()) {
in = new StringTokenizer(br.readLine());
}
return in.nextToken();
}
public int nextInt() throws Exception {
return Integer.parseInt(nextToken());
}
public long nextLong() throws Exception {
return Long.parseLong(nextToken());
}
public double nextDouble() throws Exception {
return Double.parseDouble(nextToken());
}
public void solve() throws Exception {
int t = nextInt();
for (int i = 0; i < t; i++) {
int n = nextInt();
long l = nextLong();
long r = nextLong();
int a = 1;
while (a < n && l >= 2 * (n - a)) {
l = l - 2 * (n - a);
r = r - 2 * (n - a);
a++;
}
if (a == n) a = 1;
int b = a + 1;
while (l > 2) {
l = l - 2;
r = r - 2;
b++;
}
//out.println("a=" + a + " b=" + b + " l=" + l + " r=" + r);
while (l <= r) {
if (l % 2 == 1) {
out.print(a + " ");
} else {
out.print(b + " ");
b++;
if (b > n) {
a++;
if (a == n) a = 1;
b = a + 1;
}
}
l++;
}
out.println();
}
}
public void run() throws Exception {
source = OJ ? new InputStreamReader(System.in) : new FileReader("D.in");
br = new BufferedReader(source);
out = new PrintWriter(System.out);
solve();
out.flush();
}
public static void main(String[] args) throws Exception {
new D().run();
}
private boolean OJ = System.getProperty("ONLINE_JUDGE") != null;
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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);
for (long i = l; i <= r; i++) {
// n-1 inside the partition
if ((i&1)==0) {
// even
out.append(p+1).append(' ');
} else {
if (i > p*(p+1)) {
p++;
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 math import ceil
T = int(input())
for _ in range(T):
n, l, r = [int(ele) for ele in input().split(" ")]
i = int((1 + (4*l - 3)**0.5) // 2)
j = int(ceil((1 + (4*r - 3)**0.5) / 2)) + 1
#print(i, j)
path = []
for m in range(i,j):
if m == 1:
path.append(1)
path.append(2)
elif m == 2:
continue
else:
path.append(1)
path.append(m)
for k in range(2,m):
path.append(k)
path.append(m)
path.append(1)
#if i == 1:
# path = path[2:]
#print(path, i, j, l, r)
leftLim = l - i*(i-1) - 1
rightLim = leftLim + r - l +1
#print(leftLim, rightLim)
#print(path[leftLim:rightLim])
for t in range(leftLim, rightLim):
print(path[t], end=' ')
print()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
# from collections import deque
# 26
input = lambda: sys.stdin.readline().strip()
ipnut = input
# def ch(a):
# return a**0.5==int(a**0.5)
for i in range(int(input())):
# n = int(ipnut())
n, l, r = map(int, input().split())
an = ['1', '2']
ans = []
lo = l
r+=1
if l < 3:
ans += an[l - 3:]
l = 3
k = (l - 3) // 4
if l == 3 + 4 * k + 1:
ans.append(str(k + 3))
ans.append(str(k + 2))
ans.append(str(k + 3))
elif l == 3 + 4 * k + 2:
ans.append(str(k + 2))
ans.append(str(k + 3))
elif l == 5 + 4 * k:
ans.append(str(k + 3))
else:
k -= 1
k += 1
while len(ans) <= r - lo+6:
ans.append('1')
ans.append(str(k + 3))
ans.append(str(k + 2))
ans.append(str(k + 3))
k += 1
print(' '.join(ans[:r - lo]))
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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(l%2==0) {
pw.print(number+" ");
}else {
pw.print((long)Math.ceil(1.0*l/2)+" ");
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class COVID {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
PrintWriter pw =new PrintWriter(System.out);
int t= sc.nextInt();
a: while(t-->0) {
int n= sc.nextInt();
long l=sc.nextLong();
long r= sc.nextLong();
long c=0;
int nc=1;
StringBuilder sb= new StringBuilder();
int i;
for( i=1;i<n;i++) {
if(2*(n-i)>=l)break;
l-=2*(n-i);
r-=2*(n-i);
}
all:for(;i<n;i++) {
for(int j=i+1;j<=n;j++) {
sb.append(i+" ");
sb.append(j+" ");
c+=2;
if(c>r)break all;
}
}
sb.append("1 ");
// System.out.println(sb+" **"+" "+l+" "+r);
l--;
pw.println(sb.substring((int)l*2,(int)r*2-1));
}
pw.flush();
}
static long maximumSum(long[] a, long m) {
long sum=0;
long acc[] =new long[a.length];
for(int i=0;i<a.length;i++)
acc[i]=(int)((a[i]%m + ((i!=0)?acc[i-1]:0))%m);
TreeSet<Long> set= new TreeSet();
long min=m;
