QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#450862 | #6821. Another A+B Problem | PetroTarnavskyi# | RE | 0ms | 0kb | C++20 | 1.6kb | 2024-06-22 18:52:13 | 2024-06-22 18:52:14 |
answer
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) int(a.size())
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second
typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;
const int N = 301;
db dpT[N][N][N];
int n;
db dp(int path, int edge, int isol)
{
if(path < 0 || edge < 0 || isol < 0)
return 0;
if(path == 0 && isol <= 1)
return 0;
if(dpT[path][edge][isol] > -0.5)
return dpT[path][edge][isol];
int sum = 0;
db res = n * n;
res += isol * (isol - 1) * dp(path, edge + 1, isol - 2);
sum += isol * (isol - 1);
res += 4 * isol * edge * dp(path + 1, edge - 1, isol - 1);
sum += 4 * isol * edge;
res += 4 * isol * path * dp(path, edge, isol - 1);
sum += 4 * isol * path;
res += 4 * edge * (edge - 1) * dp(path + 1, edge - 2, isol);
sum += 4 * edge * (edge - 1);
res += 4 * path * edge * dp(path, edge - 1, isol);
sum += 4 * path * edge;
res += 4 * path * (path - 1) * dp(path - 1, edge, isol);
sum += 4 * path * (path - 1);
res += 2 * path * dp(path - 1, edge, isol);
sum += 2 * path;
assert(0 < sum && sum <= n * n - 2 * n);
return dpT[path][edge][isol] = res / sum;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
FOR(i, 0, N)
FOR(j, 0, N)
FOR(k, 0, N)
dpT[i][j][k] = -1;
cin >> n;
cout << fixed << setprecision(9) << dp(0, 0, n) << "\n";
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 0
Runtime Error
input:
40+11=51 PBGPPGGB