結果

問題 No.2125 Inverse Sum
ユーザー fumofumofuni
提出日時 2022-11-18 23:38:22
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 625 ms / 2,000 ms
コード長 7,806 bytes
コンパイル時間 2,428 ms
コンパイル使用メモリ 210,720 KB
最終ジャッジ日時 2025-02-08 22:30:16
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 30
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>
using namespace std;
#pragma GCC optimize("Ofast")
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define pb push_back
#define ins insert
#define pqueue(x) priority_queue<x,vector<x>,greater<x>>
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={0,1,-1,0,1,1,-1,-1,0};
const ll dx[9]={1,0,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template<class T> inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
const int mod = MOD9;
const int max_n = 200005;
struct mint {
ll x; // typedef long long ll;
mint(ll x=0):x((x%mod+mod)%mod){}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
mint pow(ll t) const {
if (!t) return 1;
mint a = pow(t>>1);
a *= a;
if (t&1) a *= *this;
return a;
}
bool operator==(const mint &p) const { return x == p.x; }
bool operator!=(const mint &p) const { return x != p.x; }
// for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
using vm=vector<mint>;
using vvm=vector<vm>;
struct combination {
vector<mint> fact, ifact;
combination(int n):fact(n+1),ifact(n+1) {
assert(n < mod);
fact[0] = 1;
for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
}
mint operator()(int n, int k) {
if (k < 0 || k > n) return 0;
return fact[n]*ifact[k]*ifact[n-k];
}
}comb(max_n);
template <typename X, typename M>//faint len
struct SegTreeLazy {// (updateaffine(1,0))
using FX = function<X(X, X)>;
using FA = function<X(X, M, int)>;
using FM = function<M(M, M)>;
int n;
FX fx;
FA fa;
FM fm;
const X ex;
const M em;
vector<X> dat;
vector<M> lazy;
SegTreeLazy(int n_, FX fx_, FA fa_, FM fm_, X ex_, M em_)
: n(), fx(fx_), fa(fa_), fm(fm_), ex(ex_), em(em_), dat(n_ * 4, ex), lazy(n_ * 4, em) {
int x = 1;
while (n_ > x) x *= 2;
n = x;
}
void set(int i, X x) { dat[i + n - 1] = x; }
void build() {
for (int k = n - 2; k >= 0; k--) dat[k] = fx(dat[2 * k + 1], dat[2 * k + 2]);
}
/* lazy eval */
void eval(int k, int len) {
if (lazy[k] == em) return; //
if (k < n - 1) { //
lazy[k * 2 + 1] = fm(lazy[k * 2 + 1], lazy[k]);
lazy[k * 2 + 2] = fm(lazy[k * 2 + 2], lazy[k]);
}
//
dat[k] = fa(dat[k],lazy[k],len);//fa(dat[k], fp(lazy[k], len));
lazy[k] = em;
}
void update(int a, int b, M x, int k, int l, int r) {
eval(k, r - l);
if (a <= l && r <= b) { //
lazy[k] = fm(lazy[k], x);
eval(k, r - l);
} else if (a < r && l < b) { //
update(a, b, x, k * 2 + 1, l, (l + r) / 2); //
update(a, b, x, k * 2 + 2, (l + r) / 2, r); //
dat[k] = fx(dat[k * 2 + 1], dat[k * 2 + 2]);
}
}
void update(int a, int b, M x) { update(a, b, x, 0, 0, n); }
X query_sub(int a, int b, int k, int l, int r) {
eval(k, r - l);
if (r <= a || b <= l) { //
return ex;
} else if (a <= l && r <= b) { //
return dat[k];
} else { //
X vl = query_sub(a, b, k * 2 + 1, l, (l + r) / 2);
X vr = query_sub(a, b, k * 2 + 2, (l + r) / 2, r);
return fx(vl, vr);
}
}
X query(int a, int b) { return query_sub(a, b, 0, 0, n); }
X operator[](int i){
return query(i,i+1);
}
};
//0-indexed,2
template <class T>
struct SegTree {
private:
int n;//
vector<T> data;// vector
T def; //
function<T(T, T)> operation; // 使
function<T(T, T)> change;// 使
T find(int a, int b) {
T val_left = def, val_right = def;
for (a += (n - 1), b += (n - 1); a < b; a >>= 1, b >>= 1)
{
if ((a & 1) == 0){
val_left = operation(val_left, data[a]);
}
if ((b & 1) == 0){
val_right = operation(data[--b],val_right);
}
}
return operation(val_left, val_right);
}
public:
// _n:, _def:, _operation:,
// _change:
SegTree(size_t _n, T _def, function<T(T, T)> _operation,
function<T(T, T)> _change=[](T a,T b){return b;})
: def(_def), operation(_operation), change(_change) {
n = 1;
while (n < _n) {
n *= 2;
}
data = vector<T>(2 * n - 1, def);
}
void set(int i, T x) { data[i + n - 1] = x; }
void build() {
for (int k=n-2;k>=0;k--) data[k] = operation(data[2*k+1],data[2*k+2]);
}
// i(0-indexed)x
void update(int i, T x) {
i += n - 1;
data[i] = change(data[i], x);
while (i > 0) {
i = (i - 1) / 2;
data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
}
}
T all_prod(){
return data[0];
}
// [a, b)
T query(int a, int b) {
//return _query(a, b, 0, 0, n);
return find(a,b);
}
//
T operator[](int i) {
return data[i + n - 1];
}
};
vector<long long> divisor(long long n) {
vector<long long> ret;
for (long long i = 1; i * i <= n; i++) {
if (n % i == 0) {
ret.push_back(i);
if (i * i != n) ret.push_back(n / i);
}
}
sort(ret.begin(), ret.end()); //
return ret;
}
int main(){
ll p,q;cin >> p >> q;
set<pl> st;
{
ll hoge=gcd(p,q);
p/=hoge;q/=hoge;
}
for(auto a:divisor(q)){
for(auto b:divisor(q)){
if(q%(a*b)!=0)continue;
ll gx=(a+b)*q,gy=a*b*p;
if(gx%gy!=0)continue;
ll g=gx/gy;
st.insert({g*a,g*b});
}
}
cout << st.size() << endl;
for(auto p:st){
cout << p.first <<" " << p.second << endl;
}
}
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