description stringlengths 35 9.39k | solution stringlengths 7 465k |
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This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m;
vector<long long> bs;
int cnt[(1 << 16) + 1];
long long inv2 = (998244353 + 1) / 2;
int popcount(long long x) { return __builtin_popcountll(x); }
bool add(long long x) {
for (auto& t : bs) x = min(x, t ^ x);
if (!x) return false;
f... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
int v = 0, f = 1;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-')
f = -1;
else
v = (c & 15);
while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
x = v * f;
}
inline void read(lo... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int maxN = 200005, maxM = 60, p = 998244353;
void MOD(int &x) {
if (x >= p) x -= p;
}
int N, M, K;
long long A[maxM], B[maxM];
int pow2[maxN], ipow2[maxM];
void Insert(long long x) {
for (int i = M - 1; ~i; i--)
if ((x >> i) & 1) {
if (A[i])
x ^=... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 100;
const long long inf = 0x3f3f3f3f;
const long long iinf = 1 << 30;
const long long linf = 2e18;
const long long mod = 998244353;
const double eps = 1e-7;
template <class T = long long>
T chmin(T &a, T b) {
return a = min(a, b);
}
template ... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
const int mod =... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m;
vector<long long> bs;
int cnt[(1 << 16) + 1];
long long inv2 = (998244353 + 1) / 2;
int popcount(long long x) {
return cnt[x & 65535] + cnt[x >> 16 & 65535] + cnt[x >> 32 & 65535] +
cnt[x >> 48 & 65535];
}
bool add(long long x... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmin(T &x, T y) {
return x > y ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
return x < y ? x = y, 1 : 0;
}
long long readint() {
long long x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
const int MAXN = 53 + 5;
const int ha = 998244353;
const int inv2 = 499122177;
int n, m;
long long b[MAXN];
inline void insert(long long x) {
for (int i = m - 1; i >= 0; --i) {
if ((x >> i) & 1) {
if (b[i])
x ^= b[i];
else {
b[i] = x;
break;
}
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int maxN = 223456;
const int P = 998244353;
int n, m, rnk;
i64 base[maxN], p[maxN], dp[60][60][2];
int cnt[maxN], ans[maxN];
void dfs1(int d, i64 x) {
if (d == rnk) {
cnt[__builtin_popcountll(x)]++;
} else {
dfs1(d + 1, x);
dfs1(... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m;
vector<long long> bs;
int cnt[(1 << 16) + 1];
long long inv2 = (998244353 + 1) / 2;
int popcount(long long x) { return __builtin_popcountll(x); }
bool add(long long x) {
for (auto& t : bs) x = min(x, t ^ x);
if (!x) return false;
f... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
for (const auto &x : v) os << sep << x, sep = ", ";
return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
re... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int M = 55;
const int mod = 998244353;
const int inv2 = (mod + 1) / 2;
inline int pls(int a, int b) {
a += b - mod;
return a + (a >> 31 & mod);
}
inline int mns(int a, int b) {
a -= b;
return a + (a >> 31 & mod);
}
inline void inc(int& a, int b) {
a += b - m... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long read() {
long long x = 0;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
;
for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
const int Mod = 998244353, inv2 = Mod + 1 >> 1, MAXW = (1 << 18) - 1;
int upd(i... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
namespace mine {
long long qread() {
long long ans = 0, f = 1;
char c = getchar();
while (c < '0' or c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while ('0' <= c and c <= '9') ans = ans * 10 + c - '0', c = getchar();
return ans * f;
}
void write(lon... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const long long MOD = (long long)1e9 + 7;
const long long MOD1 = 2286661337;
const long long MOD2 = 998244353;
const int INF = (int)1e9 + 7;
const double EPS = 1e-7;
const int N = (int)2e5;
const int M = 53;
int n, m, k;
long long a[N], b[M], in_basis[M], p[M + 1], q[M + 1]... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
int n, m;
long long P[55], B[55], T[55], Comb[55][55];
void Put(long long a) {
int i;
for (i = m; i >= 0; i--) {
if ((a >> i) & 1) {
if (!P[i]) {
P[i] = a;
return;
}
a ^= P[i];
}
}
}
int C[55], Mod = 998244353;
void DFS(int pv... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m, num = 0, cnt[300010];
long long a, b[110], c[110], sum[110], pw2[200010], C[110][110];
int getcnt(long long x) {
return cnt[x & ((1 << 18) - 1)] + cnt[(x >> 18) & ((1 << 18) - 1)] +
cnt[x >> 36];
}
long long power(long long a,... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
