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name stringlengths 2 88 | description stringlengths 31 8.62k | public_tests dict | private_tests dict | solution_type stringclasses 2
values | programming_language stringclasses 5
values | solution stringlengths 1 983k |
|---|---|---|---|---|---|---|
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long inf = 1000000;
const int N = 300000;
vector<int> tree(2 * N + 4);
vector<int> ranks(2 * N + 4, 0);
char str[N + 4];
int find(int x) { return x == tree[x] ? x : tree[x] = find(tree[x]); }
void link(int x, int y) {
if (ranks[x] > ranks[y]) {
tree[y] = x;
} e... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long ans = 0;
long long val[600010];
int have[600010][2], fa[600010];
char str[600010];
int n, k;
int find(int x) { return x == fa[x] ? x : (fa[x] = find(fa[x])); }
int other(int u) { return u <= k ? u + k : u - k; }
void update(int u, long long x, long long y) {
ans... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int w = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
w = (w << 3) + (w << 1) + ch - 48;
ch = getchar();
}
return w * f;
}
int n, m, f... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, k, s, x, op[N][2];
char ch[N];
int p[N << 1], val[N << 1];
int find(int u) { return p[u] == u ? p[u] : p[u] = find(p[u]); }
void uni(int a, int b) {
if (a == b) return;
val[b] += val[a];
p[a] = b;
}
int main() {
scanf("%d%d%s... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 7;
using ll = long long;
string s;
int n, k;
int fa[maxn];
vector<int> st[maxn];
const int inf = 0x3f3f3f3f;
ll sz[maxn];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
void Union(int x, int y) {
x = find(x), y = find(y);
if (x !... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
int dp[maxn][2], fa[maxn * 2], cnt[maxn * 2], n, k;
char buf[maxn];
int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); }
int calc(int x) {
int y = (x <= k ? x + k : x - k);
x = Find(x), y = Find(y);
if (x == 0 || y == 0) return cnt... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] == u ? fa[u] : fa[u] = find(fa[u]); }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9 + 5;
const int N = 600010;
class DSU {
public:
int p[N], cost[N];
int n;
int ans = 0;
DSU(int _n) : n(_n) {
for (int i = 0; i < n; i++) {
cost[2 * i + 1] = 1;
cost[2 * i] = 0;
}
cost[2 * n] = inf;
cost[2 * n + 1]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 10;
int f[maxn], n, k, w[maxn];
vector<int> v[maxn];
int Find(int x) { return f[x] == x ? x : f[x] = Find(f[x]); }
int ask(int x) {
int y = (x <= k) ? x + k : x - k;
x = Find(x);
y = Find(y);
if (x == 0)
return w[y];
else if (y == 0)
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct DS {
int rank, cost, parent;
bool forced;
} ds[4 * 300100];
int ans;
char status[300100];
vector<int> subs[300100];
int neg(int n) { return n + 300100; }
int find(int n) {
if (n == ds[n].parent) return n;
return ds[n].parent = find(ds[n].parent);
}
void join(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long llINF = 2e18, llMOD = 1e9 + 7;
const int INF = 2e9, MOD = 1e9 + 7, P = 179, N = 2e3 + 1, K = 23000, L = 18;
const long double EPS = 1e-6, ldINF = 1e18, PI = acos(-1);
template <typename T>
inline void sort_uniq(vector<T>& v) {
sort(v.begin(), v.end());
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MX = 1e6, INF = 1e9 + 10;
vector<long long> v[MX];
long long par[MX], s0[MX], s1[MX], col[MX], ans = 0;
pair<long long, long long> GETPAR(long long v) {
if (par[v] == v) return {v, 0};
pair<long long, long long> ans = GETPAR(par[v]);
return {par[v] = a... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int test = 0;
const int MAXN = 300009;
const long long MOD = 119 << 23 | 1;
class {
public:
vector<int> V[MAXN];
int fa[MAXN * 2], cnt[MAXN * 2];
int n, m;
int findfa(int x) {
if (fa[x] == x) return x;
return fa[x] = findfa(fa[x]);
}
void merge(int x, i... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-10;
const int MOD = 998857459;
const int INF = 0x3f3f3f3f;