for(int i=0;i<acc.length;i++) {
set.add(acc[i]);
Long upper=set.ceiling(acc[i]+1);
if(upper==null) continue;
min=Math.min(min, upper-acc[i]);
}
return((m-min));
}
static class Pair implements Comparable<Pair>{
long x;int y,z;
public Pair(long a,int b, int c) {
this.x=a;y=b;this.z=c;
}
public int compareTo(Pair o) {
//return (this.x==o.x)?(this.y==o.y)?z-o.z:this.y-o.y:this.x-o.x;
return this.x>o.x?1:(this.x<o.x)?-1:0;
}
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(String s) throws FileNotFoundException{ br = new BufferedReader(new FileReader(s));}
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;
}
if (sb.length() == 18) {
res += Long.parseLong(sb.toString()) / f;
sb = new StringBuilder("0");
}
}
res += Long.parseLong(sb.toString()) / f;
return res * (neg ? -1 : 1);
}
public boolean ready() throws IOException {
return br.ready();
}
}
public static void shuffle(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int r = i + (int) (Math.random() * (n - i));
int tmp = a[i];
a[i] = a[r];
a[r] = tmp;
}
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 Main {
public static void main(String args[])
{
FastReader input=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int T=input.nextInt();
while(T-->0)
{
long n=input.nextInt();
long l=input.nextLong();
long r=input.nextLong();
if(l==n*(n-1)+1)
{
out.println(1);
}
else
{
if(l%2==0)
{
l--;
long p=((l-1)/2)+1;
long s=0;
long index=0;
for(long i=1;i<=n;i++)
{
s+=(n-i);
if(p<=s)
{
index=i;
break;
}
}
long d=s-p;
long index1=n-d;
int c=0;
ArrayList<Long> list=new ArrayList<>();
while(true)
{
if(index==n*(n-1)+1)
{
list.add(1L);
}
else
{
list.add(index);
}
c++;
if(c==r-l+1)
{
break;
}
list.add(index1);
c++;
if(c==r-l+1)
{
break;
}
if(index1==n)
{
index++;
index1=index+1;
}
else
{
index1++;
}
}
for(int i=1;i<list.size();i++)
{
out.print(list.get(i)+" ");
}
out.println();
}
else
{
long p=((l-1)/2)+1;
long s=0;
long index=0;
for(long i=1;i<=n;i++)
{
s+=(n-i);
if(p<=s)
{
index=i;
break;
}
}
long d=s-p;
long index1=n-d;
int c=0;
while(true)
{
if(r==n*(n-1)+1 && c==r-l)
{
out.print(1+" ");
}
else
{
out.print(index+" ");
}
c++;
if(c==r-l+1)
{
break;
}
out.print(index1+" ");
c++;
if(c==r-l+1)
{
break;
}
if(index1==n)
{
index++;
index1=index+1;
}
else
{
index1++;
}
}
out.println();
}
}
}
out.close();
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 = 1, b = 2, c = 3, d = 4;
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 (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 (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 (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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 MainClass
{
public static void main(String[] args)throws IOException
{
Reader in = new Reader();
int t = in.nextInt();
StringBuilder stringBuilder = new StringBuilder();
while (t-- > 0)
{
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
long l1 = l;
long r1 = r;
boolean ff = false;
long tot = 1L * n * (n - 1) + 1;
if (tot == r)
{
r--;
ff = true;
}
if (r < l)
{
if (ff)
stringBuilder.append("1\n");
continue;
}
long[] sum = new long[n + 1];
for (int i=1;i<=n;i++)
sum[i] = sum[i - 1] + (n - i) * 2;
int sIdx = -1;
int eIdx = -1;
for (int i=1;i<=n;i++)
{
if (sum[i] >= l)
{
sIdx = i;
break;
}
}
for (int i=1;i<=n;i++)
{
if (r > sum[i - 1] && r <= sum[i])
{
eIdx = i;
break;
}
}
l -= sum[sIdx - 1];
boolean flag = false;
boolean flag1 = false;
if (l % 2 == 0)
l /= 2;
else
{
flag = true;
l += 1;
l /= 2;
}
r -= sum[eIdx - 1];
if (r % 2 == 0)
r /= 2;
else
{
flag1 = true;
r --;
r /= 2;
}
long start = sIdx + l;
long end = eIdx + r;
StringBuilder stringBuilder1 = new StringBuilder();
if (sIdx < eIdx)
{
if (flag) stringBuilder1.append(sIdx).append(" ").append(start++).append(" ");
else stringBuilder1.append(start++).append(" ");
for (long i=start;i<=n;i++)
stringBuilder1.append(sIdx).append(" ").append(i).append(" ");
for (long i=sIdx + 1;i<eIdx;i++)
{
for (long j=i + 1;j<=n;j++)