for (const auto &x : v) os << sep << x, sep = ", ";
return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
re... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int P = 998244353;
int pw2(int y) {
if (y < 0) y += P - 1;
int x = 2;
int s = 1;
for (; y; y >>= 1, x = (long long)x * x % P)
if (y & 1) s = (long long)s * x % P;
return s;
}
int k, c[100];
long long a[100];
void calc(int t, long long x) {
if (t > k) {... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <int MOD_>
struct modnum {
static constexpr int MOD = MOD_;
static_assert(MOD_ > 0, "MOD must be positive");
private:
using ll = long long;
int v;
static int minv(int a, int m) {
a %= m;
assert(a);
return a == 1 ? 1 : int(m - ll(minv(m, a)) ... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using std::max;
using std::min;
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x3f3f3f3f3f3f3f3f;
std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
return !b ? a : gcd(b, a % b);
}... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T &t1, T t2) {
t1 = max(t1, t2);
}
template <typename T>
void mintt(... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
for (; !isdigit(ch); ch = getchar()) {
if (ch == '-') f = -1;
}
for (; isdigit(ch); ch = getchar()) {
x = x * 10 + ch - 48;
}
return x * f;
}
const int mxN = 1 << 19;
const int mxM = 53;
const ... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
template <int MOD>
struct Integral {
int v_ = 0;
template <typename T>
Integral(T v) : v_(norm(v)) {
static_assert(std::is_integral<T>::value, "input should be an integral.");
}
Integral() = default;
~Integral() = default;
template <typename T>
T norm(T v) const {
if con... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long a[200010], b[100], v[100], p[100], r[100], t[100], pw[200010],
c[100][100], g[100][100], n, rk, m, ct;
void ins(long long x) {
for (long long i = m; ~i; i--)
if (x & (1ll << i)) {
if (b[i])
x ^= b[i];
else {
b[i] = x;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long inf;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T, class T2>
long long chkmin(T &a, T2 b) {
return a > b ? a = b, 1 : 0;
}
template <class T, class T2>
long long chkmax(T &a, T2 b) {
return a < b ? a = b, 1 : 0;
}
template <class T>... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int N = 2e5 + 5;
const int M = 110;
const int inv2 = (mod + 1) >> 1;
int n, m;
int bin[M][M];
struct LinarBase {
long long b[M];
int rk;
bool side[M];
void insert(long long s) {
for (int i = m - 1; ~i; i--)
if (1 & (s >> i)... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long read() {
long long x = 0;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
;
for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
const int Mod = 998244353, inv2 = Mod + 1 >> 1, MAXW = (1 << 18) - 1;
int upd(i... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int N = 60;
const int mod = 998244353;
long long gi() {
long long x = 0, o = 1;
char ch = getchar();
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') o = -1, ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return ... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
struct MI {
private:
char bb[1 << 14];
FILE* f;
char *bs, *be;
char e;
bool o, l;
public:
MI() : f(stdin), bs(0), be(0) {}
inline char get() {
if (o) {
o = 0;
return e;
}
if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
if (bs == be) {
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int M = 55, P = 998244353;
int n, m, c[M][M], g[M][M];
long long b[M];
vector<long long> B, G;
int f[60], p[60];
long long pw(long long a, long long m) {
long long res = 1;
while (m) m & 1 ? res = res * a % P : 0, a = a * a % P, m >>= 1;
return res;
}
void inser... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
int const p = 998244353;
int mod(int x) { return x >= p ? x - p : x; }
int pw(int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = 1ll * res * x % p;
x = 1ll * x * x % p;
y >>= 1;
}
return res;
}
int n, m, k, ans[55], d[55][55], C[55][55];
long long... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m, rk, r, c;
int C[65][65], h[65];
long long p[65], t[65], s[65];
inline int ADD(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int SUB(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline void insert(long l... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline void cmin(T &a, const T &b) {
((a > b) && (a = b));
}
template <class T>
inline void cmax(T &a, const T &b) {
((a < b) && (a = b));
}
char IO;
template <class T = int>
T rd() {
T s = 0;
int f = 0;
while (!isdigit(IO = getchar()))
if (... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
namespace IO {
char gc() { return getchar(); }
template <typename Tp>
