#70146 (C++11（廃止可能性あり）) No.319 happy b1rthday 2 me

• 問題: No.319 happy b1rthday 2 me
• ユーザー: okaduki
• 提出日時: 2016-01-05 01:01:39
• 言語: C++11（廃止可能性あり） (gcc 13.3.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 1,902 bytes
• コンパイル時間: 1,282 ms
• コンパイル使用メモリ: 161,256 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-09-19 11:20:18
• 合計ジャッジ時間: 2,383 ms
• ジャッジサーバーID （参考情報）: judge3 / judge4

ファイルパターン	結果
sample	AC * 4
other	AC * 29

ソースコード

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#include <bits/stdc++.h>
using namespace std;
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI  = acos(-1.0);
LL dp[100][2];
LL g(LL x){
  if(x < 12) return 0;
  stringstream ss; ss << x;
  string s; ss >> s;
  LL res = 0;
  LL p = 1;
  for(int i=SZ(s)-2;i>=0;--i){
    LL y = atoll(s.substr(i,2).c_str());
    LL a = atoll(s.substr(0,i).c_str());
    LL b = atoll(s.substr(i+2).c_str());
    if(y < 12)
      res += a * p;
    else if(y == 12)
      res += a * p + b + 1;
    else
      res += (a+1) * p;
    p *= 10;
  }
  return res;
}
LL f(LL x){
  LL res = g(x);
  if(x >= 2) ++res;
  LL p;
  for(p=1;p*10<=x;p*=10)
    res += p / 10;
  if(p*3 <= x) res += p / 10;
  else if(2*p+2 <= x){
    if(x % 10 >= 2)
      res += x % p / 10 + 1;
    else
      res += x % p / 10;
  }
  return res;
}
LL naive(LL x){
  stringstream ss;
  for(LL i=1;i<=x;++i)
    ss << i;
  string s; ss >> s;
  LL res = 0;
  REP(i,SZ(s)-1)
    if(s[i] == '1' && s[i+1] == '2')
      ++res;
  return res;
}
int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  LL A, B; cin >> A >> B;
  LL ans = f(B) - f(A-1);
  LL x = A; while(x >= 10) x /= 10;
  if(x == 2 && A%10 == 2) --ans;
  cout << ans << endl;
  return 0;
}
 
 
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