結果

問題 No.186 中華風 (Easy)
ユーザー sune232002
提出日時 2015-04-20 00:03:53
言語 C++11
(gcc 4.8.5)
結果
AC  
実行時間 4 ms
コード長 3,975 Byte
コンパイル時間 1,399 ms
使用メモリ 1,536 KB
最終ジャッジ日時 2019-10-09 09:18:18

テストケース

テストケース表示
入力 結果 実行時間
使用メモリ
0.in AC 4 ms
1,532 KB
1.in AC 4 ms
1,524 KB
2.in AC 3 ms
1,536 KB
3.in AC 4 ms
1,524 KB
4.in AC 4 ms
1,528 KB
5.in AC 3 ms
1,532 KB
6.in AC 4 ms
1,528 KB
7.in AC 4 ms
1,532 KB
8.in AC 3 ms
1,524 KB
9.in AC 3 ms
1,524 KB
A.in AC 4 ms
1,524 KB
B.in AC 3 ms
1,528 KB
C.in AC 4 ms
1,524 KB
D.in AC 4 ms
1,524 KB
E.in AC 3 ms
1,528 KB
F.in AC 3 ms
1,532 KB
G.in AC 3 ms
1,524 KB
system_test1.txt AC 3 ms
1,532 KB
system_test2.txt AC 4 ms
1,524 KB
system_test3.txt AC 3 ms
1,528 KB
system_test4.txt AC 3 ms
1,532 KB
system_test5.txt AC 3 ms
1,532 KB
テストケース一括ダウンロード

ソースコード

diff #
#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for(int i=0;i<(int)(n);++i)
#define REPR(i,n) for (int i=(int)(n)-1;i>=0;--i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()
#define valid(y,x,h,w) (0<=y&&y<h&&0<=x&&x<w)
#define tpl(...) make_tuple(__VA_ARGS__)
const int INF = 0x3f3f3f3f;
const double EPS = 1e-8;
const double PI = acos(-1);
const int dy[] = {-1,0,1,0};
const int dx[] = {0,1,0,-1};
typedef long long ll;
typedef pair<int,int> pii;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename Ch,typename Tr,typename C,typename=decltype(begin(C()))>basic_ostream<Ch,Tr>& operator<<(basic_ostream<Ch,Tr>&os,
const C& c){os<<'[';for(auto i=begin(c);i!=end(c);++i)os<<(i==begin(c)?"":" ")<<*i;return os<<']';}
template<class S,class T>ostream&operator<<(ostream &o,const pair<S,T>&t){return o<<'('<<t.first<<','<<t.second<<')';}
template<int N,class Tp>void output(ostream&,const Tp&){}
template<int N,class Tp,class,class ...Ts>void output(ostream &o,const Tp&t){if(N)o<<',';o<<get<N>(t);output<N+1,Tp,Ts...>(o,t);}
template<class ...Ts>ostream&operator<<(ostream&o,const tuple<Ts...>&t){o<<'(';output<0,tuple<Ts...>,Ts...>(o,t);return o<<')';}
template<class T>void output(T t,char z=10){if(t<0)t=-t,putchar(45);int c[20];
int k=0;while(t)c[k++]=t%10,t/=10;for(k||(c[k++]=0);k;)putchar(c[--k]^48);putchar(z);}
template<class T>void outputs(T t){output(t);}
template<class S,class ...T>void outputs(S a,T...t){output(a,32);outputs(t...);}
template<class T>void output(T *a,int n){REP(i,n)cout<<a[i]<<(i!=n-1?',':'\n');}
template<class T>void output(T *a,int n,int m){REP(i,n)output(a[i],m);}
template<class T>bool input(T &t){int n=1,c;for(t=0;!isdigit(c=getchar())&&~c&&c-45;);
if(!~c)return 0;for(c-45&&(n=0,t=c^48);isdigit(c=getchar());)t=10*t+c-48;t=n?-t:t;return 1;}
template<class S,class ...T>bool input(S&a,T&...t){input(a);return input(t...);}
template<class T>bool inputs(T *a, int n) { REP(i,n) if(!input(a[i])) return 0; return 1;}

ll mulmod(ll a, ll n, ll m) {
  ll res = 0;
  while(n>=1) {
    if (n % 2 == 1)
      res = (res + a) % m;
    a = a*2%m;
    n >>= 1;
  }
  return res;
}

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

ll ChineseRemainderTheorem(const vector<ll> &B, const vector<ll> &M) {
  ll all = 1;
  FOR(it, M) all*=*it;
  ll res = 0;
  REP(i,B.size()) {
    ll x,y;
    extgcd(M[i], all/M[i], x, y);
    y = (y+all)%all;
    res = (res + mulmod(mulmod(B[i],y,all),(all/M[i]),all) + all) % all;
  }
  return res;
}

map<ll, int> factorize(ll n) {  // O(√n)
  map<ll, int> res;
  for (ll i=2; i*i<=n; ++i) {
    int cnt = 0;
    while (n%i==0) {
      n /= i;
      cnt++;
    }
    if (cnt) {
      res[i] = cnt;
    }
  }
  if (n!=1) res[n] = 1;
  return res;
}

ll lcm(ll a, ll b) {
  return a/gcd(a,b)*b;
}
ll lcm(vector<ll> v) {
  ll res = 1;
  FOR(it, v)
    res = lcm(res, *it);
  return res;
}

int main() {
  while(1) {
    vector<ll> x(3),y(3);
    bool eof = 0;
    REP(i,3) {
      if (!(cin>>x[i]>>y[i])) eof = 1;
    }
    if (eof) break;
    map<int,pair<ll,ll> > mp;
    REP(i,3) {
      map<ll,int> pf = factorize(y[i]);
      for (auto p : pf) {
        ll t = 1;
        REP(i,p.second) t *= p.first;
        if (mp[p.first].first < t) {
          mp[p.first] = make_pair(t, x[i] % t);
        }
      }
    }
    vector<ll> B,M;
    for (auto p : mp) {
      B.push_back(p.second.second);
      M.push_back(p.second.first);
    }
    ll res = ChineseRemainderTheorem(B,M);
    if (res == 0) res = lcm(y);
    bool ok = 1;
    REP(i,3) {
      ok &= (res % y[i] == x[i]);
    }
    if (ok) cout << res << endl;
    else cout << -1 << endl;
  }
}
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