結果
| 問題 |
No.493 とても長い数列と文字列(Long Long Sequence and a String)
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-09-19 19:48:52 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 3 ms / 800 ms |
| コード長 | 3,812 bytes |
| コンパイル時間 | 2,364 ms |
| コンパイル使用メモリ | 198,968 KB |
| 最終ジャッジ日時 | 2025-01-14 18:53:12 |
|
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 115 |
ソースコード
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 0xcccccccccccccccLL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}
const unsigned int mod = 1000000007;
//const unsigned int mod = 998244353;
struct mint {
unsigned int x;
mint():x(0) {}
mint(int64_t x_) {
int64_t v = int64_t(x_ % mod);
if(v < 0) v += mod;
x = (unsigned int)v;
}
static mint row(int v) {
mint v_;
v_.x = v;
return v_;
}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) {
uint64_t z = x;
z *= a.x;
x = (unsigned int)(z % mod);
return *this;
}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
friend bool operator==(const mint &a, const mint &b) {return a.x == b.x;}
friend bool operator!=(const mint &a, const mint &b) {return a.x != b.x;}
mint &operator++() {
x++;
if(x == mod) x = 0;
return *this;
}
mint &operator--() {
if(x == 0) x = mod;
x--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint pow(int64_t t) const {
mint x_ = *this, r = 1;
while(t) {
if(t&1) r *= x_;
x_ *= x_;
t >>= 1;
}
return r;
}
//for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) {return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
#define U 61
//head
int k;
ll l, r, u[U], sum[U];
mint prod[U];
pair<ll, mint> calc(ll C, int k) {
if(!C) return {0, 1};
ll *itr = lower_bound(u, u+U, C);
int z = itr-u;
if(z > k) return {-1, -1};
string s = to_string(z*z);
int siz = s.size();
pair<ll, mint> res = {sum[z-1], prod[z-1]};
C -= u[z-1];
if(C < siz) {
rep(i, C) {
int h = (s[i] == '0'?10:s[i]-'0');
res.first += h;
res.second *= h;
}
return res;
}
rep(i, siz) {
int h = (s[i] == '0'?10:s[i]-'0');
res.first += h;
res.second *= h;
}
auto [_, __] = calc(C-siz, k);
res.first += _;
res.second *= __;
return res;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> k >> l >> r;
fill(prod, prod+U, mint(1));
for(int i = 1; i < U; i++) {
string z = to_string(i*i);
int siz = z.size();
u[i] = siz + u[i-1]*2;
rep(j, siz) sum[i] += (z[j] == '0'?10:z[j]-'0');
sum[i] += sum[i-1]<<1;
rep(j, siz) prod[i] *= (z[j] == '0'?10:z[j]-'0');
prod[i] *= prod[i-1]*prod[i-1];
}
auto [s1, p1] = calc(l-1, k);
auto [s2, p2] = calc(r, k);
if(s2 == -1) cout << s2 << endl;
else cout << make_pair(s2-s1, p2/p1) << endl;
}