結果

問題 No.260 世界のなんとか3
コンテスト
ユーザー hibit_at
提出日時 2024-01-23 09:45:12
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 201 ms / 2,000 ms
コード長 5,963 bytes
コンパイル時間 1,941 ms
コンパイル使用メモリ 132,676 KB
最終ジャッジ日時 2025-02-18 22:15:38
ジャッジサーバーID
（参考情報） 		judge2 / 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 # 

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <limits.h>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>
#include <stack>
#include <complex>

using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i < n + 1; i++)
#define rev(i, n) for (int i = n - 1; i >= 0; i--)
#define all(A) A.begin(), A.end()
#define itr(A, l, r) A.begin() + l, A.begin() + r
#define debug(var) cout << #var << " = " << var << endl;
typedef long long ll;

template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p)
{
    os << "(" << p.first << "," << p.second << ")";
    return os;
}

template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p)
{
    is >> p.first >> p.second;
    return is;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << (i + 1 != (int)v.size() ? " " : "");
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<vector<T>> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << endl;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, const vector<vector<vector<T>>> &v)
{
    int n = v.size();
    int m = v[0].size();
    int p = v[0][0].size();
    rep(k, p)
    {
        os << "k = " << k << endl;
        rep(i, n)
        {
            rep(j, m)
            {
                os << v[i][j][k];
                if (j < m - 1)
                {
                    os << " ";
                }
                else
                {
                    os << endl;
                }
            }
        }
    }
    return os;
}

template <typename T>
istream &operator>>(istream &is, vector<T> &v)
{
    for (T &in : v)
        is >> in;
    return is;
}

template <typename T, typename S>
ostream &operator<<(ostream &os, map<T, S> &mp)
{
    for (auto &[key, val] : mp)
    {
        os << key << ":" << val << " ";
    }
    cout << endl;
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, set<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, multiset<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, queue<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, deque<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop_front();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, stack<T> st)
{
    while (st.size())
    {
        os << st.top() << " ";
        st.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, priority_queue<T> pq)
{
    while (pq.size())
    {
        os << pq.top() << " ";
        pq.pop();
    }
    return os;
}

template <typename T>
ostream &operator<<(ostream &os, priority_queue<T, vector<T>, greater<T>> mpq)
{
    while (mpq.size())
    {
        os << mpq.top() << " ";
        mpq.pop();
    }
    return os;
}

#include <atcoder/modint>
using namespace atcoder;
using mint = modint1000000007;
ostream &operator<<(ostream &os, mint &i)
{
    os << i.val();
    return os;
}

ostream &operator<<(ostream &os, const vector<mint> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i].val() << (i + 1 != (int)v.size() ? " " : "");
    }
    return os;
}

mint solve(string s)
{
    vector<int> a;
    for (char c : s)
    {
        a.push_back(c - '0');
    }
    // debug(a);
    int m = s.size();
    using VI = vector<mint>;
    using VVI = vector<VI>;
    using VVVI = vector<VVI>;
    using VVVVI = vector<VVVI>;
    using VVVVVI = vector<VVVVI>;
    VVVVI dp(3,VVVI(8, VVI(2, VI(2))));
    dp[0][0][0][0] = 1;
    vector<int> factor(m,1);
    rep(i,m-1){
        factor[i+1] = factor[i]*10;
        factor[i+1] %= 8;
    }
    // debug(factor);
    reverse(all(factor));
    rep(i, m)
    {
        VVVVI ndp(3,VVVI(8, VVI(2, VI(2))));
        rep(mod3, 3)
        {
            rep(mod8, 8)
            {
                rep(contains, 2)
                {
                    rep(small, 2)
                    {
                        rep(next, 10)
                        {
                            ndp[(mod3 + next) % 3][(mod8 + next*factor[i]) % 8][contains || next == 3][small || next < a[i]] += dp[mod3][mod8][contains][small] * (small || next <= a[i]);
                        }
                    }
                }
            }
        }
        dp = ndp;
    }
    mint ans = 0;
    rep(mod3, 3)
    {
        rep(mod8, 8)
        {
            rep(contains, 2)
            {
                rep(small, 2)
                {
                    if((mod3==0||contains) && mod8>0){
                        ans += dp[mod3][mod8][contains][small];
                    }
                }
            }
        }
    }
    return ans;
}
int main()
{
    string a, b;
    cin >> a >> b;
    vector<int> v;
    for (char c : a)
    {
        v.push_back(c - '0');
    }
    reverse(all(v));
    // debug(v);
    int n = v.size();
    v[0]--;
    rep(i, n - 1)
    {
        if (v[i] < 0)
        {
            v[i] = 9;
            v[i + 1]--;
        }
    }
    // debug(v);
    string c;
    for (int i : v)
    {
        c.push_back('0' + i);
    }
    reverse(all(c));
    // debug(c);
    mint ans = solve(b) - solve(c);
    cout << ans << endl;
}
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