제출 #110848

#	제출 시각UTC-0	아이디	문제	언어	결과	실행 시간	메모리
110848	2019-05-12 13:14:22	mosesmayer	Kangaroo (CEOI16_kangaroo)	C++17	100 / 100	82 ms	31992 KiB

kangaroo

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#include <iostream>
#include <iomanip>
#include <cstring>
using namespace std;
typedef long long LL;
const LL mod = 1e9 + 7;
int n, s, t;
LL dp[2005][2005];
LL f(int pos, int cc){
  if (pos > n) return cc == 1; // base case, reached if cc = 1 when we reach end position
  if (pos > 1 && cc < 1) return 0; // once pos after 1, impossible as we have at least 1 contiguous component
  LL &ret = dp[pos][cc];
  if (ret != -1) return ret;
  ret = 0;
  //special case --> endpoints
  if (!(pos ^ s) || !(pos ^ t)){ // UwU ಠ_ಠ
    (ret += f(pos + 1, cc)) %= mod; // already ada.
    (ret += f(pos + 1, cc + 1)) %= mod; // desired endpoint not used
  } else {
    int ends = 2 - (pos > s) - (pos > t);
    //new component --> two adjacent will be smaller
    (ret += f(pos + 1, cc + 1) * (cc - 1 + ends)) %= mod; // create new component, ends == extra empty spaces due to ends
    //merge component --> larger than two adjacent
    (ret += f(pos + 1, cc - 1) * (cc - 1)) %= mod;
  }
  return ret;
}
 
 
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• Subtask 1: 6 / 6
• Subtask 2: 30 / 30
• Subtask 3: 15 / 15
• Subtask 4: 49 / 49