結果
問題 | No.2527 H and W |
ユーザー |
|
提出日時 | 2023-11-03 22:57:33 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 85 ms / 2,000 ms |
コード長 | 2,967 bytes |
コンパイル時間 | 4,327 ms |
コンパイル使用メモリ | 256,172 KB |
最終ジャッジ日時 | 2025-02-17 18:45:30 |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 23 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>#include <time.h>using namespace atcoder;using ll = long long;using vll = vector<ll>;using vvll = vector<vll>;using vvvll = vector<vvll>;using vb = vector<bool>;using vvb = vector<vb>;using vvvb = vector<vvb>;#define all(A) A.begin(),A.end()#define rep(i, n) for (ll i = 0; i < (ll) (n); i++)template<class T>bool chmax(T& p, T q, bool C = 1) {if (C == 0 && p == q) {return 1;}if (p < q) {p = q;return 1;}else {return 0;}}template<class T>bool chmin(T& p, T q, bool C = 1) {if (C == 0 && p == q) {return 1;}if (p > q) {p = q;return 1;}else {return 0;}}ll modPow(long long a, long long n, long long p) {if (n == 0) return 1; // 0乗にも対応する場合if (n == 1) return a % p;if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;long long t = modPow(a, n / 2, p);return (t * t) % p;}ll cnt = 0;ll gcd(ll(a), ll(b)) {cnt++;if (a == 0)return b;if (b == 0)return a;ll c = a;while (a % b != 0) {c = a % b;a = b;b = c;}return b;}ll sqrtz(ll N) {ll L = 0;ll R = sqrt(N) + 10000;while (abs(R - L) > 1) {ll mid = (R + L) / 2;if (mid * mid <= N)L = mid;else R = mid;}return L;}ll nzkon(ll N, ll K) {//return 0;}using mint = modint998244353;using vm = vector<mint>;using vvm = vector<vm>;using vvvm = vector<vvm>;vector<mint> fact, factinv, inv;const ll mod = 998244353;void prenCkModp(ll n) {fact.resize(n + 5);factinv.resize(n + 5);inv.resize(n + 5);fact[0] = fact[1] = 1;factinv[0] = factinv[1] = 1;inv[1] = 1;for (ll i = 2; i < n + 5; i++) {fact[i] = (fact[i - 1] * i);inv[i] = (mod - ((inv[mod % i] * (mod / i))));factinv[i] = (factinv[i - 1] * inv[i]);}}mint nCk(ll n, ll k) {if (n < k || k < 0) return 0;return (fact[n] * ((factinv[k] * factinv[n - k])));}bool DEB = 0;vvm mul(vvm A, vvm B) {ll n = A.size();vvm res(n, vm(n, 0));rep(i, n)rep(j, n)rep(k, n)res[i][j] += A[i][k] * B[k][j];return res;}bool C(ll H, ll M) {if (H % 10 > 5)return 0;if (H / 10 <= 1) {return 1;}if (M / 10 <= 3)return 1;return 0;}int main() {cin.tie(nullptr);ios::sync_with_stdio(false);ll H,W,K;cin>>H>>W>>K;prenCkModp(H+W);mint an=0;for(ll h=1;h*h<=K;h++){if(K%h==0){ll e=h;ll f=K/h;if(1<=e&&e<=H&&1<=f&&f<=W){an+=nCk(H,e)*nCk(W,f);}if(e!=f){swap(e,f);if(1<=e&&e<=H&&1<=f&&f<=W){an+=nCk(H,e)*nCk(W,f);}}}}cout<<an.val()<<endl;}