結果

問題 No.314 ケンケンパ
ユーザー tancahn2380
提出日時 2017-11-14 21:58:52
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
RE  
実行時間 -
コード長 1,446 bytes
コンパイル時間 679 ms
コンパイル使用メモリ 87,636 KB
実行使用メモリ 15,884 KB
最終ジャッジ日時 2024-11-25 02:15:44
合計ジャッジ時間 2,280 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 * 2 WA * 1
other AC * 5 WA * 9 RE * 3
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp:35:22: warning: overflow in conversion from ‘double’ to ‘int’ changes value from ‘1.0000000007e+10’ to ‘2147483647’ [-Woverflow]
   35 | const int MOD = 10e9 + 7;
      |                 ~~~~~^~~

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 <bitset>
# include <complex>
# include <numeric>
# include <tuple>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
constexpr int INF = 2000000000;
constexpr int HINF = INF / 2;
constexpr double DINF = 100000000000000000.0;
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 };
#define ALL(x)    (x).begin(),(x).end()
#define UNIQ(c)   (c).erase(unique(ALL((c))), end((c)))
#define mp        make_pair
#define eb        emplace_back
typedef pair<LL, double> P;
typedef pair<P, P> PP;
const int MOD = 10e9 + 7;
LL dp[101010][4][4];
int main() {
    dp[0][0][0] = 1;
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        if (i == 1) {
            dp[1][1][0] += dp[0][0][0];
        }
        else if (i == 2) {
            dp[2][1][1] += dp[1][1][0];
            dp[2][2][1] += dp[1][1][0];
        }
        else {
            dp[i][1][1] = (dp[i - 1][1][2]) % MOD;
            dp[i][1][2] = (dp[i - 1][2][1]) % MOD;
            dp[i][2][1] = (dp[i - 1][1][1] + dp[i - 1][1][2]) % MOD;
        }
    }
    LL ans = 0;
    for (int i = 0; i <= 2; i++)for (int j = 0; j <= 2; j++)ans = (ans + dp[n][i][j]) % MOD;
    cout << ans%MOD << endl;
}
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