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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 |
---|---|---|---|---|---|---|---|---|---|
#116776 | #5547. Short Function | berarchegas# | WA | 1ms | 3508kb | C++17 | 2.7kb | 2023-06-30 03:50:29 | 2023-06-30 03:50:30 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
mt19937 rng((int) chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 998244353;
const int MAXN = 1e5 + 5;
const ll INF = 2e18;
int v[MAXN], n, k;
int fexp(ll b, int e, int mod) {
ll ans = 1;
while (e) {
if (e&1) ans = (ans * b) % mod;
b = (b * b) % mod;
e >>= 1;
}
return ans;
}
pll fexp2(ll b, int e, int x) {
if (!e) {
return {0, 1};
}
pll at = fexp2(b, e / 2, x);
ll pri = at.first * at.first % (MOD - 1);
pri = (pri * x) % (MOD - 1);
pri = (pri + 2ll * at.first * at.second) % (MOD - 1);
ll sec = at.second * at.second;
pri = (pri + sec / x) % (MOD - 1);
sec %= x;
if (e & 1) {
sec *= 2;
pri = (2 * pri + sec / x) % (MOD - 1);
sec %= x;
}
return {pri, sec};
}
vector<int> simulate(int rounds) {
vector<int> a(n), b(n);
for (int i = 0; i < n; i++) a[i] = v[i];
b = a;
for (int i = 0; i < rounds; i++) {
a = b;
for (int j = 0; j < n; j++) {
b[j] = (1ll * a[j] * a[(j + (1ll << i)) % n]) % MOD;
}
}
return b;
}
int mul(ll a, ll b) {
return (a * b) % MOD;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n >> k;
ll tudo = 1;
for (int i = 0; i < n; i++) {
cin >> v[i];
tudo = (tudo * v[i]) % MOD;
}
int cnt = 0, x = n;
while (x % 2 == 0) {
cnt++;
x /= 2;
}
if (x == 1) {
// potencia de 2
if (k < cnt) {
// simular o processo
vector<int> ans = simulate(k);
for (int x : ans) cout << x << ' ';
cout << '\n';
return 0;
}
int ans = fexp(tudo, fexp(2, k - cnt, MOD - 1), MOD);
for (int i = 0; i < n; i++) {
cout << ans << ' ';
}
cout << '\n';
return 0;
}
vector<int> ans(n);
int qtd = fexp(2, k, n);
int elevate = fexp2(2, k, n).first;
cout << qtd << ' ' << elevate << '\n';
vector<int> pref(2 * n + 1);
pref[0] = 1;
int at = 1;
for (int i = 0; i < 2 * n; i++) {
if (i < n)
pref[at] = (1ll * pref[at - 1] * v[i]) % MOD;
else
pref[at] = (1ll * pref[at - 1] * v[i - n]) % MOD;
at++;
}
at = 1;
for (int i = 0; i < n; i++) {
cout << mul(mul(pref[at + qtd - 1], fexp(pref[at - 1], MOD - 2, MOD)), fexp(tudo, elevate, MOD)) << ' ';
at++;
}
cout << '\n';
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 3508kb
input:
5 2 1 2 3 4 5
output:
4 0 24 120 60 40 30
result:
wrong answer 1st numbers differ - expected: '24', found: '4'