結果
| 問題 |
No.3038 シャッフルの再現
|
| ユーザー |
 qwewe
|
| 提出日時 | 2025-05-14 13:01:33 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 8,980 bytes |
| コンパイル時間 | 1,055 ms |
| コンパイル使用メモリ | 86,548 KB |
| 実行使用メモリ | 7,848 KB |
| 最終ジャッジ日時 | 2025-05-14 13:03:47 |
| 合計ジャッジ時間 | 3,938 ms |
|
ジャッジサーバーID （参考情報） |
judge1 / judge4 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | RE * 1 |
| other | RE * 21 |
ソースコード
#include <iostream>
#include <vector>
#include <map>
#include <utility> // For std::pair
// Use __int128 for intermediate calculations in modular exponentiation and matrix multiplication
// to prevent overflow, especially with large moduli or intermediate values.
// This type is supported by GCC and Clang. Check platform compatibility if using other compilers.
#ifdef __GNUC__
using LargeInt = __int128;
#else
// Fallback type if __int128 is not available. This might lead to incorrect results due to overflow.
using LargeInt = long long;
#warning "__int128 not supported, using long long as fallback. Potential overflow issues."
#endif
// Modular exponentiation: computes (base^exp) % modulus
long long power_mod(long long base, long long exp, long long modulus) {
long long res = 1;
base %= modulus;
// Ensure base is non-negative
if (base < 0) base += modulus;
while (exp > 0) {
if (exp % 2 == 1) res = (LargeInt)res * base % modulus;
base = (LargeInt)base * base % modulus;
exp /= 2;
}
return res;
}
// Computes the Legendre symbol (a/p), assuming p is an odd prime.
// Returns 1 if a is a quadratic residue modulo p, -1 if a quadratic non-residue, 0 if a is divisible by p.
long long legendre(long long a, long long p) {
a %= p;
// Ensure a is in the range [0, p-1]
if (a < 0) a += p;
if (a == 0) return 0;
// Use Euler's criterion: (a/p) = a^((p-1)/2) mod p
long long res = power_mod(a, (p - 1) / 2, p);
// The result must be 1 or p-1 (which represents -1 mod p)
if (res == p - 1) return -1;
// If res is not p-1, it must be 1 due to properties of modular arithmetic.
return 1;
}
// Structure to represent a 2x2 matrix
struct Matrix {
long long mat[2][2];
};
// Multiplies two 2x2 matrices A and B modulo modulus
Matrix multiply(Matrix A, Matrix B, long long modulus) {
Matrix C;
for(int i=0; i<2; ++i) {
for(int j=0; j<2; ++j) {
C.mat[i][j] = 0;
for(int k=0; k<2; ++k) {
// Use LargeInt for intermediate product to avoid overflow
C.mat[i][j] = (C.mat[i][j] + (LargeInt)A.mat[i][k] * B.mat[k][j]) % modulus;
}
}
}
return C;
}
// Computes matrix A raised to the power exp modulo modulus using binary exponentiation
Matrix matrix_power(Matrix A, long long exp, long long modulus) {
Matrix res;
// Initialize result matrix as Identity matrix
res.mat[0][0] = 1; res.mat[1][1] = 1;
res.mat[0][1] = 0; res.mat[1][0] = 0;
// Reduce initial matrix A elements modulo modulus. This ensures values are in range [0, modulus-1].