bool get1(Tp &x) {
bool neg = 0;
char c = gc();
while (c != EOF && (c < '0' || c > '9') && c != '-') c = gc();
if (c == '-') c = gc(), neg = 1;
if (c == EOF) return false;
x = 0;
for (; c >= '0' && c <= '9'; c =... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
namespace IO {
char gc() { return getchar(); }
template <typename Tp>
bool get1(Tp &x) {
bool neg = 0;
char c = gc();
while (c != EOF && (c < '0' || c > '9') && c != '-') c = gc();
if (c == '-') c = gc(), neg = 1;
if (c == EOF) return false;
x = 0;
for (; c >= '0' && c <= '9'; c =... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const long long N = 57;
const long long M = 262144;
const long long INF = 1e9 + 7;
const long long mod = 998244353;
long long Pow(long long x, long long y) {
long long ans = 1, now = x;
while (y) {
if (y & 1) ans = ans * now % mod;
now = now * now % mod;
y >... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 1, M = 64, mod = 998244353;
inline void check(int& x) { x -= mod, x += x >> 31 & mod; }
int n, k, m, tot;
long long int a[N], f[M], g[M], t[M];
int ans[M], tmp[M], pw[N], c[M][M], w[M][M];
inline void ins(long long int x) {
for (int i = m; i >= 0; i--)... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
struct Z2Basis {
long long int a[53];
int sz;
void add(long long int x) {
for (int i = 0; i < 53; i++)
if ((x >> i) & 1) {
if ((a[i] >> i) & 1)
x ^= a[i];
else {
a[i] = x;
sz++;
return;
}
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long qpow(long long x, long long y) {
while (y < 0) y += 998244353 - 1;
long long res = 1;
while (y) {
if (y & 1) res = res * x % 998244353;
x = x * x % 998244353;
y = y / 2;
}
return res;
}
int n, m;
long long vis[200], ypa;
void ins(long long x)... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long qpow(long long a, long long b) {
long long r = 1;
while (b) {
if (b & 1) r = r * a % 998244353;
a = a * a % 998244353, b >>= 1;
}
return r;
}
long long read() {
long long x = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long pow_mod(long long x, int k) {
long long ans = 1;
while (k) {
if (k & 1) ans = ans * x % 998244353;
x = x * x % 998244353;
k >>= 1;
}
return ans;
}
long long C[60][60];
void pre(int n) {
for (int i = 0; i <= n; i++) C[i][0] = 1;
for (int i =... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
long long n, m, a[1100005], cntA, A[1100005], C[105][105], G[105][105],
ans[105], S[1100005], E[1100005], cnt[1100005];
long long read() {
char c = getchar();
long long ans = 0;
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') ans = ans * 10 + c - '0', c = getc... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int M = 55, P = 998244353;
int n, m, c[M][M], g[M][M];
long long b[M];
vector<long long> B, G;
int f[60], p[60];
long long pw(long long a, long long m) {
long long res = 1;
while (m) m & 1 ? res = res * a % P : 0, a = a * a % P, m >>= 1;
return res;
}
void inser... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int maxm = 54;
int n, m, k;
long long first[maxm], a[maxm], b[maxm], p[maxm];
long long c[maxm][maxm];
int ipow(long long b, int e) {
long long ret = 1;
for (; e; e >>= 1) {
if (e & 1) ret = ret * b % mod;
b = b * b % mod;
}
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long read() {
long long x = 0;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
;
for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
const int Mod = 998244353, inv2 = Mod + 1 >> 1, MAXW = (1 << 18) - 1;
int upd(i... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int MAXM = 60;
int N, M;
int B;
vector<long long> basis;
int nc[MAXM];
long long ans[MAXM];
void gogo() {
long long cv = 0;
for (int i = 1; i < (1 << B); i++) {
nc[__builtin_popcountll(cv)]++;
cv ^= basis[__builtin_ctz(i)];... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int RLEN = 1 << 20 | 1;
inline char gc() {
static char ibuf[RLEN], *ib, *ob;
(ib == ob) && (ob = (ib = ibuf) + fread(ibuf, 1, RLEN, stdin));
return (ib == ob) ? EOF : *ib++;
}
inline int read() {
char ch = gc();
int res = 0;
bool f = 1;
while (!isdigit(c... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int n, m, num = 1, rnk, k, ans[100], res[100], C[100][100];
long long b[100], a[100], c[100];
void insert(long long x) {
for (int i = m - 1; ~i; i--)
if (x >> i & 1) {
if (b[i])
x ^= b[i];
else {
for (int j = i - ... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
using namespace std;
long long read() {
long long x = 0;
char ch = getchar();
for (; !isdigit(ch); ch = getchar())
;
for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
const int Mod = 998244353, inv2 = Mod + 1 >> 1, MAXW = (1 << 18) - 1;
int upd(i... |
This is the hard version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.
