結果

問題 No.1629 Sorting Integers (SUM of M)
ユーザー tanimani364tanimani364
提出日時 2021-07-30 21:37:24
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
TLE  
実行時間 -
コード長 4,020 bytes
コンパイル時間 2,056 ms
コンパイル使用メモリ 201,016 KB
実行使用メモリ 8,704 KB
最終ジャッジ日時 2023-10-14 03:49:14
合計ジャッジ時間 17,899 ms
ジャッジサーバーID
(参考情報)
judge11 / judge12
このコードへのチャレンジ
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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
8,284 KB
testcase_01 AC 2 ms
4,348 KB
testcase_02 AC 1 ms
4,348 KB
testcase_03 AC 1 ms
4,348 KB
testcase_04 AC 284 ms
4,352 KB
testcase_05 AC 58 ms
4,348 KB
testcase_06 AC 80 ms
4,352 KB
testcase_07 AC 1,111 ms
4,352 KB
testcase_08 AC 591 ms
4,348 KB
testcase_09 AC 1,056 ms
4,352 KB
testcase_10 AC 599 ms
4,352 KB
testcase_11 AC 471 ms
4,356 KB
testcase_12 AC 1,509 ms
4,352 KB
testcase_13 AC 1,642 ms
4,348 KB
testcase_14 AC 1,068 ms
4,352 KB
testcase_15 AC 1,739 ms
4,352 KB
testcase_16 AC 1,629 ms
4,352 KB
testcase_17 TLE -
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>
//#include<boost/multiprecision/cpp_int.hpp>
//#include<boost/multiprecision/cpp_dec_float.hpp>
//#include <atcoder/all>
using namespace std;
#define rep(i, a) for (int i = (int)0; i < (int)a; ++i)
#define rrep(i, a) for (int i = (int)a; i > -1; --i)
#define REP(i, a, b) for (int i = (int)a; i < (int)b; ++i)
#define RREP(i, a, b) for (int i = (int)a; i > b; --i)
#define repl(i, a) for (ll i = (ll)0; i < (ll)a; ++i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll
#define fi first
#define se second
using ll = long long;
constexpr ll mod = 1e9 + 7;
constexpr ll mod_998244353 = 998244353;
constexpr ll INF = 1LL << 60;

// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

// using lll=boost::multiprecision::cpp_int;
// using
// Double=boost::multiprecision::number<boost::multiprecision::cpp_dec_float<128>>;//仮数部が1024桁
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;
}

ll mypow(ll x, ll n, const ll &p = -1)
{ // x^nをmodで割った余り

  if (p != -1)
  {
    x = (x % p + p) % p;
  }
  ll ret = 1;
  while (n > 0)
  {
    if (n & 1)
    {
      if (p != -1)
        ret = (ret * x) % p;
      else
        ret *= x;
    }
    if (p != -1)
      x = (x * x) % p;
    else
      x *= x;
    n >>= 1;
  }
  return ret;
}

struct myrand{
  random_device seed;
  mt19937 mt;
  myrand():mt(seed()){}
  int operator()(int a,int b){//[a,b)
    uniform_int_distribution<int>dist(a,b-1);
    return dist(mt);
  }
};

//using namespace atcoder;

//------------------------
//------------------------
//------------------------
//------------------------
//------------------------

template<int mod>
struct Modint{
    int x;
    Modint():x(0){}
    Modint(int64_t y):x((y%mod+mod)%mod){}

    Modint &operator+=(const Modint &p){
			if((x+=p.x)>=mod)
				x -= mod;
			return *this;
		}

		Modint &operator-=(const Modint &p){
			if((x+=mod-p.x)>=mod)
				x -= mod;
			return *this;
		}

		Modint &operator*=(const Modint &p){
			x = (1LL * x * p.x) % mod;
			return *this;
		}

		Modint &operator/=(const Modint &p){
			*this *= p.inverse();
			return *this;
		}

		Modint operator-() const { return Modint(-x); }
		Modint operator+(const Modint &p) const{
			return Modint(*this) += p;
		}
		Modint operator-(const Modint &p) const{
			return Modint(*this) -= p;
		}
		Modint operator*(const Modint &p) const{
			return Modint(*this) *= p;
		}
		Modint operator/(const Modint &p) const{
			return Modint(*this) /= p;
		}

		bool operator==(const Modint &p) const { return x == p.x; }
		bool operator!=(const Modint &p) const{return x != p.x;}

		Modint inverse() const{//非再帰拡張ユークリッド
			int a = x, b = mod, u = 1, v = 0;
			while(b>0){
				int t = a / b;
				swap(a -= t * b, b);
				swap(u -= t * v, v);
			}
			return Modint(u);
		}

		Modint pow(int64_t n) const{//繰り返し二乗法
			Modint ret(1), mul(x);
			while(n>0){
				if(n&1)
					ret *= mul;
				mul *= mul;
				n >>= 1;
			}
			return ret;
		}

		friend ostream &operator<<(ostream &os,const Modint &p){
			return os << p.x;
		}
};

using modint = Modint<mod>;
using modint2= Modint<mod_998244353>;


void solve()
{
  ll n;
  cin>>n;
  ll c[9];
  rep(i,9)cin>>c[i];

  vector<modint>fact(n+1,1);
  for(ll i=1;i<=n;++i)fact[i]=fact[i-1]*i;
  modint ans=0;
  rep(i,n){
    rep(j,9){
      modint add=fact[n-1];
      rep(k,9){
        if(j==k)add/=fact[c[k]-1];
        else add/=fact[c[k]];
      }
      add*=j+1;
      add*=mypow(10,i,mod);
      ans+=add;
    }
  }
  cout<<ans<<"\n";
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  solve();
  return 0;
}
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