結果
問題 | No.1627 三角形の成立 |
ユーザー | null |
提出日時 | 2021-05-02 04:05:50 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 6,122 bytes |
コンパイル時間 | 6,599 ms |
コンパイル使用メモリ | 470,740 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-07-18 16:04:29 |
合計ジャッジ時間 | 7,530 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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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 | WA | - |
testcase_01 | WA | - |
testcase_02 | AC | 1 ms
6,940 KB |
testcase_03 | AC | 1 ms
6,940 KB |
testcase_04 | AC | 2 ms
6,940 KB |
testcase_05 | WA | - |
testcase_06 | WA | - |
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 | - |
ソースコード
/* このコード、と~おれ! Be accepted! ∧_∧ (。・ω・。)つ━☆・*。 ⊂ ノ ・゜+. しーJ °。+ *´¨) .· ´¸.·*´¨) ¸.·*¨) (¸.·´ (¸.·'* ☆ */ #include <cstdio> #include <algorithm> #include <string> #include <cmath> #include <cstring> #include <vector> #include <numeric> #include <iostream> #include <random> #include <map> #include <unordered_map> #include <queue> #include <regex> #include <functional> #include <complex> #include <list> #include <cassert> #include <iomanip> #include <set> #include <stack> #include <bitset> #include <array> #include <chrono> //#pragma GCC target("arch=skylake-avx512") #pragma GCC target("avx2") //#pragma GCC optimize("O3") #pragma GCC optimize("Ofast") #pragma GCC target("sse4") #pragma GCC optimize("unroll-loops") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #define repeat(i, n, m) for(int i = n; i < (m); ++i) #define rep(i, n) for(int i = 0; i < (n); ++i) #define printynl(a) printf(a ? "yes\n" : "no\n") #define printyn(a) printf(a ? "Yes\n" : "No\n") #define printYN(a) printf(a ? "YES\n" : "NO\n") #define printim(a) printf(a ? "possible\n" : "imposible\n") #define printdb(a) printf("%.50lf\n", a) #define printLdb(a) printf("%.50Lf\n", a) #define printdbd(a) printf("%.16lf\n", a) #define prints(s) printf("%s\n", s.c_str()) #define all(x) (x).begin(), (x).end() #define deg_to_rad(deg) (((deg)/360.0L)*2.0L*PI) #define rad_to_deg(rad) (((rad)/2.0L/PI)*360.0L) #define Please return #define AC 0 #define manhattan_dist(a, b, c, d) (abs(a - c) + abs(b - d)) using ll = long long; using ull = unsigned long long; constexpr int INF = 1073741823; constexpr int MINF = -1073741823; constexpr ll LINF = ll(4661686018427387903); constexpr ll MOD = 1e9 + 7; constexpr ll mod = 998244353; constexpr long double eps = 1e-6; const long double PI = acosl(-1.0L); using namespace std; void scans(string& str) { char c; str = ""; scanf("%c", &c); if (c == '\n')scanf("%c", &c); while (c != '\n' && c != -1 && c != ' ') { str += c; scanf("%c", &c); } } void scanc(char& str) { char c; scanf("%c", &c); if (c == -1)return; while (c == '\n') { scanf("%c", &c); } str = c; } double acot(double x) { return PI / 2 - atan(x); } ll LSB(ll n) { return (n & (-n)); } template<typename T> inline T chmin(T& a, const T& b) { if (a > b)a = b; return a; } template<typename T> inline T chmax(T& a, const T& b) { if (a < b)a = b; return a; } //cpp_int #if __has_include(<boost/multiprecision/cpp_int.hpp>) #include <boost/multiprecision/cpp_int.hpp> #include <boost/multiprecision/cpp_dec_float.hpp> using namespace boost::multiprecision; #else using cpp_int = ll; #endif //atcoder library #if __has_include(<atcoder/all>) #include <atcoder/all> //using namespace atcoder; #endif /* random_device seed_gen; mt19937 engine(seed_gen()); uniform_int_distribution dist(1, 100); */ /*----------------------------------------------------------------------------------*/ /* * @title modint * @docs kyopro/docs/modint.md */ template<int mod> struct modint { int val; modint() : val(0) {}; modint(ll x) : val(x >= 0 ? x % mod : (mod + x % mod) % mod) {}; modint& operator=(const modint& x) { val = x.val; return *this; } modint& operator+=(const modint& x) { val += x.val; if (val >= mod)val -= mod; return *this; } modint& operator-=(const modint& x) { val += mod - x.val; if (val >= mod)val -= mod; return *this; } modint& operator*=(const modint& x) { val = (int)((ll)val * (ll)x.val % mod); return *this; } modint& operator/=(const modint& x) { int a = x.val, b = mod, u = 1, v = 0, t; while (b > 0) { t = a / b; swap(a -= t * b, b); swap(u -= t * v, v); } *this *= modint(u); return *this; } modint operator++() { val = (val + 1 == mod ? 0 : val + 1); return *this; } modint operator--() { val = (val == 0 ? mod - 1 : val - 1); return *this; } modint operator+(const modint& x) const { return (modint(*this) += x); } modint operator-(const modint& x) const { return (modint(*this) -= x); } modint operator*(const modint& x) const { return (modint(*this) *= x); } modint operator/(const modint& x) const { return (modint(*this) /= x); } bool operator==(const modint& x)const { return (val == x.val); } bool operator!=(const modint& x)const { return (val != x.val); } bool operator<(const modint& x)const { return (val < x.val); } bool operator>(const modint& x)const { return (val > x.val); } modint pow(ll n) { modint ret(1), a(val); while (n > 0) { if (n % 2) ret *= a; a *= a; n /= 2; } return ret; } static int getmod() { return mod; }; }; using ModInt = modint<MOD>; using Modint = modint<mod>; void f1(vector<ModInt>& a) { vector<bool> sieve(a.size(), true); for (size_t i = 2; i < a.size(); i++) { if (!sieve[i]) continue; for (size_t j = (a.size() - 1) / i; j > 0; j--) { sieve[i * j] = false; a[j] += a[i * j]; } } } void f2(vector<ModInt>& a) { vector<bool> sieve(a.size(), true); for (size_t i = 2; i < a.size(); i++) { if (!sieve[i]) continue; for (size_t j = 1; i * j < a.size(); j++) { sieve[i * j] = false; a[j] -= a[i * j]; } } } int main() { int n, m; scanf("%d%d", &n, &m); ModInt ans = n, cnt = 0; ans *= m; ans = ans * (ans - 1) * (ans - 2) / 2 / 3; vector<ModInt> a(max(n, m) + 1), b(max(n, m) + 1), c(max(n, m) + 1); for (int i = 1; i <= n; ++i)a[i] += (n - i) * 2; for (int i = 1; i <= m; ++i)b[i] += (m - i) * 2; f1(a); f1(b); rep(i, max(n, m) + 1)c[i] = a[i] * b[i]; f2(c); rep(i, max(n, m) + 1)cnt += c[i] * (i - 1); cnt /= 2; ans -= cnt; printf("%d\n", ans.val); Please AC; } //O(NM)