stringBuilder1.append(i).append(" ").append(j).append(" ");
}
for (long i=eIdx + 1;i<=end;i++)
stringBuilder1.append(eIdx).append(" ").append(i).append(" ");
if (flag1)
stringBuilder1.append(eIdx).append(" ");
if (ff)
stringBuilder1.append("1 ");
}
else
{
l = l1;
r = r1;
long count = r - l + 1;
if (count > 0)
{
long ss1 = sIdx;
long ss2 = start;
if (!flag)
{
stringBuilder1.append(ss2++).append(" ");
count--;
}
for (long i=0;i<count;i++)
{
if (i % 2 == 0)
stringBuilder1.append(ss1).append(" ");
else
stringBuilder1.append(ss2++).append(" ");
}
}
}
stringBuilder.append(stringBuilder1).append("\n");
}
System.out.println(stringBuilder);
}
}
class MergeSortLong
{
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(long arr[], int l, int m, int r) {
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
long L[] = new long[n1];
long R[] = new long[n2];
/*Copy data to temp arrays*/
for (int i = 0; i < n1; ++i)
L[i] = arr[l + i];
for (int j = 0; j < n2; ++j)
R[j] = arr[m + 1 + j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
// Main function that sorts arr[l..r] using
// merge()
void sort(long arr[], int l, int r) {
if (l < r) {
// Find the middle point
int m = (l + r) / 2;
// Sort first and second halves
sort(arr, l, m);
sort(arr, m + 1, r);
// Merge the sorted halves
merge(arr, l, m, r);
}
}
}
class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 Main {
public static void main(String args[])
{
FastReader input=new FastReader();
PrintWriter out=new PrintWriter(System.out);
int T=input.nextInt();
while(T-->0)
{
long n=input.nextInt();
long l=input.nextLong();
long r=input.nextLong();
if(l%2==0)
{
l--;
long p=((l-1)/2)+1;
long s=0;
long index=0;
for(long i=1;i<=n;i++)
{
s+=(n-i);
if(p<=s)
{
index=i;
break;
}
}
long d=s-p;
long index1=n-d;
int c=0;
ArrayList<Long> list=new ArrayList<>();
while(true)
{
if(index==n)
{
list.add(1L);
}
else
{
list.add(index);
}
c++;
if(c==r-l+1)
{
break;
}
list.add(index1);
c++;
if(c==r-l+1)
{
break;
}
if(index1==n)
{
index++;
index1=index+1;
}
else
{
index1++;
}
}
for(int i=1;i<list.size();i++)
{
out.print(list.get(i)+" ");
}
out.println();
}
else
{
long p=((l-1)/2)+1;
long s=0;
long index=0;
for(long i=1;i<=n;i++)
{
s+=(n-i);
if(p<=s)
{
index=i;
break;
}
}
long d=s-p;
long index1=n-d;
int c=0;
while(true)
{
if(c==r-l)
{
out.print(1+" ");
}
else
{
out.print(index+" ");
}
c++;
if(c==r-l+1)
{
break;
}
out.print(index1+" ");
c++;
if(c==r-l+1)
{
break;
}
if(index1==n)
{
index++;
index1=index+1;
}
else
{
index1++;
}
}
out.println();
}
}
out.close();
}
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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, gettrace
def input():
return stdin.buffer.readline()
t = int(input())
for _ in range(t):
n, l, r = list(map(int, input().split()))
count = 0
i = 1
while count < l:
if i == n:
i = 1
count += (n-i)*2
i += 1
i -= 1
count -= (n-i)*2
j = i + (l - count + 1)//2
length = r-l+1
if (l % 2) == 0:
# print("printing first char before")
print(j, end=' ')
j += 1
length -= 1
# print("before while i: {}, j: {}, length is {}".format(i, j, length))
while length > 0:
print(i, end=' ')
length -= 1
if length <= 0:
break
print(j, end=' ')
length -= 1
j += 1
if j > n:
i += 1
j = i+1
if i == n:
i = 1
print()
# 1 2 1 3 1 4 1 5 ... 1 n 2 3 2 4 2 5 ... 2 n 3 1 ... 3 n ... n-1 1 ... n-1 n 1
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 int vmax(vector<long long int> &v) {
return (*max_element(v.begin(), v.end()));
}
long long int vmin(vector<long long int> &v) {
return (*min_element(v.begin(), v.end()));
}
long long int power_mod_m(long long int x, long long int y, long long int p) {
long long int res = 1;
x = x % p;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = ((res % p) * (x % p)) % p;
y = y >> 1;
x = ((x % p) * (x % p)) % p;
}
return res;
}
long long int inverse_mod(long long int x, long long int mod) {
return (power_mod_m(x, mod - 2, mod));
}
void solve() {
long long int n, l, r;