const int maxn = 3e5 + 10;
const int maxm = 5e6 + 10;
int n, k, op[maxn][2], p[maxn << 2];
long long val[maxn << 2];
char str[maxn];
int find(int first) {
return... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9 + 7;
class DSU {
private:
vector<int> comp, siz;
vector<int> tp;
vector<int> onc, offc;
public:
DSU(int n) : comp(n), siz(n, 1), tp(n, 0), onc(n, 0), offc(n, 1) {
for (int i = 0; i < n; ++i) comp[i] = i;
offc[n - 1] = INF;
}
p... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] == u ? fa[u] : fa[u] = find(fa[u]); }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, const T& b) {
return a > b ? a = b, true : false;
}
template <class T>
bool umax(T& a, const T& b) {
return a < b ? a = b, true : false;
}
template <long long sz>
using tut = array<long long, sz>;
const long long N = 3e5 + 5;
const lon... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 7;
int n, k;
string s;
int l[maxn][2], r[maxn], cnt[maxn];
int getroot(int x) { return r[x] == x ? x : r[x] = getroot(r[x]); }
int calc(int x) {
int y = x <= k ? x + k : x - k;
x = getroot(x);
y = getroot(y);
if (x == 0 || y == 0) {
return... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
vector<int> g[maxn];
int f[maxn], d[maxn], L[maxn], R[maxn], fix[maxn], gt[maxn], ans = 0;
int n, k;
char str[maxn];
int calc(int x) {
if (fix[x])
return gt[x] ? R[x] : L[x];
else
return min(L[x], R[x]);
}
int find(int x) {
if (f[x] ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using ll = long long;
using namespace std;
const int maxn = 6e5 + 5;
int fa[maxn], val[maxn], ans;
vector<int> V[maxn];
int n, k;
string s;
void init() {
for (int i = 0; i < maxn; ++i) fa[i] = i;
}
int getfa(int x) { return x == fa[x] ? x : fa[x] = getfa(fa[x]); }
void merge(int x, int y) {
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
enum class LampConstraint { Distinct, Same };
enum class LampSetState { On, Off, PhantomA, PhantomB };
bool isLampSetStateKnown(LampSetState st) {
return st == LampSetState::On || st == LampSetState::Off;
}
typedef struct Edge {
int nodeId;
LampConstraint cons;
Edge(int nodeId_, LampCon... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
int n, k, m, c, vas, vvas, cc = 1000000, kl;
string s;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);
cin >> n >> k >> s;
vector<vector<int>> la(n, v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct DS {
int n;
vector<int> s;
vector<int> count;
DS(int n) : n(n), s(n, -1), count(n, 0) {}
int find(int i) { return s[i] < 0 ? i : (s[i] = find(s[i])); }
int counted(int i) { return count[find(i)]; }
void mark(int i, int x) { count[find(i)] += x; }
int ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int inf = 0x3f3f3f3f;
const int MAXN = 0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
void file() {}
const int N = 3e5 + 5;
int fat[N * 2], Size[N * 2], n, k;
vector<int> vec[N];
int find(int x) { return fat[x] == x ? x : fat[x] ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int lim = 1000 * 1000 + 5;
int nbElem, nbSub;
vector<int> subset[lim];
int side[lim];
int isIn[lim][2];
string ini;
int rep = 0;
int drepr[lim];
int dsz[lim];
int cnt[lim][2];
int dfind(int x) {
if (drepr[x] != x) drepr[x] = dfind(drepr[x... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 6e5 + 5;
int pre[N], si[N], num[N / 2], n, m;
bool vis[N];
char s[N];
vector<vector<int> > v(N / 2);
int findd(int x) {
if (pre[x] == x) return x;
pre[x] = findd(pre[x]);
if (vis[x]) vis[pre[x]] = 1;
return pre[x];
}
inline int id(int x) { return x > m... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 300005;
int dp[maxn][2], fa[maxn * 2], cnt[maxn * 2], n, k;
char buf[maxn];
int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); }
int calc(int x) {
int y;
if (x <= k)
y = x + k;
else
y = x - k;
x = Find(x), y = Find(y);
if (x ==... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<int> v[600005];
int k;
int mx[600005], a[300005], fa[600005];
int find(int x) {
if (fa[x] != x) {
fa[x] = find(fa[x]);
}
return fa[x];
}