Chiori loves dolls and now she is going to decorate her bedroom!
<image>
As a doll collector, Chiori has got n dolls. The i-th doll has a non-negati... | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math,inline")
#pragma GCC target( \
"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << '{';
string sep;
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | # abs((desired*(2*n + 1) - ((n+1)*hot + n*cold))/(2*n + 1))
#EXPERIMENTING WITH LOSS DEFINTION - float ops ko aage peeche kar diya
import sys
def input():
return sys.stdin.readline().rstrip()
testcases = int(input())
answers = []
def loss(two_n_plus_one, hot, cold, desired):
n = two_n_plus_one//2
# if n == 0:
#... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | # Nilfer
for _ in range(int(input())):
p,a,t=map(int,input().split())
if t<=(p+a)/2:print(2)
else:
k=(p-t)//(2*t-p-a)
if abs((2*k+3)*t-k*p-2*p-k*a-a)*(2*k+1)<abs((2*k+1)*t-k*p-p-k*a)*(2*k+3):print(2*k+3)
else:print(2*k+1) |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | eps = 1e-10
t = int(input())
for _ in range(t):
h, c, t = map(int, input().split())
l = 0
r = 100000000
while l + 1 < r:
mid = (l + r) // 2
if mid % 2 == 0:
if mid + 1 < r:
mid += 1
elif mid - 1 > l:
mid -= 1
else:
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | tt=int(input())
while(tt):
h,c,t=map(int,input().split())
if h == t:
print(1)
elif t <= (h + c) / 2:
print(2)
else:
k = (t - c - 1) // (2 * t - h - c)
ans = 2 * k + 1
if(((4 * k * k - 1) * (2 * t - h - c)) >= (2 * (h - c) * k)):
ans -= 2
prin... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import math
T = int(input())
for test in range(T):
h, c, t = map(int, input().split())
if h + c - 2 * t >= 0:
print(2)
elif h <= t:
print(1)
else:
x = int((h - t) / (2 * t - h - c))
xp1 = x + 1
if abs((h + c) * x + h - 2 * x * t - t) * (2 * xp1 + 1) <= abs((h + ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | for _ in range(int(input())):
h,c,t=[float(i) for i in raw_input().split()]
if t<=(h+c)/2:
print(2)
else:
val=(t-c)/(2*t-h-c)
x1=int(val)
diff=h*x1+c*(x1-1)
diff=abs(t*(2*x1-1)-diff)
x1+=1
diff2=h*x1+c*(x1-1)
diff2=abs(t*(2*x1-1)-diff2)
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import os
import sys
from atexit import register
from io import BytesIO
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
raw_input = lambda: sys.stdin.readline().rstrip('\r\n')
from decim... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | for _ in range(int(input())):
h, c, t = map(int, input().split())
if 2 * t <= h + c:
print(2)
continue
x = (h - t) // (2*t - h - c)
k = 2*x + 1
val1 = abs(k//2 * c + (k+1)//2 * h - t*k)
val2 = abs((k+2)//2 * c + (k+3)//2 * h - t*(k+2))
print(k if val1 * (k... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | //package learning;
import java.util.*;
import java.io.*;
import java.lang.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class NitsLocal {
static ArrayList<String> s1;
static boolean[] prime;
static int n = 200001;
static void sieve() {
Arrays.fill(prime , true);
prim... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import sys
input = sys.stdin.buffer.readline
def print(val):
sys.stdout.write(str(val) + '\n')
def temp(x,t,h,c):
return abs((t*(2*x+1) - ((x+1)*h + x*c))/(2*x+1))
def prog():
for _ in range(int(input())):
h,c,t = map(int,input().split())
if t <= (h+c)/2:
print(2)
else:
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #8:47
MOD=10**9+7
INT_MAX=10**20+7
def INPUT():return list(float(i) for i in input().split())
def LIST_1D_ARRAY(n):return [0 for _ in range(n)]
def LIST_2D_ARRAY(m,n):return [[0 for _ in range(n)]for _ in range(m)]
def power(a,x):
res=1
while(x>0):
if x&1:
res=(res*a)%MOD
a=(a*a)%MOD... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
c b, e;
};
template <class c>
rge<c> range(c i, c j) {
