結果

問題 No.616 へんなソート
ユーザー tancahn2380
提出日時 2018-03-29 15:02:46
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
AC  
実行時間 232 ms / 2,000 ms
コード長 1,918 bytes
コンパイル時間 826 ms
コンパイル使用メモリ 105,088 KB
実行使用メモリ 310,148 KB
最終ジャッジ日時 2024-06-25 23:47:21
合計ジャッジ時間 2,822 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 27
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

# include <iostream>
# include <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <tuple>
# include <unordered_map>
# include <numeric>
# include <complex>
# include <bitset>
# include <random>
# include <chrono>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
constexpr int INF = 2147483647;
constexpr int HINF = INF / 2;
constexpr double DINF = 100000000000000000.0;
constexpr double HDINF = 50000000000000000.0;
constexpr long long LINF = 9223372036854775807;
constexpr long long HLINF = 4500000000000000000;
const double PI = acos(-1);
int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); };
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); };
# define ALL(x)      (x).begin(),(x).end()
# define UNIQ(c)     (c).erase(unique(ALL((c))),(c).end())
# define LOWER(s)    transform(ALL((s)),(s).begin(),TL<char>)
# define UPPER(s)    transform(ALL((s)),(s).begin(),TU<char>)
# define mp          make_pair
# define eb          emplace_back
# define FOR(i,a,b)  for(LL i=(a);i<(b);i++)
# define RFOR(i,a,b) for(LL i=(a);i>=(b);i--)
# define REP(i,n)    FOR(i,0,n)
# define INIT        std::ios::sync_with_stdio(false);std::cin.tie(0)
LL dp[330][101010], sum[330][101010];
const int MOD = 1e9 + 7;
int main() {
    int n, k; cin >> n >> k;
    int a;
    REP(i, n) cin >> a;
    dp[0][0] = 1;
    REP(j, k + 1) sum[0][j] = 1LL;
    for (LL i = 1; i < n; i++) {
        REP(j, k + 1) {
            if (min(i, j) <= j - 1) dp[i][j] = (sum[i - 1][j] - sum[i - 1][j - min(i, j) - 1] + MOD) % MOD;
            else dp[i][j] = sum[i - 1][j];
        }
        sum[i][0] = 1LL;
        for (int j = 1; j < k + 1; j++) sum[i][j] = (sum[i][j - 1] + dp[i][j]) % MOD;
    }
    cout << sum[n - 1][k] << endl;
}
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