A.mat[0][0] %= modulus; if(A.mat[0][0] < 0) A.mat[0][0] += modulus;
A.mat[0][1] %= modulus; if(A.mat[0][1] < 0) A.mat[0][1] += modulus;
A.mat[1][0] %= modulus; if(A.mat[1][0] < 0) A.mat[1][0] += modulus;
A.mat[1][1] %= modulus; if(A.mat[1][1] < 0) A.mat[1][1] += modulus;
while (exp > 0) {
if (exp % 2 == 1) res = multiply(res, A, modulus);
A = multiply(A, A, modulus);
exp /= 2;
}
return res;
}
// Computes the prime factorization of a positive integer n using trial division
// Returns a vector of pairs {prime, exponent}
std::vector<std::pair<long long, int>> prime_factorize(long long n) {
std::vector<std::pair<long long, int>> factors;
for (long long i = 2; i * i <= n; ++i) {
if (n % i == 0) {
int count = 0;
while (n % i == 0) {
n /= i;
count++;
}
factors.push_back({i, count});
}
}
// If n is still greater than 1, it must be a prime factor itself
if (n > 1) {
factors.push_back({n, 1});
}
return factors;
}
// Computes the Pisano period pi(p) for a prime p
long long compute_pisano_prime(long long p) {
// Handle base cases for p=2 and p=5
if (p == 2) return 3;
if (p == 5) return 20;
// Determine the upper bound B for pi(p) based on Legendre symbol (5/p)
long long B;
if (legendre(5, p) == 1) {
// If 5 is a quadratic residue mod p, pi(p) divides p-1
B = p - 1;
} else {
// If 5 is a quadratic non-residue mod p, pi(p) divides 2(p+1)
B = 2 * (p + 1);
}
// Find the prime factorization of the bound B
std::vector<std::pair<long long, int>> factors = prime_factorize(B);
// Start with pi = B and reduce it by prime factors q until A^(pi/q) is not Identity
long long pi = B;
// The Fibonacci matrix M = {{1,1},{1,0}}
Matrix A_base;
A_base.mat[0][0] = 1; A_base.mat[0][1] = 1;
A_base.mat[1][0] = 1; A_base.mat[1][1] = 0;
// Identity matrix I
Matrix I;
I.mat[0][0] = 1; I.mat[0][1] = 0;
I.mat[1][0] = 0; I.mat[1][1] = 1;
// Iterate through each distinct prime factor q of B
for (const auto& factor_pair : factors) {
long long q = factor_pair.first;
// Repeatedly divide pi by q as long as pi is divisible by q and A^(pi/q) mod p is the Identity matrix
while (pi > 0 && pi % q == 0) {
long long exponent_check = pi / q;
// Prevent checking exponent 0, which should not happen with correct logic but safe check
if (exponent_check == 0) break;
// Compute A^(pi/q) mod p
Matrix A_pow = matrix_power(A_base, exponent_check, p);
// Check if A_pow is the Identity matrix
bool is_identity = true;
for(int r=0; r<2; ++r) {
for(int c=0; c<2; ++c) {
if (A_pow.mat[r][c] != I.mat[r][c]) {
is_identity = false;
break;
}
}
if (!is_identity) break;
}
if (is_identity) {
// If it is Identity, then the period is smaller. Reduce pi by factor q.
pi /= q;
} else {
// If it's not Identity, we found the correct power for this prime factor q.
// Stop dividing pi by q and move to the next prime factor.
break;
}
}
}
// The final value of pi is the smallest period.
return pi;
}
int main() {
// Use faster I/O operations
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
int N; // Number of prime factors of M
std::cin >> N;
// Read the prime factorization of M: pairs of {prime p_i, exponent k_i}
std::vector<std::pair<long long, int>> PK(N);
for (int i = 0; i < N; ++i) {
std::cin >> PK[i].first >> PK[i].second;
}
// Map to store the maximum exponent for each prime factor found in the LCM calculation
std::map<long long, int> max_powers;
// The final modulus for the result
long long P_mod = 1000000007;
// Process each prime power factor p_i^k_i of M
for (int i = 0; i < N; ++i) {
long long p = PK[i].first;
int k = PK[i].second;
// Compute the Pisano period for the prime base p
long long pi_p = compute_pisano_prime(p);
// Use the property pi(p^k) = p^(k-1) * pi(p) (assuming Wall's conjecture holds, widely believed true)
// Compute the prime factorization of pi(p^k)
std::map<long long, int> current_factors_map;
// Add factors from pi(p)
std::vector<std::pair<long long, int>> factors_pi_p = prime_factorize(pi_p);
for(const auto& factor : factors_pi_p) {
current_factors_map[factor.first] += factor.second;
}
// Add the factor p^(k-1)
if (k > 1) {
current_factors_map[p] += (k - 1);
}
// Update the global max_powers map with the maximum exponents encountered so far for each prime factor
for(const auto& pair : current_factors_map) {
long long prime_factor = pair.first;
int exponent = pair.second;
// Update if this prime factor is new or has a larger exponent than previously seen
if (max_powers.find(prime_factor) == max_powers.end() || exponent > max_powers[prime_factor]) {
max_powers[prime_factor] = exponent;
}
}
}
// Compute the final LCM by multiplying together prime factors raised to their maximum required exponents, modulo P_mod
long long final_lcm = 1;
for (const auto& pair : max_powers) {
long long q = pair.first;
int exponent = pair.second;
// Calculate q^exponent mod P_mod
if (exponent > 0) { // Only consider factors with positive exponent
long long term = power_mod(q, exponent, P_mod);
// Multiply into the final LCM result modulo P_mod
final_lcm = (LargeInt)final_lcm * term % P_mod;
}
}
// Output the final result
std::cout << final_lcm << std::endl;
return 0;
}