void mer(int x, int y) {
int fx = find(x), fy = find(y);
if (fx != fy) {
mx[fx] += mx[fy];
fa[fy] = fx;
}
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 6e5 + 10;
const long long inf = 1e9 + 10;
int f[N], vist[N];
char s[N];
long long val[N];
vector<int> vec[N];
int find(int x) {
if (f[x] == x) return x;
f[x] = find(f[x]);
return f[x];
}
void myunion(int x, int y) {
if (x < y) {
f[y] = x;
val[x... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] ? fa[u] = find(fa[u]) : u; }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)] = find(b... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, x;
vector<int> vec[300010];
char c[300010];
int fa[300010 << 1], sz[300010 << 1];
int lim[300010 << 1];
int find(int x) {
if (fa[x] == x) return x;
return fa[x] = find(fa[x]);
}
int main() {
scanf("%d%d%s", &n, &k, c + 1);
for (int i = 1; i <= k; i++) {... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int val[300000 + 100], sum[300000 + 100][2], head[300000 + 100];
int f[300000 + 100];
vector<int> V[300000 + 100];
int n, k;
int ans;
int findd(int x) {
if (f[x] == x) return x;
int F = f[x];
f[x] = findd(f[x]);
val[x] = val[x] ^ val[F];
return f[x];
}
char s[3000... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, k;
string s;
vector<int> a[300001];
int p[1000001];
long long f[1000001];
const int Max = 6e5;
const int T = 3e5;
int root(int u) {
if (u == p[u]) return u;
return p[u] = root(p[u]);
}
void Merge(int u, int v) {
u = root(u);
v = root(v);
int ru = root(u);
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 6E5 + 10;
int n, k, l[N][2];
int fa[N], sc[N];
string s;
int find(int x) {
if (x == fa[x]) return x;
return fa[x] = find(fa[x]);
}
int cal(int x) {
int y = x <= k ? x + k : x - k;
int xx = find(x), yy = find(y);
if (xx == 0 || yy == 0) return sc[xx +... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | java | import java.util.*;
import java.io.*;
import java.math.*;
public class Solution {
public static void main(String[] args) throws IOException {
//PrintWriter out = new PrintWriter(new File("out.txt"));
PrintWriter out = new PrintWriter(System.out);
//Reader in = new Reader(new FileInputStrea... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 300000 + 10;
const int apsz = 26;
const int INF = 10000007;
const unsigned long long mod = 97;
const int maxm = 10000 + 10;
struct rev {
int c0, c1, tag;
void maintain() {
c0 = min(c0, INF);
c1 = min(c1, INF);
}
};
rev info[maxn];
int dsu[maxn... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] == u ? fa[u] : fa[u] = find(fa[u]); }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] == u ? fa[u] : fa[u] = find(fa[u]); }
void uni(int a, int b) {
if (a == b) return;
val[b] += val[a];
fa[a] = b;
}
int mix(int a) { retur... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 810000;
int fa[MAXN], val[MAXN], vec[MAXN][3];
int n, k, ans;
char str[MAXN];
int getfa(int x) {
if (fa[x] == x) return x;
return fa[x] = getfa(fa[x]);
}
void merge(int x, int y) {
int p = getfa(x), q = getfa(y);
if (p == q) return;
fa[p] = q;
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 69;
int n, k;
string initial;
int dsu[N];
int sz[N];
int szsame[N];
int parity[N];
int restrict[N];
vector<int> belong[N];
int curr = 0;
int getVal(int u) {
if (restrict[u] == 0) return szsame[u];
if (restrict[u] == 1) return (sz[u] - szsame[u]);
r... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
inline long long read1(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long n, k, pa[600010], ct[600010], siz[600010], ans = 0;
vector<long long> g[300010];
string s;
long long find(long long x) {
if (x == pa[x]) return x;
pa[x] = find(pa[x]);
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = (int)3e5 + 10;
int par[maxn << 1], num[maxn << 1];
inline void makeSet(int size) {
for (int i = 1; i <= size; i++) {
par[i] = i;
}
}
inline int find(int x) {
int k, j, r;
r = x;
while (r != par[r]) {
r = par[r];
}
k = x;
while (k != ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, inf = 0x3f3f3f3f;