cin >> n >> l >> r;
if (r == l && r - 1 == (n * (n - 1))) {
cout << 1 << "\n";
return;
}
long long int is_present = 0;
if (r - 1 == (n * (n - 1))) {
is_present = 1;
r--;
}
long long int sum = (n - 1) * 2;
vector<long long int> v;
while (sum != 0) {
v.push_back(sum);
sum = sum - 2;
}
for (int j = 1; j < v.size(); j += 1) {
v[j] = v[j - 1] + v[j];
}
long long int i = lower_bound(v.begin(), v.end(), l) - v.begin();
long long int element_prev = 0;
if (i >= 1) {
element_prev = v[i - 1];
}
if (l - element_prev % 2 != 0) {
long long int add = (l - element_prev + 1) / 2;
long long int start = i + 1;
add = add + start;
if (l == r) {
cout << start << "\n";
return;
} else {
cout << start << " " << add << " ";
long long int cnt = 2;
long long int x = start;
long long int in = 1;
while (cnt < r - l + 1) {
add++;
if (add == n + 1) {
start = x + in;
in++;
add = start + 1;
}
cout << start << " ";
cnt++;
if (cnt < r - l + 1) {
cout << add << " ";
cnt++;
} else {
break;
}
}
}
} else {
long long int start = i + 1;
long long int add = (l - element_prev) / 2;
add = add + start;
if (l == r) {
cout << add << "\n";
} else {
long long int cnt = 1;
long long int x = start;
long long int in = 1;
while (cnt < r - l + 1) {
cout << add << " ";
cnt++;
if (cnt < r - l + 1) {
add++;
if (add == n + 1) {
start = x + in;
in++;
add = start + 1;
}
cout << start << " ";
cnt++;
} else {
break;
}
}
}
}
if (is_present == 1) {
cout << 1 << "\n";
return;
}
cout << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
mt19937 rand(chrono::steady_clock::now().time_since_epoch().count());
long long int t = 1;
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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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, gettrace
def input():
return stdin.buffer.readline()
t = int(input())
for _ in range(t):
n, l, r = list(map(int, input().split()))
count = 0
i = 1
while count < l:
if i == n:
i = 1
count += (n-i)*2
i += 1
i -= 1
count -= (n-i)*2
j = i + (l - count + 1)//2
length = r-l+1
if (l % 2) == 0:
print("printing first char before")
print(j, end=' ')
j += 1
length -= 1
print("before while i: {}, j: {}, length is {}".format(i, j, length))
while length > 0:
print(i, end=' ')
length -= 1
if length <= 0:
break
print(j, end=' ')
length -= 1
j += 1
if j > n:
i += 1
j = i+1
if i == n:
i = 1
print()
# 1 2 1 3 1 4 1 5 ... 1 n 2 3 2 4 2 5 ... 2 n 3 1 ... 3 n ... n-1 1 ... n-1 n 1
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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
readline = sys.stdin.readline
read = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')
def make(x):
if x == 1:
return [1]
f = [x]*(x-1)*2
for i in range(x-1):
f[i*2+1] = i + 2
f[-1] = 1
return f
def solve():
n, l, r = nm()
ok, ng, t = 0, l, 1
if l > 1:
while ng - ok > 1:
mid = (ng + ok) // 2
if 1 + mid * (mid - 1) < l:
ok = mid
else:
ng = mid
t = ok + 1
l -= (ok-1)*ok + 1
r -= (ok-1)*ok + 1
# print(t, l, r)
g = list()
while len(g) < r:
g += make(t)
t += 1
# print(g[:20])
print(*g[l-1:r])
return
# solve()
T = ni()
for _ in range(T):
solve()
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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 MainClass
{
public static void main(String[] args)throws IOException
{
Reader in = new Reader();
int t = in.nextInt();
StringBuilder stringBuilder = new StringBuilder();
while (t-- > 0)
{
int n = in.nextInt();
long l = in.nextLong();
long r = in.nextLong();
long l1 = -1;
long r1 = -1;
boolean ff = false;
long tot = 1L * n * (n - 1) + 1;
if (tot == r)
{
r--;
l1 = l;
r1 = r;
ff = true;
}
if (r < l)
{
if (ff)
stringBuilder.append("1\n");
continue;
}
long[] sum = new long[n + 1];
for (int i=1;i<=n;i++)
sum[i] = sum[i - 1] + (n - i) * 2;
int sIdx = -1;
int eIdx = -1;
for (int i=1;i<=n;i++)
{
if (sum[i] >= l)
{
sIdx = i;
break;
}
}
for (int i=1;i<=n;i++)
{
if (r > sum[i - 1] && r <= sum[i])
{
eIdx = i;
break;
}
}
l -= sum[sIdx - 1];
boolean flag = false;
