| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 332073 | 2020-12-01T10:58:55 Z | alextodoran | Fishing Game (RMI19_fishing) | C++17 | 2000 ms | 99188 KB |
/**
____ ____ ____ ____ ____
||a |||t |||o |||d |||o ||
||__|||__|||__|||__|||__||
|/__\|/__\|/__\|/__\|/__\|
**/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N_MAX = 102;
const int MOD = 1e9+7;
int n, t;
int dp[3 * N_MAX][3 * N_MAX][3 * N_MAX];
bool a[3 * N_MAX], b[3 * N_MAX], c[3 * N_MAX];
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> t;
dp[0][0][0] = 1;
for(int s = 1; s <= 6 * n; s++)
for(int i = 0; i <= 3 * n && i <= s; i++)
for(int j = 0; j <= 3 * n && i + j <= s; j++)
{
int k = s - i - j;
if(k > 3 * n)
continue;
ll sum = 0;
int x1 = 1, y1 = 1, z1 = 1;
if(i + k > 0)
x1 = 2;
if(j + i - 1 > 0)
y1 = 2;
if(k + j - 2 > 0)
z1 = 2;
for(int x = x1; x <= 3; x++)
for(int y = y1; y <= 3; y++)
for(int z = z1; z <= 3; z++)
{
int i1 = i, j1 = j, k1 = k;
bool ok = false;
int prod = 1;
if(x == 1)
{
if(i1 + k1 > 0)
continue;
}
else if(x == 2)
{
if(i1 == 0)
continue;
prod *= i1;
i1--;
ok = true;
}
else
{
if(k1 == 0)
continue;
prod *= k1;
j1++;
k1--;
}
if(y == 1)
{
if(j1 + i1 > 0)
continue;
}
else if(y == 2)
{
if(j1 == 0)
continue;
prod *= j1;
j1--;
ok = true;
}
else
{
if(i1 == 0)
continue;
prod *= i1;
k1++;
i1--;
}
if(z == 1)
{
if(k1 + j1 > 0)
continue;
}
else if(z == 2)
{
if(k1 == 0)
continue;
prod *= k1;
k1--;
ok = true;
}
else
{
if(j1 == 0)
continue;
prod *= j1;
i1++;
j1--;
}
if(ok == false)
continue;
sum += 1LL * dp[i1][j1][k1] * prod;
}
dp[i][j][k] = sum % MOD;
}
while(t--)
{
for(int i = 1; i <= n * 3; i++)
a[i] = b[i] = c[i] = false;
for(int i = 1; i <= n * 2; i++)
{
int card;
cin >> card;
a[card] = true;
}
for(int i = 1; i <= n * 2; i++)
{
int card;
cin >> card;
b[card] = true;
}
for(int i = 1; i <= n * 2; i++)
{
int card;
cin >> card;
c[card] = true;
}
int I = 0, J = 0, K = 0;
for(int i = 1; i <= n * 3; i++)
I += (a[i] && b[i]);
for(int i = 1; i <= n * 3; i++)
J += (b[i] && c[i]);
for(int i = 1; i <= n * 3; i++)
K += (c[i] && a[i]);
cout << dp[I][J][K] << "\n";
}
return 0;
}
| # | 결과 | 실행 시간 | 메모리 | Grader output |
|---|---|---|---|---|
| 1 | Correct | 1 ms | 364 KB | Output is correct |
| 2 | Correct | 1 ms | 492 KB | Output is correct |
| 3 | Correct | 3 ms | 1644 KB | Output is correct |
| 4 | Correct | 12 ms | 4972 KB | Output is correct |
| 5 | Correct | 227 ms | 28268 KB | Output is correct |
| 6 | Correct | 605 ms | 40272 KB | Output is correct |
| 7 | Correct | 1125 ms | 54456 KB | Output is correct |
| 8 | Correct | 1525 ms | 70764 KB | Output is correct |
| 9 | Execution timed out | 2097 ms | 88940 KB | Time limit exceeded |
| 10 | Execution timed out | 2071 ms | 99188 KB | Time limit exceeded |