int n, K, m, x, pos[N][2], ans, fa[N], val[N];
char s[N];
int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]); }
void add(int x, int y) {
int fx = getfa(x), fy = getfa(y);
if (fx != fy) fa[fx] = fy, val[fy] += val[fx... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
long long t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 1000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.ti... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class t>
inline t read(t &x) {
x = 0;
char c = getchar();
bool f = 0;
while (!isdigit(c)) f |= c == '-', c = getchar();
while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
if (f) x = -x;
return x;
}
template <class t>
inline void wr... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] ? fa[u] = find(fa[u]) : u; }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)] = find(b... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 300000 + 10;
char s[N];
int par[N], diff[N];
pair<int, int> find(int x) {
if (par[x] == x) return {x, 0};
auto r = find(par[x]);
par[x] = r.first, diff[x] ^= r.second;
return {par[x], diff[x]};
}
int f[N][2], col[N], ans;
int get(int x) {
if (col[x] ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | python3 | from sys import stdin
input = stdin.readline
n , k = [int(i) for i in input().split()]
pairs = [i + k for i in range(k)] + [i for i in range(k)]
initial_condition = list(map(lambda x: x == '1',input().strip()))
data = [i for i in range(2*k)]
constrain = [-1] * (2*k)
h = [0] * (2*k)
L = [1] * k + [0] * k
dp1 = [-1 for... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int test = 0;
const int MAXN = 300009;
const long long MOD = 119 << 23 | 1;
class {
public:
vector<int> V[MAXN];
int fa[MAXN * 2], cnt[MAXN * 2];
int n, m;
int curset[MAXN * 2], ans;
int findfa(int x) {
if (fa[x] == x) return x;
return fa[x] = findfa(fa[x... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
int n, k, m, c, vas, cc = 1000000, kl;
vector<int> vv;
string s;
vector<vector<int>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int INF = 1e9 + 7;
struct node {
int l, r;
node(int _l = 0, int _r = 0) : l(_l), r(_r){};
void operator+=(node x) {
l = min(l + x.l, INF);
r = min(r + x.r, INF);
}
int get() { return min(l, r); }
} val[N];
int root[N], rev[N];... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] ? fa[u] = find(fa[u]) : u; }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)] = find(b... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | CORRECT | cpp | #include <bits/stdc++.h>
template <class V>
int get_p(V& p, int j) {
if (p[j] == j) return j;
if (p[j] < 0) return p[j] = -get_p(p, -p[j]);
return p[j] = get_p(p, p[j]);
}
int get_min(bool a_j, int S_j, bool coa_j, int coS_j) {
if (!coa_j) return S_j;
if (!a_j) return coS_j;
return std::min(S_j, coS_j);
}
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long maxn = 300005;
const long long inf = 1ll << 50;
int ans = 0;
bool init[maxn];
int par[maxn], sz[maxn], cnt[maxn][2], col[maxn];
vector<int> vec[maxn];
int find(int x) {
if (par[x] == x) {
return x;
}
par[x] = find(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
int n, k, m, c, vas, cc = 1000000, kl;
vector<int> vv;
string s;
vector<vector<int>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9 + 5;
const int N = 600010;
class DSU {
public:
int p[N], cost[N];
int n;
int ans = 0;
DSU(int _n) : n(_n) {
for (int i = 0; i < N / 2; i++) {
cost[2 * i + 1] = 1;
cost[2 * i] = 0;
}
cost[2 * n] = inf;
cost[2 * n ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
long long t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 1000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.ti... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = (int)1e6 + 100;
const int mod = (int)1e9 + 7;
int n, k, fa[maxn], tot[maxn], ans;
char s[maxn];
vector<int> v[maxn];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
void join(int x, int y) {
x = find(x);
y = find(y);
if (x != y) fa[x]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct UF {
