Codeforces Round #730 (Div. 2)A. Exciting Bets
给两个个数字A,B. 求是否通过同时增加两个数或者同时减少两个数, 得到最大的gcd. 如果能, 需要同时加.减几?
典型的求gcd缺和gcd无关的题, 因为同增加/减少, 所以差值一样, 所以gcd最大就是差值, 因为gcd(0,max(A,B))最大.
然后几步嘛, 就是求比较小的数到差值的倍数的大小, 求余即可.
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#include "bits/stdc++.h" using namespace std; // fast read const auto fr = [](){ std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout << std::fixed << std::setprecision(12); return 1; }(); template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v); template<typename A, typename B> ostream& operator<<(ostream &cout, pair<A, B> const &p) { return cout << "(" << p.first << ", " << p.second << ")"; }; template<typename A> ostream& operator<<(ostream &cout, vector<A> const &v) { cout << "["; for(int i = 0; i < v.size(); i++) {if (i) cout << ", "; cout << v[i];} return cout << "]"; } template<typename A, typename B> istream& operator>>(istream& cin, pair<A, B> &p) { cin >> p.first; return cin >> p.second; } // vars: using ll = long long; using ull = unsigned long long; using ld = long double; using vi = std::vector<int>; using vl = std::vector<ll>; using vvi = std::vector<vi>; using vvl = std::vector<vl>; using pii = std::pair<int,int>; using pil = std::pair<int,ll>; using pli = std::pair<ll,int>; using pll = std::pair<ll,ll>; using vpii = std::vector<pii>; using vvpii = std::vector<vpii>; // consts ll M = 0; // ksm (kuai su mi) ll ksm(ll a,ll p){ll res=1;while(p){if(p&1){res=res*a%M;}a=a*a%M;p>>=1;}return res;} ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a % b);} ll lcm(ll a, ll b){return a * b / gcd(a, b);} int main() { fr; int T; cin >> T; while (T--) { ll A,B; cin >> A >> B; if(A > B) { swap(A, B); } if(A == B) { cout << 0 << " " << 0 << endl; continue; } ll C = B - A; ll D = min(A % C, C - A % C); cout << C << " " << D << endl; } return 0; } |