boolean flag1 = false;
if (l % 2 == 0)
l /= 2;
else
{
flag = true;
l += 1;
l /= 2;
}
r -= sum[eIdx - 1];
if (r % 2 == 0)
r /= 2;
else
{
flag1 = true;
r --;
r /= 2;
}
long start = sIdx + l;
long end = eIdx + r;
StringBuilder stringBuilder1 = new StringBuilder();
if (sIdx < eIdx)
{
if (flag) stringBuilder1.append(sIdx).append(" ").append(start++).append(" ");
else stringBuilder1.append(start++).append(" ");
for (long i=start;i<=n;i++)
stringBuilder1.append(sIdx).append(" ").append(i).append(" ");
for (long i=sIdx + 1;i<eIdx;i++)
{
for (long j=i + 1;j<=n;j++)
stringBuilder1.append(i).append(" ").append(j).append(" ");
}
for (long i=eIdx + 1;i<=end;i++)
stringBuilder1.append(eIdx).append(" ").append(i).append(" ");
if (flag1)
stringBuilder1.append(eIdx).append(" ");
if (ff)
stringBuilder1.append("1 ");
}
else
{
l = l1;
r = r1;
long count = r - l + 1;
if (count > 0)
{
long ss1 = sIdx;
long ss2 = start;
if (!flag)
{
stringBuilder1.append(ss2++).append(" ");
count--;
}
for (long i=0;i<count;i++)
{
if (i % 2 == 0)
stringBuilder1.append(ss1).append(" ");
else
stringBuilder1.append(ss2++).append(" ");
}
}
if (ff)
stringBuilder1.append("1 ");
}
stringBuilder.append(stringBuilder1).append("\n");
}
System.out.println(stringBuilder);
}
}
class MergeSortLong
{
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(long arr[], int l, int m, int r) {
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
long L[] = new long[n1];
long R[] = new long[n2];
/*Copy data to temp arrays*/
for (int i = 0; i < n1; ++i)
L[i] = arr[l + i];
for (int j = 0; j < n2; ++j)
R[j] = arr[m + 1 + j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
// Main function that sorts arr[l..r] using
// merge()
void sort(long arr[], int l, int r) {
if (l < r) {
// Find the middle point
int m = (l + r) / 2;
// Sort first and second halves
sort(arr, l, m);
sort(arr, m + 1, r);
// Merge the sorted halves
merge(arr, l, m, r);
}
}
}
class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
|
1334_D. Minimum Euler Cycle
|
You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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-x*(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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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;
bool isPrime(long long int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (long long int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
long long int gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long int test;
cin >> test;
while (test--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int i = 1;
for (i = 1; i <= n; i++) {
if (i * (i - 1) < l && i * (i + 1) >= l) break;
}
long long int j = 1;
for (j = 1; j <= n; j++) {
if (j * (j - 1) < r && j * (j + 1) >= r) break;
}
if (l == n * (n - 1) + 1) {
cout << "1" << endl;
continue;
}
l = i * (i + 1) - (l - 1);
if (i == j) {
r = i * (i + 1) - r;
vector<long long int> v;
long long int t = i + 1, p = i;
for (long long int x = 0; x < l; x++) {
if (x % 2 == 0)
v.push_back(t);
else {
v.push_back(p);
p--;
}
}
for (long long int y = v.size() - r - 1; y >= 0; y--) cout << v[y] << " ";
cout << endl;
continue;
}
vector<long long int> v;
long long int t = i + 1, p = i;
for (long long int x = 0; x < l; x++) {
if (x % 2 == 0)
v.push_back(t);
else {
v.push_back(p);
p--;
}
}
for (long long int y = v.size() - 1; y >= 0; y--) cout << v[y] << " ";
for (long long int y = i + 1; y < j; y++) {
for (long long int k = 1; k < y + 1; k++)
cout << k << " " << y + 1 << " ";
}
if (r == n * (n - 1) + 1) {
cout << "1" << endl;
continue;
}
r -= j * (j - 1);
long long int c = 1;
for (long long int y = 0; y < r; y++) {
if (y % 2 == 0) {
cout << c << " ";
c++;
} else
cout << j + 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 ≠ 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 — 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 ≤ T ≤ 100) — the number of test cases.
Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — 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--) {
long long int n, l, r;
cin >> n >> l >> r;
long long int sum = 0;
long long int start = 1;
int done = 0;
while (sum <= l) {
sum += 2 * (n - start);
start++;
if (start > n) {
cout << 1 << " " << endl;
done = 1;
break;
}
}
if (done) continue;
start--;
sum -= 2 * (n - start);
long long int next = start + 1;
long long int cur = 0;
for (long long int i = sum + 1; i <= r; i++) {
if (i < l) {
if (i % 2 == 0) next++;
} else {
if (i % 2 == 1)
cout << start << " ";
else {
cout << next << " ";
next++;
}
}
cur++;
if (cur == 2 * (n - start)) {
cur = 0;
start = start + 1;
next = start + 1;
if (start == n) start = 1;
}
}
cout << endl;
}
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
struct Edge {
long long to, dis, next, cost;
} edge[24050];
long long num = -1;
bool vis[10010];
long long mincost;
long long pre[10010], head[10010], cost[10010], last[10010], flow[10010], n, k,
a[110], b[110], s, t, maxflow;
long long to[110];
void add(long long f, long long t, long long dis, long long cost) {
edge[++num].to = t;
edge[num].dis = dis;
edge[num].next = head[f];
edge[num].cost = cost;
head[f] = num;
edge[++num].to = f;
edge[num].dis = 0;
edge[num].cost = -cost;
edge[num].next = head[t];
head[t] = num;
}
queue<long long> q;
bool spfa(long long s, long long t) {
memset(cost, 0x3f3f3f3f, sizeof cost);
memset(flow, 0x3f3f3f3f, sizeof flow);
memset(vis, 0, sizeof vis);
q.push(s);
vis[s] = 1;
cost[s] = 0;
pre[t] = -1;
while (!q.empty()) {
long long nowp = q.front();
q.pop();
vis[nowp] = 0;
for (long long i = head[nowp]; i != -1; i = edge[i].next) {
if (edge[i].dis > 0 && cost[edge[i].to] > cost[nowp] + edge[i].cost) {
cost[edge[i].to] = cost[nowp] + edge[i].cost;
pre[edge[i].to] = nowp;
last[edge[i].to] = i;
flow[edge[i].to] = min(flow[nowp], edge[i].dis);
if (!vis[edge[i].to]) {
vis[edge[i].to] = 1;
q.push(edge[i].to);
}
}
}
}
return pre[t] != -1;
}
void MCMF(long long s, long long t) {
while (spfa(s, t)) {
long long now = t;
maxflow += flow[t];
mincost += flow[t] * cost[t];
while (now != s) {
edge[last[now]].dis -= flow[t];
edge[last[now] ^ 1].dis += flow[t];
now = pre[now];
}
}
}
signed main() {
long long T;
cin >> T;
while (T--) {
num = -1;
memset(to, -1, sizeof to);
memset(head, -1, sizeof head);
cin >> n >> k;
for (long long i = 1; i <= n; i++) cin >> a[i] >> b[i];
maxflow = mincost = 0;
s = 0;
t = 2 * n + 1;
for (long long i = 1; i <= n; i++) add(s, i, 1, 0), add(i + n, t, 1, 0);
for (long long i = 1; i <= n; i++) {
for (long long j = 1; j <= n; j++) {
long long nc = 0;
if (j <= k - 1)
nc = a[i] + b[i] * (j - 1ll);
else if (j != n)
nc = b[i] * (k - 1ll);
else
nc = a[i] + b[i] * (k - 1ll);
nc = 0x3f3f3f3f - nc;
add(i, j + n, 1, nc);
}
}
MCMF(s, t);
long long nowi = -1;
for (long long i = n * 4; i <= num; i += 2) {
nowi++;
if (edge[i].dis == 0) to[1 + nowi % n] = 1 + nowi / n;
}
cout << 2 * n - k << endl;
for (long long i = 1; i <= k - 1; i++) cout << to[i] << " ";
for (long long i = k; i < n; i++) cout << to[i] << " " << -to[i] << " ";
cout << to[n];
cout << endl;
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
void read(int &val) {
int x = 0;
int bz = 1;
char c;
for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
;
if (c == '-') {
bz = -1;
c = getchar();
}
for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - 48;
val = x * bz;
}
const int maxn = 1e5 + 101;
const int mod = 1e9 + 7;
struct Edge {
int from, to, cap, flow;
long long cost;
};
int a[maxn], b[maxn];
bool cmp(int x, int y) { return b[x] < b[y]; }
int n, k;
int N;
struct MCMF {
int n, m;
vector<Edge> edge;
vector<int> G[maxn];
int inq[maxn], p[maxn], a[maxn], To[maxn];
long long d[maxn];
int tag[maxn];
void init(int _n) {
n = _n;
for (int i = 0; i <= n; i++) G[i].clear();
edge.clear();
}
void add(int from, int to, int cap, long long cost, int flag) {
edge.push_back(Edge{from, to, cap, 0, cost});
edge.push_back(Edge{to, from, 0, 0, -cost});
m = edge.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
tag[m - 2] = tag[m - 1] = flag;
To[m - 1] = To[m - 2] = to - N;
}
bool BellFord(int s, int t, int &flow, long long &cost) {
