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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#136288 | #6417. Classical Summation Problem | ethening | WA | 1ms | 3536kb | C++17 | 4.5kb | 2023-08-07 19:08:11 | 2023-08-07 19:08:13 |
Judging History
answer
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using pii = pair<int, int>;
template <int32_t MOD = 1000000007>
struct Mint {
int32_t value;
Mint() : value(0) {}
Mint(int64_t value_) {
value_ %= MOD;
if (value_ < 0) value_ += MOD;
value = value_;
}
inline Mint &operator += (const Mint &other) { this->value += other.value; if (this->value >= MOD) this->value -= MOD; return *this; }
inline Mint &operator -= (const Mint &other) { this->value -= other.value; if (this->value < 0) this->value += MOD; return *this; }
inline Mint &operator *= (const Mint &other) { this->value = (int64_t)this->value * other.value % MOD; if (this->value < 0) this->value += MOD; return *this; }
inline Mint &operator /= (const Mint &other) { return *this = *this * other.inv(); }
inline Mint operator + () const { return *this; }
inline Mint operator - () const { return Mint(this->value ? MOD - this->value : 0); }
inline friend Mint operator + (const Mint& lhs, const Mint& rhs) { return Mint(lhs) += rhs; }
inline friend Mint operator - (const Mint& lhs, const Mint& rhs) { return Mint(lhs) -= rhs; }
inline friend Mint operator * (const Mint& lhs, const Mint& rhs) { return Mint(lhs) *= rhs; }
inline friend Mint operator / (const Mint& lhs, const Mint& rhs) { return Mint(lhs) /= rhs; }
Mint pow(uint64_t k) const {
Mint x = *this, y = 1;
for (; k; k >>= 1) { if (k & 1) y *= x; x *= x; }
return y;
}
Mint sqrt() const {
if (value == 0) return 0;
if (MOD == 2) return 1;
if (pow((MOD - 1) >> 1) == MOD - 1) return 0; // does not exist, it should be -1, but kept as 0 for this program
unsigned int Q = MOD - 1, M = 0, i;
Mint zQ; while (!(Q & 1)) Q >>= 1, M++;
for (int z = 1; ; z++) {
if (Mint(z).pow((MOD - 1) >> 1) == MOD - 1) {
zQ = Mint(z).pow(Q); break;
}
}
Mint t = pow(Q), R = pow((Q + 1) >> 1), r;
while (true) {
if (t == 1) { r = R; break; }
for (i = 1; Mint(t).pow(1 << i) != 1; i++);
Mint b = Mint(zQ).pow(1 << (M - 1 - i));
M = i, zQ = b * b, t = t * zQ, R = R * b;
}
return min(r, - r + MOD);
}
Mint inv() const { return pow(MOD - 2); } // MOD must be a prime
friend bool operator==(const Mint& lhs, const Mint& rhs) { return lhs.value == rhs.value; }
friend bool operator!=(const Mint& lhs, const Mint& rhs) { return lhs.value != rhs.value; }
friend bool operator<(const Mint& lhs, const Mint& rhs) { return lhs.value < rhs.value; }
friend bool operator>(const Mint& lhs, const Mint& rhs) { return lhs.value > rhs.value; }
friend Mint operator * (int32_t value, Mint n) { return Mint(value) * n; }
friend Mint operator * (int64_t value, Mint n) { return Mint(value) * n; }
friend ostream &operator << (std::ostream &os, const Mint &n) { return os << n.value; }
};
template <typename T, int64_t MOD = 1000000007>
struct Combin {
int _n;
vector<T> _fac;
vector<T> _invfac;
vector<T> _inv;
Combin() : _n{0}, _fac{1}, _invfac{1}, _inv{0} {}
Combin(int n) : Combin() {
init(n);
}
void init(int m) {
if (m <= _n) return;
_fac.resize(m + 1);
_invfac.resize(m + 1);
_inv.resize(m + 1);
for (int i = _n + 1; i <= m; i++) {
_fac[i] = _fac[i - 1] * i;
if (i == 1) {
_inv[i] = _invfac[i] = 1;
}
else {
_inv[i] = _inv[MOD % i] * (MOD - MOD / i);
_invfac[i] = _invfac[i - 1] * _inv[i];
}
}
_n = m;
}
T fac(int m) {
if (m > _n) init(2 * m);
return _fac[m];
}
T invfac(int m) {
if (m > _n) init(2 * m);
return _invfac[m];
}
T inv(int m) {
if (m > _n) init(2 * m);
return _inv[m];
}
T binom(int n, int m) {
if (n < m || m < 0) return 0;
return fac(n) * invfac(m) * invfac(n - m);
}
T catalan(int n) {
return binom(2 * n, n) * inv(n + 1);
}
};
using mint = Mint<998244353>;
using combin = Combin<mint, 998244353>;
void solve() {
ll n, k;
cin >> n >> k;
mint ans = 0;
combin cb;
mint denom = mint(n).pow(k);
if (k % 2 == 1) {
ans = (n + 1) / 2;
}
else {
mint Ed = 0;
for (int i = 1; i <= n - 1; i++) {
mint pi = mint(i).pow(k / 2) * mint(n - i).pow(k / 2) * cb.binom(k, k / 2) / denom;
Ed += pi;
}
ans = (n + 1) / 2 - Ed / 2;
}
ans *= denom;
cout << ans.value << "\n";
}
int main() {
cin.tie(0)->sync_with_stdio(0);
// int t;
// cin >> t;
// while (t--) {
solve();
// }
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3492kb
input:
3 2
output:
14
result:
ok 1 number(s): "14"
Test #2:
score: 0
Accepted
time: 1ms
memory: 3536kb
input:
5 3
output:
375
result:
ok 1 number(s): "375"
Test #3:
score: -100
Wrong Answer
time: 0ms
memory: 3536kb
input:
2 2
output:
3
result:
wrong answer 1st numbers differ - expected: '5', found: '3'