結果
問題 | No.616 へんなソート |
ユーザー |
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提出日時 | 2019-07-16 00:35:53 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 344 ms / 2,000 ms |
コード長 | 3,086 bytes |
コンパイル時間 | 1,605 ms |
コンパイル使用メモリ | 113,644 KB |
実行使用メモリ | 215,680 KB |
最終ジャッジ日時 | 2024-11-28 03:42:02 |
合計ジャッジ時間 | 3,672 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 27 |
ソースコード
#pragma GCC optimize ("-O3","unroll-loops")#include<iostream>#include<string>#include<algorithm>#include<vector>#include<queue>#include<map>#include<math.h>#include<iomanip>#include<set>#include<numeric>#include<cstring>#include<cstdio>#include<functional>#include<bitset>#include<limits.h>#include<cassert>#include<iterator>#include<complex>#include<stack>#define REP(i, n) for(int i = 0;i < n;i++)#define REPR(i, n) for(int i = n;i >= 0;i--)#define FOR(i, m, n) for(int i = m;i < n;i++)#define FORR(i, m, n) for(int i = m;i >= n;i--)#define SORT(v, n) sort(v, v+n);#define VSORT(v) sort(v.begin(), v.end());#define REVERSE(v,n) reverse(v,v+n);#define VREVERSE(v) reverse(v.begin(), v.end());#define ll long long#define pb(a) push_back(a)#define m0(x) memset(x,0,sizeof(x))#define print(x) cout<<x<<'\n';#define pe(x) cout<<x<<" ";#define lb(v,n) lower_bound(v.begin(), v.end(), n);#define ub(v,n) upper_bound(v.begin(), v.end(), n);#define int long long#define all(x) (x).begin(), (x).end()#define double long doubleusing namespace std;template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}int MOD = (ll)1000000000 + 7;const ll INF = 1e17;const double pi = acos(-1);const double EPS = 1e-10;typedef pair<int, int>P;const int MAX = 200020;long long fac[MAX], finv[MAX], inv[MAX];// テーブルを作る前処理void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++) {fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k) {if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}ll add(ll x, ll y) {x += y;if (x >= MOD) return x - MOD;return x;}ll sub(ll x, ll y) {x -= y;if (x < 0) return x + MOD;return x;}ll mult(ll x, ll y) {return (x * y) % MOD;}ll bin_pow(ll x, ll p) {if (p == 0) return 1;if (p & 1) return mult(x, bin_pow(x, p - 1));return bin_pow(mult(x, x), p / 2);}ll dp[303][50000];vector<int>v;int N, K;ll cum[303][500000];signed main() {cin.tie(0);ios::sync_with_stdio(false);cin >> N >> K;REP(i, N) {int a; cin >> a;v.pb(a);}VSORT(v); VREVERSE(v);dp[1][0] = 1;FOR(j,1, K + 2)cum[1][j] = 1;int cnt = 0;FOR(i, 1, N) {if (v[i] < v[i - 1])cnt++;REP(j, K + 1) {int res = 0;int counter = min(cnt + 1, j+1);if (counter == j + 1) {res = cum[i][j+1];}else {res = sub(cum[i][j+1], cum[i][j - cnt]);}/*REP(k, counter) {res =add(res, dp[i][j - k]);}*/dp[i + 1][j] = res;//pe(i + 1)pe(j)print(res);}cum[i + 1][1] = dp[i + 1][0];FOR(j, 1, K + 1) {cum[i + 1][j+1] = cum[i + 1][j ] + dp[i + 1][j];cum[i + 1][j+1] %= MOD;}}ll ans = 0;REP(j, K + 1){ans = add(ans,dp[N][j]);}print(ans);}