return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
template <class c>
debug& operator<<(const c&... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import java.nio.charset.Charset;
import java.util.*;
public class Test {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
for (int i = 0; i < t; i++) {
double h = sc.nextInt();
double c = sc.nextInt();
do... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #include <bits/stdc++.h>
using namespace std;
int main() {
int te;
cin >> te;
for (int q = 0; q < te; q++) {
long long h, c, t;
cin >> h >> c >> t;
if (c + h - 2 * t >= 0) {
cout << 2 << '\n';
continue;
}
long long x = (h - t) / (2 * t - h - c);
long double tb1 = double((x + 2)... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | from fractions import Fraction
for _ in range(int(input())):
h,c,t=map(Fraction,input().split())
if 2*t<=(h+c):
print(2)
else:
x1=(h-t)//(2*t-h-c)
x2=x1+1
x1,x2=Fraction(x1),Fraction(x2)
y1,y2=Fraction(h*(x1+1)+c*x1)/Fraction(2*x1+1),Fraction(h*(x2+1)+c*x2)/Fraction(2... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | def main(h, c, t):
lowtmp = (h + c) / 2
if t == h: return 1
if t <= lowtmp: return 2
def tempSum(n):
# note, here we cannot get temp but only to get tempSum
# otherwise the tmp we get will have not enough floating precision
return (n * h + (n - 1) * c) # / (2* n -1)
l = 1
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | from __future__ import division
from sys import stdin
from decimal import Decimal
ceil1 = lambda a, b: (a + b - 1) // b
out = []
for _ in range(int(input())):
h, c, t = map(int, stdin.readline().split())
d1, d2 = h - t, t - c
if d1 >= d2:
out.append(2)
else:
lo = (c - t) // (h + c - 2 ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | def f(h: int, c: int, t: int) -> int:
if h + c >= t * 2:
return 2
else:
x = (h - t) // (t * 2 - h - c)
lhs = (h * (x + 1) + x * c) * (x * 2 + 3) - \
t * (x * 2 + 1) * (x * 2 + 3)
rhs = t * (x * 2 + 1) * (x * 2 + 3) - \
(h * (x + 2) + (x + 1) * c) * (x * 2 ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | from math import ceil ,floor
for _ in range(int(input())):
h,c,t=map(int,input().split())
if t==h:
print(1)
elif t<=(h+c)/2:
print(2)
else:
f=(h-t)/(2*t-h-c)
# print(f,abs((c*ceil(f)+(ceil(f)+1)*h)/(2*ceil(f)+1) -t ),abs((c*floor(f)+(floor(f)+1)*h)/(2*floor(f)+1)-t))
val1=((c*ceil(f)+(ceil(f)+1)*h)-t*(2*... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | // Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
// Please name your class Main
public class Main {
public static void main (String[] args) throws java.lang.Exception {
Scanner in = new Scanner(System.in);
int T =in.nextInt();
for(int i=0;i... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import sys
input = sys.stdin.readline
def print(val):
sys.stdout.write(str(val) + '\n')
def val(x,h,c):
return (h-c)/(2*(x*2 -1))
def prog():
for _ in range(int(input())):
L = 1
R = 10**12
h,c,t = map(int,input().split())
if t <= (h+c)/2:
print(2)
else:
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import math
for _ in range(int(input())):
h, c, t = map(int, input().split())
if 2 * t <= h + c:
print(2)
continue
if t >= h:
print(1)
continue
n = int((h - t) / (2 * t - h - c))
t1 = abs(((h + c) * n + h) - t * (2 * n + 1)) * (2 * n + 3)
t2 = abs(((h + c) * (n + ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import java.io.*;
import java.math.*;
import java.util.*;
public class Main{
static long MOD = 1000000007L;
static long [] fac;
static int[][] dir = new int[][]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
static long lMax = 0x3f3f3f3f3f3f3f3fL;
static int iMax = 0x3f3f3f3f;
static HashMap <Long, Long> ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigDecimal;
import java.util.StringTokenizer;
/*
1
50 10 49
2
99999 0 50000
99999 0 50002
*/
public class C2 {
static double EPS=1e-8;
