結果

問題 No.1417 100の倍数かつ正整数(2)
ユーザー kanra824kanra824
提出日時 2021-03-05 21:41:56
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 33 ms / 3,000 ms
コード長 4,033 bytes
コンパイル時間 2,173 ms
コンパイル使用メモリ 206,716 KB
最終ジャッジ日時 2025-01-19 10:46:17
ジャッジサーバーID
(参考情報)
judge5 / 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 * 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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i=0; i<(n); ++i)
#define RREP(i, n) for(int i=(n);i>=0;--i)
#define FOR(i, a, n) for (int i=(a); i<(n); ++i)
#define RFOR(i, a, b) for(int i=(a);i>=(b);--i)
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()
#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl;
template<class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
REP(i, SZ(v)) {
if (i) os << " ";
os << v[i];
}
return os;
}
template <class T>
void debug(const vector<T> &v) {
cout << "[";
REP(i, SZ(v)) {
if(i) cout << ", ";
cout << v[i];
}
cout << "]" << endl;
}
template<class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
return os << p.first << " " << p.second;
}
template <class T, class U>
void debug(const pair<T, U> &p) {
cout << "(" << p.first << " " << p.second << ")" << endl;
}
template<class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template<class T>
bool chmin(T &a, const T &b) {
if (b < a) {
a = b;
return true;
}
return false;
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
const ll MOD = 1e9 + 7;
const ll MOD998 = 998244353;
const int INF = INT_MAX;
const ll LINF = LLONG_MAX;
const int inf = INT_MIN;
const ll linf = LLONG_MIN;
const ld eps = 1e-9;
template<int m>
struct mint {
int x;
mint(ll x = 0) : x(((x % m) + m) % m) {}
mint operator-() const { return x ? m - x : 0; }
mint &operator+=(mint r) {
if ((x += r.x) >= m) x -= m;
return *this;
}
mint &operator-=(mint r) {
if ((x -= r.x) < 0) x += m;
return *this;
}
mint &operator*=(mint r) {
x = ((ll) x * r.x) % m;
return *this;
}
mint inv() const { return pow(m - 2); }
mint &operator/=(mint r) { return *this *= r.inv(); }
friend mint operator+(mint l, mint r) { return l += r; }
friend mint operator-(mint l, mint r) { return l -= r; }
friend mint operator*(mint l, mint r) { return l *= r; }
friend mint operator/(mint l, mint r) { return l /= r; }
mint pow(ll n) const {
mint ret = 1, tmp = *this;
while (n) {
if (n & 1) ret *= tmp;
tmp *= tmp, n >>= 1;
}
return ret;
}
friend bool operator==(mint l, mint r) { return l.x == r.x; }
friend bool operator!=(mint l, mint r) { return l.x != r.x; }
friend ostream &operator<<(ostream &os, mint a) {
return os << a.x;
}
friend istream &operator>>(istream &is, mint &a) {
ll x;
is >> x;
a = x;
return is;
}
};
using Int = mint<MOD>;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(10);
string s; cin >> s;
int n = SZ(s);
auto dp = vector(n+1, vector(3, vector(3, vector(2, vector(2, Int(0))))));
dp[0][0][0][1][1] = 1;
REP(i, n) {
REP(j, 3) {
REP(k, 3) {
dp[i+1][j][k][0][1] += dp[i][j][k][1][1] + dp[i][j][k][0][1];
FOR(l, 1, 10) {
int dj = 0, dk = 0;
if(l == 2) dj = 1;
if(l == 4) dj = 2;
if(l == 5) dk = 1;
if(l == 6) dj = 1;
if(l == 8) dj = 3;
int nj = min(2, j + dj);
int nk = min(2, k + dk);
dp[i+1][nj][nk][0][0] += dp[i][j][k][0][0] + dp[i][j][k][0][1];
if(l == s[i] - '0') {
dp[i+1][nj][nk][1][0] += dp[i][j][k][1][1] + dp[i][j][k][1][0];
} else if(l < s[i] - '0') {
dp[i+1][nj][nk][0][0] += dp[i][j][k][1][1] + dp[i][j][k][1][0];
}
}
}
}
}
cout << dp[n][2][2][0][0] + dp[n][2][2][1][0] << endl;
return 0;
}
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