結果

問題 No.1232 2^x = x
ユーザー Haar
提出日時 2020-09-18 22:09:55
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,142 bytes
コンパイル時間 1,927 ms
コンパイル使用メモリ 200,264 KB
最終ジャッジ日時 2025-01-14 17:15:58
ジャッジサーバーID
(参考情報)
judge3 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 3
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#ifdef DEBUG
#include <Mylib/Debug/debug.cpp>
#else
#define dump(...) ((void)0)
#endif
template <typename T, typename U>
bool chmin(T &a, const U &b){
return (a > b ? a = b, true : false);
}
template <typename T, typename U>
bool chmax(T &a, const U &b){
return (a < b ? a = b, true : false);
}
template <typename T, size_t N, typename U>
void fill_array(T (&a)[N], const U &v){
std::fill((U*)a, (U*)(a + N), v);
}
template <typename T, size_t N, size_t I = N>
auto make_vector(const std::array<int, N> &a, T value = T()){
static_assert(I >= 1);
static_assert(N >= 1);
if constexpr (I == 1){
return std::vector<T>(a[N - I], value);
}else{
return std::vector(a[N - I], make_vector<T, N, I - 1>(a, value));
}
}
template <typename T>
std::ostream& operator<<(std::ostream &s, const std::vector<T> &a){
for(auto it = a.begin(); it != a.end(); ++it){
if(it != a.begin()) s << " ";
s << *it;
}
return s;
}
template <typename T>
std::istream& operator>>(std::istream &s, std::vector<T> &a){
for(auto &x : a) s >> x;
return s;
}
namespace haar_lib {
std::tuple<int64_t, int64_t, int64_t> ext_gcd(int64_t a, int64_t b){
if(b == 0) return std::make_tuple(a, 1, 0);
int64_t d, p, q;
std::tie(d, q, p) = ext_gcd(b, (a + b) % b);
return std::make_tuple(d, p, q - a / b * p);
}
}
namespace haar_lib {
bool chinese_remainder_algorithm(int64_t b1, int64_t m1, int64_t b2, int64_t m2, int64_t &r, int64_t &m){
int64_t p, q, d;
std::tie(d, p, q) = ext_gcd(m1, m2);
if((b2 - b1) % d != 0) return false;
m = m1 * m2 / d;
int64_t t = ((b2 - b1) * p / d) % (m2 / d);
r = (b1 + m1 * t + m) % m;
return true;
}
bool chinese_remainder_algorithm(const std::vector<int64_t> &bs, const std::vector<int64_t> &ms, int64_t &r, int64_t &m){
int64_t R = 0, M = 1;
for(int i = 0; i < (int)bs.size(); ++i){
if(not chinese_remainder_algorithm(R, M, bs[i], ms[i], r, m)) return false;
R = r;
M = m;
}
return true;
}
}
namespace haar_lib {
int64_t mod_pow(int64_t n, int64_t p, int64_t m){
int64_t ret = 1;
while(p > 0){
if(p & 1) (ret *= n) %= m;
(n *= n) %= m;
p >>= 1;
}
return ret;
}
}
namespace haar_lib {}
namespace solver {
using namespace haar_lib;
constexpr int m1000000007 = 1000000007;
constexpr int m998244353 = 998244353;
void init(){
std::cin.tie(0);
std::ios::sync_with_stdio(false);
std::cout << std::fixed << std::setprecision(12);
std::cerr << std::fixed << std::setprecision(12);
std::cin.exceptions(std::ios_base::failbit);
}
void solve(){
int N; std::cin >> N;
while(N--){
int64_t p; std::cin >> p;
if(p == 2) std::cout << 2 << "\n";
else{
int64_t ans, m;
chinese_remainder_algorithm(0, p - 1, 1, p, ans, m);
dump(mod_pow(2, ans, p) == ans % p);
std::cout << ans << "\n";
}
}
}
}
int main(){
solver::init();
while(true){
try{
solver::solve();
}catch(const std::istream::failure &e){
break;
}
}
return 0;
}
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