結果
| 問題 | No.3478 XOR-Folding Primes |
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2026-03-20 22:13:41 |
| 言語 | C++17 (gcc 15.2.0 + boost 1.89.0) |
| 結果 |
AC
|
| 実行時間 | 749 ms / 4,000 ms |
| コード長 | 4,399 bytes |
| 記録 | |
| コンパイル時間 | 1,568 ms |
| コンパイル使用メモリ | 229,248 KB |
| 実行使用メモリ | 10,016 KB |
| 最終ジャッジ日時 | 2026-03-20 22:13:50 |
| 合計ジャッジ時間 | 4,477 ms |
|
ジャッジサーバーID (参考情報) |
judge2_1 / judge3_0 |
(要ログイン)
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 8 |
ソースコード
#include <bits/stdc++.h>
using namespace std;
//入力が必ず-mod<=a<modの時.
template<const int mod> //mod<2^30.
struct modint{ //mod変更が不可能.
public:
long long v;
static void setmod(int m){} //飾り.
static constexpr long long getmod(){return mod;}
modint():v(0){}
template<typename T>
modint(T a):v(a){if(v < 0) v += mod;}
long long val()const{return v;}
modint &operator=(const modint &b) = default;
modint &operator+()const{return (*this);}
modint operator-()const{return modint(0)-(*this);}
modint operator+(const modint b)const{return modint(v)+=b;}
modint operator-(const modint b)const{return modint(v)-=b;}
modint operator*(const modint b)const{return modint(v)*=b;}
modint operator/(const modint b)const{return modint(v)/=b;}
modint &operator+=(const modint b){
v += b.v; if(v >= mod) v -= mod;
return *this;
}
modint &operator-=(const modint b){
v -= b.v; if(v < 0) v += mod;
return *this;
}
modint &operator*=(const modint b){v = v*b.v%mod; return *this;}
modint &operator/=(modint b){ //b!=0 mod素数が必須.
assert(b.v != 0);
(*this) *= b.pow(mod-2);
return *this;
}
modint pow(long long n)const{
modint ret = 1,p = v;
if(n < 0) p = p.inv(),n = -n;
while(n){
if(n&1) ret *= p;
p *= p; n >>= 1;
}
return ret;
}
modint inv()const{return pow(mod-2);} //素数mod必須.
modint &operator++(){*this += 1; return *this;}
modint &operator--(){*this -= 1; return *this;}
modint operator++(int){modint ret = *this; *this += 1; return ret;}
modint operator--(int){modint ret = *this; *this -= 1; return ret;}
friend bool operator==(const modint a,const modint b){return a.v==b.v;}
friend bool operator!=(const modint a,const modint b){return a.v!=b.v;}
friend bool operator<(const modint a,const modint b){return a.v<b.v;}
friend bool operator<=(const modint a,const modint b){return a.v<=b.v;}
friend bool operator>=(const modint a,const modint b){return a.v>=b.v;}
friend bool operator>(const modint a,const modint b){return a.v>b.v;}
friend ostream &operator<<(ostream &os,const modint a){return os<<a.v;}
friend istream &operator>>(istream &is,modint &a){ //入力はmodをとってくれる.
long long x; is >> x; x %= mod;
a = modint(x); return is;
}
};
using mint = modint<998244353>; const long long mod = 998244353;
template<typename T>
vector<vector<T>> multima(vector<vector<T>> &A,vector<vector<T>> &B){
assert(A.at(0).size() == B.size() && B.size());
int H = A.size(),W = B.at(0).size();
vector<vector<T>> ret(H,vector<T>(W));
for(int i=0; i<H; i++) for(int k=0; k<W; k++){
for(int l=0; l<A.at(i).size(); l++){
ret.at(i).at(k) += A.at(i).at(l)*B.at(l).at(k);
}
}
return ret; //計算できる行列のみ.
}
template<typename T>
vector<vector<T>> Powmat(vector<vector<T>> A,vector<vector<T>> B,long long N){ //return A^N*B ABは行列;
//n*n行列A n*1行列Bと想定 O(n^3*logN).
assert(A.size() && A.size() == A.at(0).size() && N >= 0);
while(N){
if(N&1) B = multima(A,B);
A = multima(A,A); N >>= 1;
}
return B;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int n = 1'000'000'0;
vector<bool> prime(n+1,true);
prime.at(0) = false; prime.at(1) = false;
for(int i=2; i*i<=n; i++){
if(!prime.at(i)) continue;
for(int k=i*i; k<=n; k+=i) prime.at(k) = false;
}
vector<int> twin,allp = {2};
for(int i=3; i<=n; i+=2){
if(prime.at(i)) allp.push_back(i);
if(prime.at(i) && prime.at(i-2) && ((i-2)^2) == i) twin.push_back(i);
}
int T; cin >> T;
while(T--){
int N,M; cin >> N >> M;
if(N == 1){
int pos = upper_bound(allp.begin(),allp.end(),M)-allp.begin();
cout << pos << "\n"; continue;
}
int mul = upper_bound(twin.begin(),twin.end(),M)-twin.begin();
vector<vector<mint>> A = {
{0,1,1},
{mul,0,1},
{mul,1,0}
},B = {{1},{mul},{mul}};
B = Powmat(A,B,N-1);
mint answer = B.at(0).at(0)+B.at(1).at(0)+B.at(2).at(0);
cout << answer << "\n";
}
}