# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
811186 | maomao90 | Ancient Machine 2 (JOI23_ancient2) | C++17 | 1461 ms | 496 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "ancient2.h"
#include <bits/stdc++.h>
using namespace std;
#define REP(i, j, k) for (int i = (j); i < (k); i++)
#define RREP(i, j, k) for (int i = (j); i >= (k); i--)
template <class T>
inline bool mnto(T &a, const T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T &a, const T b) {return a < b ? a = b, 1 : 0;}
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int) x.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef tuple<int, int, int> iii;
typedef vector<iii> viii;
#ifndef DEBUG
#define cerr if (0) cerr
#endif
const int INF = 1000000005;
const ll LINF = 1000000000000000005;
const int MAXN = 500005;
namespace {
int n;
char findPrefix(int p) {
int m = p + 3;
vi a(m);
iota(ALL(a), 1);
a[p + 1] = p + 1;
a[p + 2] = p + 2;
vi b = a;
a[p] = p + 1;
b[p] = p + 2;
int r = Query(m, a, b);
return '0' + (r == p + 2);
}
char findSuffix(string s) {
s = "1" + s;
int n = SZ(s);
s += "?";
int m = n + 1;
vi a(m), b(m);
string ps = "";
REP (i, 0, n + 1) {
auto get = [&] (char c) {
string ts = ps + c;
REP (j, 0, SZ(ts)) {
bool pos = 1;
REP (k, 0, SZ(ts) - j) {
if (ts[j + k] != s[k]) {
pos = 0;
}
}
if (pos) {
return SZ(ts) - j;
}
}
return 0;
};
a[i] = get('0');
b[i] = get('1');
ps += s[i];
}
int r = Query(m, a, b);
return '0' + (r == m - 1);
}
}
string Solve(int N) {
n = N;
string ans(n, '0');
REP (i, 0, n / 2) {
ans[i] = findPrefix(i);
}
string suf = "";
RREP (i, n - 1, n / 2) {
ans[i] = findSuffix(suf);
suf = ans[i] + suf;
}
cerr << ans << '\n';
return ans;
}
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