結果

問題 No.1417 100の倍数かつ正整数(2)
ユーザー tanimani364tanimani364
提出日時 2021-04-16 18:32:48
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 34 ms / 3,000 ms
コード長 4,906 bytes
コンパイル時間 2,900 ms
コンパイル使用メモリ 198,920 KB
最終ジャッジ日時 2025-01-20 18:16:36
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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ファイルパターン 結果
sample AC * 3
other AC * 36
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>
#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;
}

using namespace std;
//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>;

modint dp[100005][2][3][3][2];
void solve()
{	
	string s;
	cin>>s;
	int n=s.size();
	rep(i,n+1)rep(j,2)rep(k,3)rep(a,3)rep(z,2)dp[i][j][k][a][z]=0;

	dp[0][0][0][0][1]=1;
	rep(i,n){
		rep(j,2){//真に小さいかどうか
			rep(two,3){
				rep(five,3){
					rep(leading_zero,2){
						if(!j){
							rep(d,10){
								if(s[i]-'0'<d)continue;
								if(!leading_zero&&!d)continue;

								if(d==8)dp[i+1][s[i]-'0'>d][min(2,two+3)][five][0]+=dp[i][0][two][five][leading_zero];
								else if(d==4)dp[i+1][s[i]-'0'>d][min(2,two+2)][five][0]+=dp[i][0][two][five][leading_zero];
								else if(d==2||d==6)dp[i+1][s[i]-'0'>d][min(2,two+1)][five][0]+=dp[i][0][two][five][leading_zero];
								else if(d==5)dp[i+1][s[i]-'0'>d][two][min(2,five+1)][0]+=dp[i][0][two][five][leading_zero];
								else if(!d)dp[i+1][s[i]-'0'>d][two][five][leading_zero]+=dp[i][0][two][five][leading_zero];
								else dp[i+1][s[i]-'0'>d][two][five][0]+=dp[i][0][two][five][leading_zero];
							}
						}else{
							dp[i+1][1][min(2,two+3)][five][0]+=dp[i][1][two][five][leading_zero];
							dp[i+1][1][min(2,two+2)][five][0]+=dp[i][1][two][five][leading_zero];
							rep(x,2)dp[i+1][1][min(2,two+1)][five][0]+=dp[i][1][two][five][leading_zero];
							dp[i+1][1][two][min(2,five+1)][0]+=dp[i][1][two][five][leading_zero];
							rep(d,10){
								if((d>0&&d%2==0)||d==5)continue;
								if(!leading_zero&&!d)continue;
								if(!d)dp[i+1][1][two][five][leading_zero]+=dp[i][1][two][five][leading_zero];
								else dp[i+1][1][two][five][0]+=dp[i][1][two][five][leading_zero];
							}
						}
					}
				}
			}
		}
	}
	modint ans=0;
	rep(i,2)ans+=dp[n][i][2][2][0];
	cout<<ans<<"\n";
}

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