vector<int> e;
UF(int n) : e(n, -1) {}
int find(int x) { return e[x] < 0 ? x : e[x] = find(e[x]); }
bool join(int a, int b) {
a = find(a), b = find(b);
if (a == b) return false;
if (e[a] > e[b]) swap(a, b);
e[a] += e[b];
e[b] = a;
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long sz[300009], par[300009];
long long res;
long long n, k;
std::vector<long long> pos[300009];
long long getparent(long long x) {
if (par[x] == x) return x;
return par[x] = getparent(par[x]);
}
void merge(long long x, long long y) {
long long p1 = getparent(x);... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 6e5 + 5;
int pre[N], si[N], num[N / 2], n, m;
bool vis[N];
char s[N];
vector<vector<int> > v(N / 2);
int findd(int x) {
if (pre[x] == x) return x;
return pre[x] = findd(pre[x]);
}
inline int id(int x) { return x > m ? x - m : x; }
int un(int a, int b) {
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, x;
vector<int> vec[300010];
char c[300010];
int fa[300010 << 1], sz[300010 << 1];
int lim[300010];
int find(int x) {
if (fa[x] == x) return x;
return fa[x] = find(fa[x]);
}
int main() {
scanf("%d%d%s", &n, &k, c + 1);
for (int i = 1; i <= k; i++) {
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 10000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
long long t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, kl, cc = 1000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cou... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m;
char s[630000];
int fa[630000];
int val[630000];
vector<int> belong[630000];
int Push(int x) { return x; }
int Unpush(int x) { return m + x; }
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
void merge(int x, int y) {
int u = find(x), v = find(y... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long llINF = 2e18, llMOD = 1e9 + 7;
const int INF = 2e9, MOD = 1e9 + 7, P = 179, N = 2e3 + 1, K = 23000, L = 18;
const long double EPS = 1e-6, ldINF = 1e18, PI = acos(-1);
template <typename T>
inline void sort_uniq(vector<T>& v) {
sort(v.begin(), v.end());
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long sz[300009], par[300009];
long long res;
long long n, k;
std::vector<long long> pos[300009];
long long getparent(long long x) {
if (par[x] == x) return x;
return par[x] = getparent(par[x]);
}
void merge(long long x, long long y) {
long long p1 = getparent(x);... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5, inf = 1e9;
int n, kk, s, x, k[N][2];
char ch[N];
int fa[N << 1], val[N << 1];
int find(int u) { return fa[u] ? fa[u] = find(fa[u]) : u; }
void uni(int a, int b) {
if (find(a) == find(b)) return;
val[find(b)] += val[find(a)];
fa[find(a)] = find(b... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long llINF = 2e18, llMOD = 1e9 + 7;
const int INF = 2e9, MOD = 1e9 + 7, P = 179, N = 2e3 + 1, K = 23000, L = 18;
const long double EPS = 1e-6, ldINF = 1e18, PI = acos(-1);
template <typename T>
inline void sort_uniq(vector<T>& v) {
sort(v.begin(), v.end());
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long inf = 1000000;
const int N = 300000;
vector<int> tree(2 * N);
vector<int> ranks(2 * N, 0);
char str[N];
int find(int x) { return x == tree[x] ? x : tree[x] = find(tree[x]); }
void link(int x, int y) {
if (ranks[x] > ranks[y]) {
tr... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9 + 5;
const int N = 600010;
class DSU {
public:
int p[N], cost[N];
int n;
int ans = 0;
DSU(int _n) : n(_n) {
for (int i = 0; i < n; i++) {
cost[2 * i + 1] = 1;
cost[2 * i] = 0;
}
cost[2 * n] = inf;
cost[2 * n + 1]... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 6e5 + 5;
int pre[N], si[N], num[N / 2], n, m;
bool vis[N];
char s[N];
vector<vector<int> > v(N / 2);
int findd(int x) {
if (pre[x] == x) return x;
pre[x] = findd(pre[x]);
if (vis[x]) vis[pre[x]] = 1;
return pre[x];
}
inline int id(int x) { return x > m... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
template <class V>
int get_p(V& p, int j) {
if (p[j] == j) return j;
if (p[j] < 0) return p[j] = -get_p(p, -p[j]);