for (int i = 0; i <= n; i++) d[i] = INF, inq[i] = 0;
d[s] = 0;
inq[s] = 1;
p[s] = 0;
a[s] = INF;
queue<int> q;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (auto to : G[u]) {
Edge &e = edge[to];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
p[e.to] = to;
a[e.to] = min(a[u], e.cap - e.flow);
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
if (d[t] == INF) return false;
flow += a[t];
cost += 1LL * d[t] * a[t];
for (int u = t; u != s; u = edge[p[u]].from) {
edge[p[u]].flow += a[t];
edge[p[u] ^ 1].flow -= a[t];
}
return true;
}
void MincostMaxflow(int s, int t, long long &cost) {
int flow = 0;
cost = 0;
while (BellFord(s, t, flow, cost))
;
vector<int> ans1, ans2;
n = (n - 4) / 3;
for (int i = 1; i <= n; i++) {
for (auto to : G[i]) {
if (edge[to].flow == false) continue;
if (tag[to] == 1) {
ans1.push_back(To[to]);
} else if (tag[to] == 2) {
ans2.push_back(To[to]);
}
}
}
sort(ans1.begin(), ans1.end(), cmp);
int sz = ans1.size();
for (int i = 0; i <= sz - 2; i++) {
printf("%d ", ans1[i]);
}
for (auto to : ans2) {
printf("%d %d ", to, -to);
}
printf("%d\n", ans1[sz - 1]);
}
} M;
int main() {
int t;
read(t);
while (t--) {
read(n);
read(k);
N = n;
M.init(3 * n + 4);
int mx = 0, j = 0;
for (int i = 1; i <= n; i++) {
read(a[i]);
read(b[i]);
if (mx < a[i]) j = i;
mx = max(mx, a[i]);
}
if (k == 1) {
printf("%d\n%d\n", 1, j);
continue;
}
printf("%d\n", k + 2 * (n - k));
int s = 0, t = 3 * n + 3;
for (int i = 1; i <= n; i++) {
M.add(s, i, 1, 0, -1);
}
for (int i = 1; i <= k - 1; i++) {
for (int j = 1; j <= n; j++) {
M.add(i, j + n, 1, -(1LL * (i - 1) * b[j] + a[j]), 1);
}
}
for (int i = k; i <= n - 1; i++) {
for (int j = 1; j <= n; j++) {
M.add(i, j + n, 1, -(1LL * (k - 1) * b[j]), 2);
}
}
for (int j = 1; j <= n; j++) {
M.add(n, j + n, 1, -(1LL * (k - 1) * b[j] + a[j]), 1);
}
for (int i = n + 1; i <= n + n; i++) {
M.add(i, t, 1, 0, -1);
}
long long cost = 0;
M.MincostMaxflow(s, t, cost);
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9;
int solve() {
int n, k;
cin >> n >> k;
vector<pair<pair<int, int>, int>> arr(n);
for (int i = 0; i < n; i++)
cin >> arr[i].first.second >> arr[i].first.first, arr[i].second = i + 1;
sort(arr.begin(), arr.end());
vector<int> a(n), b(n);
for (int i = 0; i < n; i++)
a[i] = arr[i].first.second, b[i] = arr[i].first.first;
vector<vector<int>> dp(n + 1, vector<int>(k + 1, -INF));
vector<vector<bool>> op(n + 1, vector<bool>(k + 1, 0));
dp[0][0] = 0;
for (int i = 0; i < n; i++) {
dp[i + 1][0] = dp[i][0] + (k - 1) * b[i];
for (int j = 1; j < k + 1; j++) {
int v1 = dp[i][j] + (k - 1) * b[i],
v2 = dp[i][j - 1] + (j - 1) * b[i] + a[i];
if (v2 > v1) op[i + 1][j] = 1;
dp[i + 1][j] = max(v1, v2);
}
}
int curj = k;
vector<int> maj, bon;
for (int i = n; i; i--) {
int oo = op[i][curj];
if (oo) {
maj.push_back(i - 1);
curj--;
} else {
bon.push_back(i - 1);
}
}
cout << n + n - k << endl;
for (int i = k - 1; i >= 1; i--) cout << arr[maj[i]].second << " ";
for (auto i : bon) cout << arr[i].second << " " << -arr[i].second << " ";
cout << arr[maj[0]].second << endl;
return 0;
}
int32_t main() {
ios_base::sync_with_stdio(false);
int t;
cin >> t;
for (int i = 0; i < t; i++) solve();
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
template <typename T>
bool ckmax(T& a, T b) {
return a < b ? a = b, 1 : 0;
}
template <typename T>
bool ckmin(T& a, T b) {
return b < a ? a = b, 1 : 0;
}
const int MN = 80;
int N, K, T, dp[MN][MN], ans;
bool u[MN], pr[MN][MN];
struct minion {
public:
int a, b, id;
void in() { scanf("%d%d", &a, &b); }
bool operator<(minion o) const { return b < o.b; }
} a[MN];
std::vector<int> seq;
int main(void) {
scanf("%d", &T);
for (; T--;) {
scanf("%d%d", &N, &K);
for (int i = 0; i < N; ++i) a[i].in(), a[i].id = i + 1;
std::sort(a, a + N);
memset(dp, -1, sizeof dp);
dp[0][0] = 0;
for (int i = 0; i < N; ++i)
for (int j = 0; j <= K; ++j)
if (~dp[i][j]) {
if (ckmax(dp[i + 1][j], dp[i][j] + (K - 1) * a[i].b))
pr[i + 1][j] = 0;
if (j + 1 <= K &&
ckmax(dp[i + 1][j + 1], dp[i][j] + a[i].a + j * a[i].b))
pr[i + 1][j + 1] = 1;
}
seq.clear();
ans = dp[N][K];
memset(u, 0, N * sizeof u[0]);
int k = K;
for (int i = N; i > 0; --i) {
if (pr[i][k]) u[i - 1] = 1, seq.push_back(a[i - 1].id), --k;
}
assert(!k);
