結果
問題 | No.1200 お菓子配り-3 |
ユーザー | hirono999 |
提出日時 | 2020-08-28 22:46:40 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 5,472 bytes |
コンパイル時間 | 3,501 ms |
コンパイル使用メモリ | 217,608 KB |
実行使用メモリ | 16,968 KB |
最終ジャッジ日時 | 2024-11-14 16:17:11 |
合計ジャッジ時間 | 134,476 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
10,020 KB |
testcase_01 | AC | 2 ms
9,896 KB |
testcase_02 | TLE | - |
testcase_03 | TLE | - |
testcase_04 | TLE | - |
testcase_05 | TLE | - |
testcase_06 | TLE | - |
testcase_07 | TLE | - |
testcase_08 | TLE | - |
testcase_09 | TLE | - |
testcase_10 | TLE | - |
testcase_11 | TLE | - |
testcase_12 | TLE | - |
testcase_13 | TLE | - |
testcase_14 | TLE | - |
testcase_15 | TLE | - |
testcase_16 | TLE | - |
testcase_17 | TLE | - |
testcase_18 | TLE | - |
testcase_19 | TLE | - |
testcase_20 | TLE | - |
testcase_21 | TLE | - |
testcase_22 | TLE | - |
testcase_23 | TLE | - |
testcase_24 | TLE | - |
testcase_25 | TLE | - |
testcase_26 | TLE | - |
testcase_27 | AC | 2 ms
13,640 KB |
testcase_28 | TLE | - |
testcase_29 | AC | 4 ms
6,816 KB |
testcase_30 | AC | 4 ms
6,816 KB |
testcase_31 | WA | - |
testcase_32 | WA | - |
ソースコード
#pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #include <bits/stdc++.h> constexpr long long INF = 1LL << 60; constexpr long long MOD = 1000000007; double PI = acos(-1.0); #define rep(i, n) for (ll i = 0; i < (n); ++i) #define rep1(i, n) for (ll i = 1; i <= (n); ++i) #define rrep(i, n) for (ll i = (n - 1); i >= 0; --i) #define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c))) #define ALL(obj) (obj).begin(), (obj).end() #define RALL(obj) (obj).rbegin(), (obj).rend() #define pb push_back #define to_s to_string #define len(v) (ll)v.size() #define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) #define print(x) cout << (x) << '\n' #define drop(x) cout << (x) << '\n', exit(0) #define debug(x) cout << #x << ": " << (x) << '\n' using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll> P; typedef tuple<ll, ll, ll> tpl; typedef vector<ll> vec; typedef vector<vector<ll>> vec2; typedef vector<vector<vector<ll>>> vec3; template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; } template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; } inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); } inline ll devc(ll x, ll y) { return (x + y - 1) / y; } inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } inline ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); } struct IoSetup { IoSetup() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(10); cerr << fixed << setprecision(10); } } iosetup; template< typename T1, typename T2 > ostream &operator << (ostream &os, const pair< T1, T2 > &p) { os << p.first << " " << p.second; return os; } template< typename T1, typename T2 > istream &operator >> (istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; } template< typename T1, typename T2, typename T3 > ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) { os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t); return os; } template< typename T1, typename T2, typename T3 > istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return is; } template< typename T > ostream &operator << (ostream &os, const vector< T > &v){ for (int i = 0; i < (int)v.size(); ++i) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } template< typename T > istream &operator >> (istream &is, vector< T > &v){ for(T &in : v) is >> in; return is; } template< typename T > ostream &operator << (ostream &os, const set< T > &st){ int ct = 0; for(auto& s : st) cout << s << (++ct != st.size() ? " " : ""); return os; } template <typename T> constexpr set<T> &operator|= (set<T> &st1, const set<T> &st2) { for(auto& s : st2) st1.insert(s); return st1; } template <typename T> constexpr set<T> &operator-= (set<T> &st1, const set<T> &st2) { for(auto& s : st2) if(st1.count(s)) st1.erase(s); return st1; } template <typename T> constexpr set<T> &operator&= (set<T> &st1, const set<T> &st2) { auto itr = st1.begin(); while(itr != st1.end()){ if(!st2.count(*itr)) itr = st1.erase(itr); else ++itr; } return st1; } template <typename T> constexpr set<T> operator| (const set<T> &st1, const set<T> &st2) { set<T> res = st1; res |= st2; return res; } template <typename T> constexpr set<T> operator- (const set<T> &st1, const set<T> &st2) { set<T> res = st1; res -= st2; return res; } template <typename T> constexpr set<T> operator& (const set<T> &st1, const set<T> &st2) { set<T> res = st1; res &= st2; return res; } /*--------------------------------- Tools ------------------------------------------*/ template< typename T > vector<T> cumsum(const vector<T> &X){ vector<T> res(X.size() + 1, 0); for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i]; return res; } template< typename S, typename T, typename F> pair<T, T> bisearch(S left, T right, F f) { while(abs(right - left) > 1){ T mid = (right + left) / 2; if(f(mid)) right = mid; else left = mid; } return {left, right}; } template< typename S, typename T, typename F> double trisearch(S left, T right, F f, int maxLoop = 90){ double low = left, high = right; while(maxLoop--){ double mid_left = high / 3 + low * 2 / 3; double mid_right = high * 2 / 3 + low / 3; if(f(mid_left) >= f(mid_right)) low = mid_left; else high = mid_right; } return (low + high) * 0.5; } template< typename F > ll ternarySearch(ll L, ll R, F f) { //[L, R) ll lo = L - 1, hi = R - 1; while (lo + 1 != hi) { ll mi = (lo + hi) / 2; if (f(mi) <= f(mi + 1)) hi = mi; else lo = mi; } return hi; } /*------------------------------- Main Code Here -----------------------------------------*/ int main() { ll N; cin >> N; rep(i, N){ ll X, Y; cin >> X >> Y; ll ans = 0; for(ll A = 2; A <= X; ++A){ if(A * A > X * A - Y + 1 or A * A > Y * A - X + 1) break; if(Y * A - X >= 0 and X * A - Y >= 0 and (Y * A - X) % (A * A - 1) == 0) ++ans; } print(ans); } return 0; }