return p[j] = get_p(p, p[j]);
}
int get_min(bool a_j, int S_j, bool coa_j, int coS_j) {
if (!coa_j) return S_j;
if (!a_j) return coS_j;
return std::min(S_j, coS_j);
}
t... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct UF {
vector<int> e;
UF(int n) : e(n, -1) {}
int find(int x) { return e[x] < 0 ? x : e[x] = find(e[x]); }
bool join(int a, int b) {
a = find(a), b = find(b);
if (a == b) return false;
if (e[a] > e[b]) swap(a, b);
e[a] += e[b];
e[b] = a;
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define endl '\n'
#define pb push_back
#define ub upper_bound
#define lb lower_bound... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct DS {
int rank, cost, parent;
bool forced;
} ds[4 * 300100];
int ans;
char status[300100];
vector<int> subs[300100];
int neg(int n) { return n + 300100; }
int find(int n) {
if (n == ds[n].parent) return n;
return ds[n].parent = find(ds[n].parent);
}
void join(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
long long t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 1000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
fin.tie(0);
cout.ti... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
char op[300005];
int n, k, fa[300005], v[300005], num[300005], sz[300005], ans, gx[300005];
vector<int> vec[300005];
int ff(int x) {
if (fa[x] == x) return x;
int f = ff(fa[x]);
v[x] ^= v[fa[x]];
return fa[x] = f;
}
inline void merge(int x, int y, int p) {
if (ff(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long inf = 1000000;
const int N = 300002;
vector<int> tree(2 * N);
vector<int> ranks(2 * N, 0);
char str[N];
int find(int x) { return x == tree[x] ? x : tree[x] = find(tree[x]); }
void link(int x, int y) {
if (ranks[x] > ranks[y]) {
tree[y] = x;
} else {
tr... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
char op[300005];
int n, k, fa[300005], v[300005], num[300005], sz[300005], ans, gx[300005];
vector<int> vec[300005];
int ff(int x) {
if (fa[x] == x) return x;
int f = ff(fa[x]);
v[x] ^= v[fa[x]];
return fa[x] = f;
}
inline void merge(int x, int y, int p) {
if (ff(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int INF = 1e9 + 7;
struct node {
int l, r;
node(int _l = 0, int _r = 0) : l(_l), r(_r){};
void operator+=(node x) {
l = min(l + x.l, INF);
r = min(r + x.r, INF);
}
};
node val[N];
pair<int, int> root[N];
void build(int n = 0) {
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 900000;
int papa[MAXN], peso[MAXN], custo[MAXN];
int k;
bool obg[MAXN];
int find(int a) {
if (papa[a] == -1) return a;
return papa[a] = find(papa[a]);
}
bool juntos(int a, int b) { return find(a) == find(b); }
void uni(int a, int b) {
a = find(a);
b... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 1000000000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T& x) {
x = 0;
int fl = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') fl = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 1) + (x << 3) + ch - '0';
ch = getchar();
}
x ... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 10, inf = 0x3f3f3f3f;
vector<int> v[maxn];
int fa[maxn], cnt[maxn];
int n, k;
string s;
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int u, int v) {
u = find(u), v = find(v);
if (u == v) return;
cnt[u] += cnt[v];
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 7;
int n, k, fa[maxn], val[maxn], big;
char a[maxn];
vector<int> E[maxn];
int fid(int x) { return fa[x] == x ? x : fa[x] = fid(fa[x]); }
int uni(int x, int y) {
int fx = fid(x), fy = fid(y);
if (fx != fy) {
fa[fx] = fy;
val[fy] += val[fx];... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | python3 | from sys import stdin
input = stdin.readline
n , k = [int(i) for i in input().split()]
pairs = [i + n for i in range(n)] + [i for i in range(n)]
initial_condition = list(map(lambda x: x == '1',input().strip()))
data = [i for i in range(2*n)]
constrain = [-1] * (2*n)
h = [0] * (2*n)
L = [1] * n + [0] * n
dp1 = [-1 for... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, m, k, x;
vector<int> vec[300010];
char c[300010];