std::reverse(seq.begin(), seq.end());
int v = seq.back();
seq.pop_back();
for (int i = 0; i < N; ++i)
if (!u[i]) seq.push_back(a[i].id), seq.push_back(-a[i].id);
seq.push_back(v);
printf("%d\n", seq.size());
for (int i = 0; i < seq.size(); ++i)
printf("%d%c", seq[i], " \n"[i + 1 == seq.size()]);
}
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 80;
const int INF = 1000000009;
struct ban {
int i;
int a, b;
};
bool operator<(const ban& t1, const ban& t2) { return t1.b < t2.b; }
int n, k;
ban t[N];
int dp[N][N];
int p[N][N];
void solv() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; ++i) {
t[i].i = i;
scanf("%d%d", &t[i].a, &t[i].b);
}
for (int i = 0; i < N; ++i)
for (int j = 0; j < N; ++j) dp[i][j] = -INF;
sort(t + 1, t + n + 1);
dp[0][0] = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= k; ++j) {
if (j < k) {
if (dp[i][j] + t[i + 1].a + t[i + 1].b * j > dp[i + 1][j + 1]) {
dp[i + 1][j + 1] = dp[i][j] + t[i + 1].a + t[i + 1].b * j;
p[i + 1][j + 1] = 1;
}
}
if (dp[i][j] + t[i + 1].b * (k - 1) > dp[i + 1][j]) {
dp[i + 1][j] = dp[i][j] + t[i + 1].b * (k - 1);
p[i + 1][j] = 0;
}
}
}
vector<int> ans1, ans0;
int j = k;
for (int i = n; i >= 1; --i) {
if (p[i][j] == 1) {
ans1.push_back(t[i].i);
--j;
} else {
ans0.push_back(t[i].i);
}
}
reverse((ans1).begin(), (ans1).end());
printf("%d\n", ((int)(ans1).size()) + ((int)(ans0).size()) * 2);
for (int i = 0; i < ((int)(ans1).size()) - 1; ++i) printf("%d ", ans1[i]);
for (int i = 0; i < ((int)(ans0).size()); ++i)
printf("%d %d ", ans0[i], -ans0[i]);
printf("%d\n", ans1.back());
}
int main() {
int tt;
scanf("%d", &tt);
while (tt--) solv();
return 0;
}
|
1354_F. Summoning Minions
|
Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.
Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.
Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.
Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).
Help Polycarp to make up a plan of actions to summon the strongest possible army!
Input
The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.
Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.
Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.
Output
For each test case print the optimal sequence of actions as follows:
Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.
If there are multiple optimal sequences, print any of them.
Example
Input
3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1
Output
4
2 1 -1 5
1
2
5
5 4 3 2 1
Note
Consider the example test.
In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.
In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.
In the third test case, Polycarp is able to summon and control all five minions.
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
{
"input": [
"3\n5 2\n5 3\n7 0\n5 0\n4 0\n10 0\n2 1\n10 100\n50 10\n5 5\n1 5\n2 4\n3 3\n4 2\n5 1\n"
],
"output": [
"8\n2 3 -3 4 -4 1 -1 5\n3\n1 -1 2\n5\n5 4 3 2 1\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int dp[76][76];
bool taken[76][76];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t, n, m;
cin >> t;
while (t--) {
cin >> n >> m;
vector<pair<pair<int, int>, int>> data(n);
for (int i = 0; i < n; ++i) {
cin >> data[i].first.second >> data[i].first.first;
data[i].second = i + 1;
}
sort(data.begin(), data.end());
for (int i = 1; i <= n; ++i) {
int s = data[i - 1].first.first * (m - 1);
int f = data[i - 1].first.second;
f -= data[i - 1].first.first;
dp[i][min(i, m)] = 0;
taken[i][min(i, m)] = false;
for (int p = 0; p <= m && p < i; ++p) {
dp[i][p] = dp[i - 1][p] + s;
taken[i][p] = false;
}
for (int p = 0; p < m && p < i; ++p) {
f += data[i - 1].first.first;
if (dp[i - 1][p] + f > dp[i][p + 1]) {
dp[i][p + 1] = dp[i - 1][p] + f;
taken[i][p + 1] = true;
}
}
}
int r = m;
int pos = n;
vector<int> order, others;
while (pos > 0) {
if (taken[pos][r]) {
order.push_back(data[pos - 1].second);
r--;
} else
others.push_back(data[pos - 1].second);
pos--;
}
reverse(order.begin(), order.end());
int last = order.back();
order.pop_back();
cout << m + (n - m) * 2 << "\n";
for (auto &x : order) cout << x << " ";
for (auto &x : others) cout << x << " " << -x << " ";
cout << last << "\n";
}
return 0;
}
|
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