結果

問題 No.1086 桁和の桁和2
ユーザー chocopuu
提出日時 2020-06-20 18:02:24
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 419 ms / 3,000 ms
コード長 4,448 bytes
コンパイル時間 2,185 ms
コンパイル使用メモリ 204,292 KB
最終ジャッジ日時 2025-01-11 08:46:39
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 5
other AC * 31
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ソースコード

diff #
プレゼンテーションモードにする

#include "bits/stdc++.h"
using namespace std;
#define int long long
#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 FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define RFOR(i, s, n) for (int i = (int)n - 1; i >= s; --i)
#define ALL(a) a.begin(), a.end()
#define IN(a, x, b) (a <= x && x < b)
template<class T>inline void out(T t){cout << t << "\n";}
template<class T,class... Ts>inline void out(T t,Ts... ts){cout << t << " ";out(ts...);}
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;}
constexpr int INF = 1e18;
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};
 
// 二項係数ライブラリ
template<class T> struct BiCoef {
    vector<T> fact_, inv_, finv_;
    constexpr BiCoef() {}
    constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        init(n);
    }
    constexpr void init(int n) noexcept {
        fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
        int MOD = fact_[0].getmod();
        for(int i = 2; i < n; i++){
            fact_[i] = fact_[i-1] * i;
            inv_[i] = -inv_[MOD%i] * (MOD/i);
            finv_[i] = finv_[i-1] * inv_[i];
        }
    }
    constexpr T com(int n, int k) const noexcept {
        if (n < k || n < 0 || k < 0) return 0;
        return fact_[n] * finv_[k] * finv_[n-k];
    }
    constexpr T fact(int n) const noexcept {
        if (n < 0) return 0;
        return fact_[n];
    }
    constexpr T inv(int n) const noexcept {
        if (n < 0) return 0;
        return inv_[n];
    }
    constexpr T finv(int n) const noexcept {
        if (n < 0) return 0;
        return finv_[n];
    }
};
 
const int MOD = 1000000007;
// const int MOD = 
using mint = Fp<MOD>;
BiCoef<mint> bc;
// bc.init(500050);
long long mod_pow(long long x, long long n, long long MOD) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * x % MOD;
        x = x * x % MOD;
        n >>= 1;
    }
    return res;
}
long long mod_inv(long long x, long long MOD)
{
    return mod_pow(x, MOD - 2, MOD);
}
//mod_inv(x)でxの逆元をとれる
using T = tuple<int,int,int>;
signed main(){
    int N;
    cin >> N;
    vector<int>l(N),r(N),d(N);
    REP(i,N)cin >> l[i];
    REP(i,N)cin >> r[i];
    REP(i,N)cin >> d[i];
    vector<T>v(N);
    REP(i,N){
        v[i] = make_tuple(l[i], r[i], d[i]);
    }
    mint ans = 1;
    int nowD = 0;
    REP(i,N){
        auto[l,r,d] = v[i];
        mint R = modpow((mint)10, r) - 1;
        R /= 9;
        int modR = mod_pow(10, r, 9);
        vector<mint>vr(9, R);
        REP(j, modR)vr[j] += 1;
        mint L = modpow((mint)10, l) - 1;
        L /= 9;
        int modL = mod_pow(10, l, 9);
        vector<mint>vl(9, L);
        REP(j, modL)vl[j] += 1;
        vector<mint>vec(9);
        REP(j, 9)vec[j] = vr[j] - vl[j];
        if(d == 0){
            if(nowD == d)continue;
            else{
                cout << 0 << "\n";
                return 0;
            }
        }else{
            if(nowD == d)ans *= vec[0] + 1;
            else ans *= vec[0];
        }
        nowD = d;
    }
    cout << ans << "\n";
}
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