#40219 (C++11) No.187 中華風 (Hard)

• 問題: No.187 中華風 (Hard)
• ユーザー: 🐬hec
• 提出日時: 2015-07-30 03:42:26
• 言語: C++11 (gcc 11.4.0)
• 結果: AC
• 実行時間: 361 ms / 3,000 ms
• コード長: 2,039 bytes
• コンパイル時間: 1,381 ms
• コンパイル使用メモリ: 169,476 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-07-17 22:31:36
• 合計ジャッジ時間: 5,646 ms
• ジャッジサーバーID （参考情報）: judge3 / judge2

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	40 ms; 5,248 KB
testcase_01	AC	38 ms; 5,376 KB
testcase_02	AC	64 ms; 5,376 KB
testcase_03	AC	61 ms; 5,376 KB
testcase_04	AC	187 ms; 5,376 KB
testcase_05	AC	179 ms; 5,376 KB
testcase_06	AC	174 ms; 5,376 KB
testcase_07	AC	178 ms; 5,376 KB
testcase_08	AC	357 ms; 5,376 KB
testcase_09	AC	361 ms; 5,376 KB
testcase_10	AC	356 ms; 5,376 KB
testcase_11	AC	179 ms; 5,376 KB
testcase_12	AC	184 ms; 5,376 KB
testcase_13	AC	267 ms; 5,376 KB
testcase_14	AC	219 ms; 5,376 KB
testcase_15	AC	104 ms; 5,376 KB
testcase_16	AC	106 ms; 5,376 KB
testcase_17	AC	1 ms; 5,376 KB
testcase_18	AC	33 ms; 5,376 KB
testcase_19	AC	2 ms; 5,376 KB
testcase_20	AC	155 ms; 5,376 KB
testcase_21	AC	2 ms; 5,376 KB
testcase_22	AC	182 ms; 5,376 KB
testcase_23	AC	2 ms; 5,376 KB
testcase_24	AC	2 ms; 5,376 KB

ソースコード

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;

#define pb push_back
#define range(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  range(i,0,n)
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

ll extgcd(ll a,ll b,ll& x, ll& y) {
	if(b==0) return x=1,y=0,a;
	ll g=extgcd(b,a%b,y,x);
	y-=a/b*x;
	return g;
}

ll modinv(ll a,ll b){
	ll res,dummy;
	extgcd(a,b,res,dummy);
	if(res<0) res+=b, dummy-=a;
	return res;
}

const ll mod=1000000007;

ll chinese_remainder(vector<ll> b,vector<ll> m){
	int n=m.size();
	set<ll> P;
	rep(i,n){
		ll cur=m[i];
		for(ll f=2;f*f<=cur;++f){
			if(cur%f) continue;
			while(cur%f==0) cur/=f;
			P.insert(f);
		}
		if(cur>1) P.insert(cur);
	}

	vector<ll> factor(n);
	for(auto &prime:P){
		int index=0;
		rep(i,n){
			factor[i]=1LL;
			ll cur=m[i];
			while(cur%prime==0) factor[i]*=prime,cur/=prime;
			if(factor[i]>factor[index]) index=i;
		}
		rep(i,n)if(i!=index){
			if(factor[i]==1) continue;
			if(b[index]%factor[i]!=b[i]%factor[i]) return -1;
			m[i]/=factor[i],b[i]%=m[i];
		}
	}

	/*
	vector<tuple<ll,ll>> rearrange;
	rep(i,n) if(m[i]>1LL) rearrange.pb(make_tuple(m[i],b[i]));
	sort(begin(rearrange),end(rearrange));
	rep(i,rearrange.size()) tie(m[i],b[i])=rearrange[i];
	n=rearrange.size();
	*/
	vector<ll> constant(n,0LL),coef(n,1LL),v(n,0LL);


	rep(i,n){
		v[i]=(b[i]-constant[i]+m[i])%m[i];
		v[i]=(v[i]*modinv(coef[i],m[i]))%m[i];
		range(j,i+1,n){
			constant[j]=(constant[j]+v[i]*coef[j])%m[j];
			coef[j]=(coef[j]*m[i])%m[j];
		}
	}

	ll ans=0LL;
	for(int i=n-1;i>=0;--i) ans=(ans*m[i]+v[i])%mod;
	return ans;
}

int main(void){
	int N;
	cin >> N;
	vector<ll> X(N),Y(N);
	rep(i,N) cin >> X[i] >> Y[i];

	int zero=0;
	rep(i,N) zero+=(X[i]==0);

	if(zero==N){
		rep(i,N)rep(j,i) Y[i]/=__gcd(Y[i],Y[j]);
		ll ans=1LL;
		rep(i,N) ans=(ans*Y[i])%mod;
		cout << ans << endl;
		return 0;
	}else{
		cout << chinese_remainder(X,Y) << endl;
	}
}