結果

問題 No.493 とても長い数列と文字列(Long Long Sequence and a String）
ユーザー tnakao0123
提出日時 2017-03-14 12:54:09
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
AC  
実行時間 2 ms / 800 ms
コード長 2,474 bytes
コンパイル時間 633 ms
コンパイル使用メモリ 86,180 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-06-30 00:44:19
合計ジャッジ時間 3,019 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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ファイルパターン 結果
sample AC * 4
other AC * 115
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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

/* -*- coding: utf-8 -*-
 *
 * 493.cc: No.493 とても長い数列と文字列(Long Long Sequence and a String) - yukicoder
 */
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>
 
using namespace std;
/* constant */
const int MAX_N = 100;
typedef long long ll;
const ll MAX_R = 1000000000000000000LL;
const ll MOD = 1000000007;
/* typedef */
/* global variables */
ll lfs[MAX_N], sums[MAX_N], muls[MAX_N];
int lds[MAX_N], ds[MAX_N][5];;
/* subroutines */
ll sum_range(int k, ll l, ll r) {
  ll &lfk = lfs[k];
  int &ldk = lds[k];
  if (k <= 0 || r <= 0 || lfk <= l) return 0;
  if (l <= 0 && lfk <= r) return sums[k];
  ll off = lfs[k - 1] + ldk;
  ll sum =
    sum_range(k - 1, l, r) + sum_range(k - 1, l - off, r - off);
  ll i0 = max(l - lfs[k - 1], 0LL), i1 = min(r - lfs[k - 1], (ll)ldk);
  if (i0 < ldk && 0 < i1)
    for (int i = i0; i < i1; i++) sum += ds[k][i];
  return sum;
}
ll mul_range(int k, ll l, ll r) {
  ll &lfk = lfs[k];
  int &ldk = lds[k];
  if (k <= 0 || r <= 0 || lfk <= l) return 1;
  if (l <= 0 && lfk <= r) return muls[k];
  ll off = lfs[k - 1] + ldk;
  ll mul =
    (mul_range(k - 1, l, r) * mul_range(k - 1, l - off, r - off)) % MOD;
  ll i0 = max(l - lfs[k - 1], 0LL), i1 = min(r - lfs[k - 1], (ll)ldk);
  if (i0 < ldk && 0 < i1)
    for (int i = i0; i < i1; i++) mul = (mul * ds[k][i]) % MOD;
  return mul;
}
/* main */
int main() {
  lfs[0] = 0;
  lds[0] = 0;
  sums[0] = 0;
  muls[0] = 1;
  int n;
  for (n = 1;; n++) {
    sums[n] = sums[n - 1] * 2;
    muls[n] = (muls[n - 1] * muls[n - 1]) % MOD;
    int n2 = n * n, l = 0;
    for (; n2 > 0; l++, n2 /= 10) {
      int d = n2 % 10;
      if (d == 0) d = 10;
      sums[n] += d;
      muls[n] = (muls[n] * d) % MOD;
      ds[n][l] = d;
    }
    lfs[n] = lfs[n - 1] * 2 + l;
    lds[n] = l;
    reverse(ds[n], ds[n] + l);
    //printf("lfs[%d]=%lld (%d^2=%d(%d))", n, lfs[n], n, n * n, l);
    //printf(" sum=%lld, mul=%lld\n", sums[n], muls[n]);
    if (lfs[n] > MAX_R) break;
  }
  int k;
  ll l, r;
  cin >> k >> l >> r;
  l--;
  if (k > n) k = n;
  if (lfs[k] < r) {
    puts("-1");
    return 0;
  }
  ll s = sum_range(k, l, r);
  ll m = mul_range(k, l, r);
  printf("%lld %lld\n", s, m);
  return 0;
}
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