結果

問題 No.1086 桁和の桁和2
ユーザー penguinshunyapenguinshunya
提出日時 2020-06-19 23:19:33
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,152 bytes
コンパイル時間 2,121 ms
コンパイル使用メモリ 204,248 KB
実行使用メモリ 5,812 KB
最終ジャッジ日時 2023-09-16 15:31:54
合計ジャッジ時間 6,256 ms
ジャッジサーバーID
（参考情報） 		judge14 / judge15
このコードへのチャレンジ（β）

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,376 KB
testcase_01 AC 2 ms
4,376 KB
testcase_02 AC 2 ms
4,384 KB
testcase_03 WA -
testcase_04 AC 2 ms
4,376 KB
testcase_05 WA -
testcase_06 AC 1 ms
4,376 KB
testcase_07 AC 2 ms
4,376 KB
testcase_08 AC 2 ms
4,380 KB
testcase_09 AC 1 ms
4,380 KB
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
testcase_18 WA -
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 WA -
testcase_32 WA -
testcase_33 WA -
testcase_34 WA -
testcase_35 WA -
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 # 

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define all(a) (a).begin(), (a).end()
#define bit32(x) (1 << (x))
#define bit64(x) (1ll << (x))
#define sz(v) ((int) v.size())

using namespace std;

using i64 = long long;
using f80 = long double;
using vi32 = vector<int>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

void yes() { cout << "Yes" << endl; exit(0); }
void no() { cout << "No" << endl; exit(0); }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> void amax(T &x, T y) { x = max(x, y); }
template <typename T> void amin(T &x, T y) { x = min(x, y); }
template <typename T> T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

const int INF = 1001001001;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};

template <int mod>
struct ModInt {
  int x;
  ModInt(): x(0) {}
  ModInt(long long a) { x = a % mod; if (x < 0) x += mod; }
  ModInt &operator+=(ModInt that) { x = (x + that.x) % mod; return *this; }
  ModInt &operator-=(ModInt that) { x = (x + mod - that.x) % mod; return *this; }
  ModInt &operator*=(ModInt that) { x = (long long) x * that.x % mod; return *this; }
  ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
  ModInt inverse() {
    int a = x, b = mod, u = 1, v = 0;
    while (b) { int t = a / b; a -= t * b; u -= t * v; swap(a, b); swap(u, v); }
    return ModInt(u);
  }
  #define op(o, p) ModInt operator o(ModInt that) { return ModInt(*this) p that; }
    op(+, +=) op(-, -=) op(*, *=) op(/, /=)
  #undef op
  friend ostream& operator<<(ostream &os, ModInt m) { return os << m.x; }
};

using mint = ModInt<1000000007>;

void solve() {
  int n;
  cin >> n;
  vi64 L(n), R(n), D(n);
  cin >> L >> R >> D;
  vi32 d(n);
  d[0] = D[0];
  reps(i, n - 1) {
    if (D[i-1] < D[i]) d[i]=D[i]-D[i-1];
    if (D[i-1] > D[i]) d[i]=(D[i]+D[i-1]+1)%10;
  }
  int now = 0;
  mint ans = 1;
  rep(i, n) {
    if (now == 0 && d[i] == 0) {
      continue;
    }
    mint a = exp((mint) 10, L[i]);
    mint r = 10;
    i64 n = R[i] - L[i];
    mint t = (a * exp(r, n) - a) / (r - 1);
    if (d[i] == 0) {
      t += 1;
    }
    ans *= t;
    now += d[i];
  }
  cout << ans << endl;
}
0