int fa[300010 << 1], sz[300010 << 1];
int lim[300010];
int find(int x) {
if (fa[x] == x) return x;
return fa[x] = find(fa[x]);
}
int main() {
scanf("%d%d%s", &n, &k, c + 1);
for (int i = 1; i <= k; i++) {
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long maxn = 300005;
const long long inf = 1ll << 50;
int ans = 0;
bool init[maxn];
int par[maxn], sz[maxn], cnt[maxn][2], col[maxn];
vector<int> vec[maxn];
int find(int x) {
if (par[x] == x) {
return x;
}
par[x] = find(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct DS {
int rank, cost, parent;
bool forced;
} ds[4 * 300100];
int ans;
char status[300100];
vector<int> subs[300100];
int neg(int n) { return n + 300100; }
int find(int n) {
if (n == ds[n].parent) return n;
return ds[n].parent = find(ds[n].parent);
}
void join(... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n, k;
string s;
vector<int> a[300001];
int p[1000001];
int f[1000001];
const int Max = 6e5;
const int T = 3e5;
int root(int u) {
if (u == p[u]) return u;
return p[u] = root(p[u]);
}
void Merge(int u, int v) {
u = root(u);
v = root(v);
int ru = root(u);
int r... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long llINF = 2e18, llMOD = 1e9 + 7;
const int INF = 2e9, MOD = 1e9 + 7, P = 179, N = 2e3 + 1, K = 23000, L = 18;
const long double EPS = 1e-6, ldINF = 1e18, PI = acos(-1);
template <typename T>
inline void sort_uniq(vector<T>& v) {
sort(v.begin(), v.end());
v... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
template <class V>
int get_p(V& p, int j) {
if (p[j] == j) return j;
if (p[j] < 0) return p[j] = -get_p(p, -p[j]);
return p[j] = get_p(p, p[j]);
}
int get_min(bool a_j, int S_j, bool coa_j, int coS_j) {
if (!coa_j) return S_j;
if (!a_j) return coS_j;
return std::min(S_j, coS_j);
}
t... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
long long n, k, m, c, vas, cc = 250000;
vector<long long> vv;
string s;
vector<vector<long long>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct perem {
int t = 0, l = 1, p = -1, q = 0;
};
ifstream fin("AAtest.in.txt");
int n, k, m, c, vas, cc = 1000000000;
vector<int> vv;
string s;
vector<vector<int>> la;
vector<perem> pr;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cerr.tie(0);
... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long ans = 0;
long long val[600010];
int have[600010][2], fa[600010];
char str[600010];
int n, k;
int find(int x) { return x == fa[x] ? x : (fa[x] = find(fa[x])); }
int other(int u) { return u <= k ? u + k : u - k; }
void update(int u, long long x, long long y) {
ans... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
template <class V>
int get_p(V& p, int j) {
if (p[j] == j) return j;
if (p[j] < 0) return p[j] = -get_p(p, -p[j]);
return p[j] = get_p(p, p[j]);
}
int get_min(bool a_j, int S_j, bool coa_j, int coS_j) {
if (!coa_j) return S_j;
if (!a_j) return coS_j;
return std::min(S_j, coS_j);
}
t... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MAX = 3e5 + 5;
long long inf = (long long)30000;
long long mod = (long long)998244353;
long long ans = 0, vert;
long long n, k, par[MAX], sz[MAX], cnt[MAX][2], clr[MAX], pos[MAX];
vector<long long> sets[MAX];
vector<long long> b[MAX];
void x_or(long long dad... |
1291_E. Prefix Enlightenment | There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you ... | {
"input": [
"5 3\n00011\n3\n1 2 3\n1\n4\n3\n3 4 5\n",
"8 6\n00110011\n3\n1 3 8\n5\n1 2 5 6 7\n2\n6 8\n2\n3 5\n2\n4 7\n1\n2\n",
"19 5\n1001001001100000110\n2\n2 3\n2\n5 6\n2\n8 9\n5\n12 13 14 15 16\n1\n19\n",
"7 3\n0011100\n3\n1 4 6\n3\n3 4 7\n2\n2 3\n"
],
"output": [
"1\n1\n1\n1\n1\n",
"1... | {
"input": [
"1 1\n1\n1\n1\n"
],
"output": [
"0\n"
]
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 300005;
int N, K, a[MAXN];
vector<int> V[MAXN];
char S[MAXN];
int f1[MAXN], v1[MAXN], f2[MAXN], chs[MAXN], ans;
int get1(int x) {
if (f1[x] != x) f1[x] = get1(f1[x]);
return f1[x];
}
void merge1(int x, int y) {
x = get1(x), y = get1(y);
if (x != y) ... |
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