이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <list>
#include <utility>
#include <cmath>
#include <numeric>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pb push_back
#define endl "\n"
#define all(x) x.begin(), x.end()
const int M = 2007;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const double pi = acos(-1);
const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
ll n, st, en, dp[M][M][2][2];
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
cin >> n >> st >> en;
dp[0][0][0][0] = 1;
for(int i = 1; i <= n; ++i){
for(int j = 1; j <= n; ++j){
for(int on1 = 0; on1 < 2; ++on1){
for(int on2 = 0; on2 < 2; ++on2){
if(i == st){
if(on1) continue;
dp[i][j][1][on2] = (dp[i - 1][j][0][on2] + dp[i - 1][j - 1][0][on2]) % mod;
}
else if(i == en){
if(on2) continue;
dp[i][j][on1][1] = (dp[i - 1][j][on1][0] + dp[i - 1][j - 1][on1][0]) % mod;
}
else{
dp[i][j][on1][on2] = (dp[i - 1][j - 1][on1][on2] * (j - on1 - on2)) % mod;
dp[i][j][on1][on2] += (dp[i - 1][j + 1][on1][on2] * j) % mod;
}
}
}
}
}
cout << dp[n][1][1][1] << endl;
return 0;
}
// Don't forget special cases. (n = 1?)
// Look for the constraints. (Runtime array? overflow?)
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