結果

問題 No.1200 お菓子配り-3
ユーザー stoqstoq
提出日時 2020-08-28 22:34:55
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,380 bytes
コンパイル時間 2,286 ms
コンパイル使用メモリ 201,548 KB
実行使用メモリ 6,824 KB
最終ジャッジ日時 2024-11-14 15:58:46
合計ジャッジ時間 21,502 ms
ジャッジサーバーID
(参考情報)
judge1 / judge2
このコードへのチャレンジ
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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
6,820 KB
testcase_01 AC 2 ms
6,816 KB
testcase_02 WA -
testcase_03 AC 3 ms
6,820 KB
testcase_04 WA -
testcase_05 WA -
testcase_06 AC 3 ms
6,820 KB
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 AC 2 ms
6,820 KB
testcase_28 AC 1,872 ms
6,816 KB
testcase_29 AC 1,371 ms
6,816 KB
testcase_30 AC 1,372 ms
6,820 KB
testcase_31 WA -
testcase_32 WA -
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 #

#define MOD_TYPE 1

#pragma region Macros
#include <bits/stdc++.h>
using namespace std;
/*
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
*/
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
//constexpr ll MOD = 1;
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr double PI = acos(-1.0);
constexpr double EPS = 1e-11;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define Yay(n) cout << ((n) ? "Yay!" : ":(") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";

inline void init_main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(30) << setiosflags(ios::fixed);
}
template <typename T>
inline bool chmin(T &a, T b)
{
  if (a > b)
  {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b)
{
  if (a < b)
  {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b)
{
  return (a + b - 1) / b;
}
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val)
{
  fill((T *)array, (T *)(array + N), val);
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept
{
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept
{
  os << p.first << " " << p.second;
  return os;
}
#pragma endregion

ll a, b, c, d, x, y, z;
int cnt;

void solve()
{
}

int main()
{
  //init_main();
  int testcase;
  scanf("%d", &testcase);
  repi(ti, testcase)
  {
    scanf("%lld %lld", &x, &y);
    cnt = 0;
    for (d = 1; d * d <= x + y; d++)
    {
      if (((x + y) & 1) == 1 and (d & 1) == 1)
        continue;
      if ((x + y) % d != 0)
        continue;
      a = d - 1;
      if (a > 1)
      {
        z = (x + y) / (a + 1);
        if (x - z > 0 and (x - z) % (a - 1) == 0)
        {
          b = (x - z) / (a - 1);
          c = z - b;
          if (c > 0 and a * c + b == y)
            cnt++;
        }
      }
      a = (x + y) / d - 1;
      if (a > 1)
      {
        z = (x + y) / (a + 1);
        if (x - z > 0 and (x - z) % (a - 1) == 0)
        {
          b = (x - z) / (a - 1);
          c = z - b;
          if (c > 0 and a * c + b == y)
            cnt++;
        }
      }
    }
    printf("%d\n", cnt);
  }
}
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