結果
問題 | No.1189 Sum is XOR |
ユーザー |
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提出日時 | 2020-08-22 14:49:28 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 46 ms / 2,000 ms |
コード長 | 3,985 bytes |
コンパイル時間 | 2,436 ms |
コンパイル使用メモリ | 197,456 KB |
最終ジャッジ日時 | 2025-01-13 08:57:36 |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 21 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:153:16: warning: format ‘%lld’ expects argument of type ‘long long int’, but argument 2 has type ‘modint<998244353>::i64’ {aka ‘long int’} [-Wformat=] 153 | printf("%lld\n", ans.a); | ~~~^ ~~~~~ | | | | | modint<998244353>::i64 {aka long int} | long long int | %ld main.cpp:127:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 127 | scanf("%d%d", &n, &K); | ~~~~~^~~~~~~~~~~~~~~~ main.cpp:135:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 135 | scanf("%d", &a); | ~~~~~^~~~~~~~~~
ソースコード
#include <bits/stdc++.h> #define For(i, a, b) for (int(i) = (int)(a); (i) < (int)(b); ++(i)) #define rFor(i, a, b) for (int(i) = (int)(a)-1; (i) >= (int)(b); --(i)) #define rep(i, n) For((i), 0, (n)) #define rrep(i, n) rFor((i), (n), 0) #define fi first #define se second using namespace std; typedef long long lint; typedef unsigned long long ulint; typedef pair<int, int> pii; typedef pair<lint, lint> pll; template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; } template <class T> bool chmin(T &a, const T &b) { if (a > b) { a = b; return true; } return false; } template <class T> T div_floor(T a, T b) { if (b < 0) a *= -1, b *= -1; return a >= 0 ? a / b : (a + 1) / b - 1; } template <class T> T div_ceil(T a, T b) { if (b < 0) a *= -1, b *= -1; return a > 0 ? (a - 1) / b + 1 : a / b; } constexpr lint mod = 1000000007; constexpr lint INF = mod * mod; constexpr int MAX = 200010; template <int_fast64_t MOD> struct modint { using i64 = int_fast64_t; i64 a; modint(const i64 a_ = 0) : a(a_) { if (a > MOD) a %= MOD; else if (a < 0) (a %= MOD) += MOD; } modint inv() { i64 t = 1, n = MOD - 2, x = a; while (n) { if (n & 1) (t *= x) %= MOD; (x *= x) %= MOD; n >>= 1; } modint ret(t); return ret; } bool operator==(const modint x) const { return a == x.a; } bool operator!=(const modint x) const { return a != x.a; } 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; } modint operator^(const lint x) const { return modint(*this) ^= x; } modint &operator+=(const modint &x) { a += x.a; if (a >= MOD) a -= MOD; return *this; } modint &operator-=(const modint &x) { a -= x.a; if (a < 0) a += MOD; return *this; } modint &operator*=(const modint &x) { (a *= x.a) %= MOD; return *this; } modint &operator/=(modint x) { (a *= x.inv().a) %= MOD; return *this; } modint &operator^=(lint n) { i64 ret = 1; while (n) { if (n & 1) (ret *= a) %= MOD; (a *= a) %= MOD; n >>= 1; } a = ret; return *this; } modint operator-() const { return modint(0) - *this; } modint &operator++() { return *this += 1; } modint &operator--() { return *this -= 1; } bool operator<(const modint x) const { return a < x.a; } }; using mint = modint<998244353>; vector<mint> fact; vector<mint> revfact; void setfact(int n) { fact.resize(n + 1); revfact.resize(n + 1); fact[0] = 1; rep(i, n) fact[i + 1] = fact[i] * mint(i + 1); revfact[n] = fact[n].inv(); for (int i = n - 1; i >= 0; i--) revfact[i] = revfact[i + 1] * mint(i + 1); } mint getC(int n, int r) { if (n < r) return 0; return fact[n] * revfact[r] * revfact[n - r]; } int main() { int n, K; scanf("%d%d", &n, &K); if (K > 10) { printf("%d\n", 0); return 0; } mint cnt[1 << 10], dp[2][1 << 10][K + 1]; rep(i, n) { int a; scanf("%d", &a); ++cnt[a]; } dp[0][0][0] = 1; rep(i, 1 << 10) { rep(j, 1 << 10) rep(k, K + 1) if (dp[0][j][k] != 0) { dp[1][j][k] += dp[0][j][k]; if ((i ^ j) == (i + j) && k < K) dp[1][j ^ i][k + 1] += dp[0][j][k] * cnt[i]; } rep(j, 1 << 10) rep(k, K + 1) { dp[0][j][k] = dp[1][j][k]; dp[1][j][k] = 0; } } mint ans = 0; rep(j, 1 << 10) ans += dp[0][j][K]; printf("%lld\n", ans.a); }