이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*input
10 2 3
*/
#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const ll modulo = 1000000007;
int sgn(ll x)
{
if (x < 0)
return -1;
if (x > 0)
return 1;
return 0;
}
map<tuple<int, int, int, int>, ll>M;
ll f(int n, int a, int b, int s)
{
if (a == b)
return 0;
if (M.count(make_tuple(n, a, b, s)))
return M[make_tuple(n, a, b, s)];
if (n == 2)
{
if (sgn(b - a) != s)
return 0;
else
return 1;
}
ll ret = 0;
for (int c = 1; c <= n; c++)
{
if (sgn(b - c) == s)
{
ret += f(n - 1, a - (a > b), c - (c > b), -s);
}
}
ret %= modulo;
return M[make_tuple(n, a, b, s)] = ret;
}
ll f(int n, int a, int b)
{
ll ans = f(n, a, b, 1) + f(n, a, b, -1);
ans %= modulo;
return ans;
}
ll fast(int N, int A, int B)
{
ll L[N + 1][N + 1][N + 1];
ll R[N + 1][N + 1][N + 1];
for (int n = 2; n <= N; n++)
{
for (int a = 1; a <= n; a++)
{
for (int b = 1; b <= n; b++)
{
L[n][a][b] = 0;
R[n][a][b] = 0;
if (a == b)
continue;
if (n == 2)
{
if (a > b)
{
L[n][a][b] = 1;
}
if (a < b)
{
R[n][a][b] = 1;
}
}
else
{
for (int c = 1; c <= n; c++)
{
if (c > b)
{
L[n][a][b] += R[n - 1][a - (a > b)][c - (c > b)];
}
if (c < b)
{
R[n][a][b] += L[n - 1][a - (a > b)][c - (c > b)];
}
}
}
L[n][a][b] %= modulo;
R[n][a][b] %= modulo;
}
}
}
ll ret = L[N][A][B] + R[N][A][B];
ret %= modulo;
return ret;
}
int main()
{
int n, a, b;
cin >> n >> a >> b;
//cout << f(n, a, b) << "\n";
cout << fast(n, a, b) << "\n";
}
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