public static void main(String[] args) {
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #include <bits/stdc++.h>
using namespace std;
void solve() {
long long h, c, t;
cin >> h >> c >> t;
if ((h + c) / 2 >= t) {
cout << 2 << '\n';
return;
}
long long a = t - h;
long long b = h + c - 2 * t;
long long k = 2 * (a / b) + 1;
long long val2 =
abs((k + 2) / 2 * 1ll * c + (k + 3) / 2... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | // Java implementation of the approach
import java.io.*;
import java.util.*;
public class Solution {
public static class FastReader {
BufferedReader br;
StringTokenizer root;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (root == n... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | a=int(input())
from fractions import Fraction
import sys
input=sys.stdin.readline
for i in range(a):
h,c,t=map(int,input().split())
se=(h+c)/2
if(h==c):
print(1)
elif(t<=se):
print(2)
elif(t>se):
x=h-t
y=2*(t-se)
if(x%y==0):
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
import math
import collections
from sys import stdin,stdout,setrecursionlimit
import bisect as bs
setrecursionlimit(2**20)
M = 10**9+7
T = int(stdin.readline())
# T = 1
def ceil(x,y):
return x//y + (1 if x%y != 0 else 0)
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import sys,math
from collections import deque
#input = sys.stdin.buffer.readline
def solve():
return;
for _ in range(int(input())):
h,c,t = map(int,input().split())
if t<=(h+c)/2:
print(2)
else:
x = (h-t)//(2*t-h-c)
v1 = ((x+1)*h + x*c)
v2 = ((x+2)*h + (x+1)*c)
# print(v1,v2)
# print(abs(v1-(2*x+1)*... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | for _ in range(int(input())):
h,c,t = map(int,input().split())
if(h==t):print(1)
else:
xx=(h+c)/2
if(xx>=t): print(2)
else:
dif=(t-xx)
j=int(abs((c-h)//(2*dif)))
if(j%2==0):
j+=1
if(dif-(h-c)/(2*j)>=abs(dif-(h-c)/(2... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | from fractions import *
T = input()
def calc(h, total, Tmult):
mult = Tmult*2+1
return Fraction(h)/Fraction(mult) + Fraction(Tmult)*Fraction(total)/Fraction(mult)
for _ in xrange(T):
h, c, t = map(float, raw_input().split())
total = h+c
if t == h:
print 1
elif total%2 == 0 and t == ... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import java.util.Scanner;
public class C1359 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
StringBuilder output = new StringBuilder();
for (int t=0; t<T; t++) {
int H = in.nextInt();
int C = in.nextInt(... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | import sun.nio.cs.ext.MacThai;
import java.io.*;
import java.util.*;
public class Codeforces {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | #include <bits/stdc++.h>
using namespace std;
using LL = long long;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int cas;
cin >> cas;
while (cas--) {
int h, t, c;
cin >> h >> c >> t;
h -= c;
t -= c;
if (h >= 2 * t) {
cout << 2 << endl;
} else {
LL n... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | from fractions import Fraction as f
def answer():
h, c, t = list(map(int, input().split()))
if t <= (h+c)/2:
return 2
else:
if abs(t - h) <= abs(t - (h*2 + c)/3):
return 1
n = int((t - h) / (h + c - 2*t))
i = max(2,n-2)
while abs(t -f( h*i + c*(i... |
There are two infinite sources of water:
* hot water of temperature h;
* cold water of temperature c (c < h).
You perform the following procedure of alternating moves:
1. take one cup of the hot water and pour it into an infinitely deep barrel;
2. take one cup of the cold water and pour it into an infin... | t=int(input())
for _ in range(t):
h,c,t=map(int,input().split(" "))
tb = (h+c)//2
if t==h:
print(1)
else:
if t<=tb:
print(2)
else:
k=(t-h)//(h+c-2*t)
if abs((k*(h+c)+h)-t*(2*k+1))*(2*k+3)<=abs(((k+1)*(h+c)+h)-t*(2*k+3))*(2*k+1):
... |
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