結果

問題 No.229 線分上を往復する3つの動点の一致
コンテスト
ユーザー shimomire
提出日時 2015-06-20 01:49:04
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
WA  
実行時間 -
コード長 5,610 bytes
コンパイル時間 883 ms
コンパイル使用メモリ 118,756 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-07-07 04:35:04
合計ジャッジ時間 1,834 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
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 * 37 WA * 6
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 # 

#include <cassert>// c
#include <iostream>// io
#include <iomanip>
#include <fstream>
#include <sstream>
#include <vector>// container
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <algorithm>// other
#include <complex>
#include <numeric>
#include <functional>
#include <random>
#include <regex>
using namespace std;

typedef long long ll;typedef unsigned long long ull;typedef long double ld;

#define ALL(c) c.begin(),c.end()
#define IN(l,v,r) (l<=v && v < r)
template<class T> void UNIQUE(T v){v.erase(unique(ALL(v)),v.end());}
//debug
#define DUMP(x) cerr << #x <<" = " << (x)
#define LINE() cerr<< " (L" << __LINE__ << ")"

struct range{
	struct Iter{
		int v,step;
		Iter& operator++(){v+=step;return *this;}
		bool operator!=(Iter& itr){return v<itr.v;}
		int& operator*(){return v;}
	};
	Iter i, n;
	range(int i, int n,int step):i({i,step}), n({n,step}){}
	range(int i, int n):range(i,n,1){}
	range(int n):range(0,n){}
	Iter& begin(){return i;}
	Iter& end(){return n;}
};
struct rrange{
	struct Iter{
		int v,step;
		Iter& operator++(){v-=step;return *this;}
		bool operator!=(Iter& itr){return v>itr.v;}
		int& operator*(){return v;}
	};
	Iter i, n;
	rrange(int i, int n,int step):i({i-1,step}), n({n-1,step}){}
	rrange(int i, int n):rrange(i,n,1){}
	rrange(int n) :rrange(0,n){}
	Iter& begin(){return n;}
	Iter& end(){return i;}
};

//input
template<typename T1,typename T2> istream& operator >> (istream& is,pair<T1,T2>& p){return is>>p.first>>p.second;}
template<typename T1> istream& operator >> (istream& is,tuple<T1>& t){return is >> get<0>(t);}
template<typename T1,typename T2> istream& operator >> (istream& is,tuple<T1,T2>& t){return is >> get<0>(t) >> get<1>(t);}
template<typename T1,typename T2,typename T3> istream& operator >> (istream& is,tuple<T1,T2,T3>& t){return is >>get<0>(t)>>get<1>(t)>>get<2>(t);}
template<typename T1,typename T2,typename T3,typename T4> istream& operator >> (istream& is,tuple<T1,T2,T3,T4>& t){return is >> get<0>(t)>>get<1>(t)>>get<2>(t)>>get<3>(t);}
template<typename T> istream& operator >> (istream& is,vector<T>& as){for(int i:range(as.size()))is >>as[i];return is;}
//output
template<typename T> ostream& operator << (ostream& os, const set<T>& ss){for(auto a:ss){if(a!=ss.begin())os<<" "; os<<a;}return os;}
template<typename T1,typename T2> ostream& operator << (ostream& os, const pair<T1,T2>& p){return os<<p.first<<" "<<p.second;}
template<typename K,typename V> ostream& operator << (ostream& os, const map<K,V>& m){bool isF=true;for(auto& p:m){if(!isF)os<<endl;os<<p;isF=false;}return os;}
template<typename T1> ostream& operator << (ostream& os, const tuple<T1>& t){return os << get<0>(t);}
template<typename T1,typename T2> ostream& operator << (ostream& os, const tuple<T1,T2>& t){return os << get<0>(t)<<" "<<get<1>(t);}
template<typename T1,typename T2,typename T3> ostream& operator << (ostream& os, const tuple<T1,T2,T3>& t){return os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t);}
template<typename T1,typename T2,typename T3,typename T4> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4>& t){return os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t);}
template<typename T> ostream& operator << (ostream& os, const vector<T>& as){for(int i:range(as.size())){if(i!=0)os<<" "; os<<as[i];}return os;}
template<typename T> ostream& operator << (ostream& os, const vector<vector<T>>& as){for(int i:range(as.size())){if(i!=0)os<<endl; os<<as[i];}return os;}

// values
template<typename T> inline T INF(){assert(false);};
template<> inline int INF<int>(){return 1<<28;};
template<> inline ll INF<ll>(){return 1LL<<58;};
template<> inline double INF<double>(){return 1e16;};
template<> inline long double INF<long double>(){return 1e16;};

template<class T> inline T EPS(){assert(false);};
template<> inline int EPS<int>(){return 1;};
template<> inline ll EPS<ll>(){return 1LL;};
template<> inline double EPS<double>(){return 1e-8;};
template<> inline long double EPS<long double>(){return 1e-8;};

// min{2^r | n < 2^r}
template<typename T> T upper_pow2(T n){ T res=1;while(res<n)res<<=1;return res;}
// max{d | 2^d  <= n}
template<typename T> T msb(T n){ int d=62;while((1LL<<d)>n)d--;return d;}

template<typename T,typename U> T pmod(T v,U M){return (v%M+M)%M;}

ll gcd_positive(ll a,ll b) { return b == 0 ? a : gcd_positive(b,a%b); }
ll gcd(ll a,ll b) { return gcd_positive(abs(a), abs(b)); }
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}


template<class T> class Rational{
public:
	T u,d;
	Rational(T u,T d){
		T l =gcd(u,d);
		this->u = u/l;
		this->d = d/l;
	}

	static Rational<T> rlcm(Rational<T> r1,Rational<T> r2){
		T d = lcm(r1.d,r2.d);
		r1.u *= d/r1.d;r2.u *= d/r2.d;
		return {lcm(r1.u,r2.u),d};
	}

	bool operator < (const Rational<T>& r) const{
		return this->u * r.d < this->d * r.u;
	}
	bool operator > (const Rational<T>& r) const{
		return this->u * r.d > this->d * r.u;
	}

};

class Main{
	public:

	void run(){
		vector<int> ts(3);cin >> ts;
		sort(ALL(ts));

		Rational<ll> mv = {INF<ll>(),1};

		Rational<ll> a01=Rational<ll>(ts[0]*ts[1],ts[1]-ts[0]);
		Rational<ll> b01=Rational<ll>(ts[0]*ts[1],ts[1]+ts[0]);

		Rational<ll> a12=Rational<ll>(ts[1]*ts[2],ts[2]-ts[1]);
		Rational<ll> b12=Rational<ll>(ts[1]*ts[2],ts[2]+ts[1]);

		mv=min(mv,Rational<ll>::rlcm(a01,a12));
		mv=min(mv,Rational<ll>::rlcm(a01,b12));
		mv=min(mv,Rational<ll>::rlcm(b01,a12));
		mv=min(mv,Rational<ll>::rlcm(b01,b12));

		cout << mv.u <<"/"<<mv.d <<endl;
	}
};

int main(){
	cout <<fixed<<setprecision(20);
	cin.tie(0);
	ios::sync_with_stdio(false);
	Main().run();
	return 0;
}
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