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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#225175 | #2913. Archery Accuracy | PetroTarnavskyi | WA | 187ms | 5060kb | C++17 | 2.3kb | 2023-10-24 02:46:34 | 2023-10-24 02:46:34 |
Judging History
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;
vector<db> Gauss(vector<vector<db>> a)
{
int n = SZ(a);
if(n == 0)
return {};
int m = SZ(a[0]) - 1;//number of variables
assert(n >= m);
int vars = m;
FOR(i, 0, m)
{
if (a[i][i] == 0)
{
int row = i;
FOR(k, i + 1, n)
{
if (abs(a[row][i]) < abs(a[k][i]))
row = k;
}
if(row == -1)
{
vars--;
continue;
}
swap(a[i], a[row]);
}
db d = 1 / a[i][i];
FOR(k, i + 1, n)
{
db c = a[k][i] * d;
FOR(j, 0, m + 1)
a[k][j] -= c * a[i][j];
}
}
FOR(i, vars, n)
if(a[i].back() != 0)
cout << "No solution\n";
vector<db> x(m);
RFOR(i, m, 0)
{
x[i] = a[i].back();
FOR(j, i + 1, m)
x[i] -= a[i][j] * x[j];
x[i] = x[i] / a[i][i];
}
return x;
}
const int N = 17;
db dp[1 << N];
db r1[N + 1][N + 1];
db r2[N + 1][N + 1];
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout << fixed << setprecision(15);
int n = 17;
//cin >> n;
vector<db> p(n);
FOR (i, 0, n)
p[i] = 0.47;
//cin >> p[i];
VI s(n + 1, 0);
FOR (i, 0, n)
s[i + 1] = 100 - n + i + 1;
FOR (i, 0, n)
{
FOR (j, 0, n)
{
int sz = 2 * s[i + 1];
vector<vector<db>> a(sz, vector<db>(sz + 1));
FOR (k, 0, sz - 1)
{
a[k][k] = -1;
a[k][k + 1] = p[j];
if (k)
{
a[k][k - 1] = 1 - p[j];
}
a[k][sz] = 0;
}
a[sz - 1][sz - 1] = 1;
a[sz - 1][sz] = 1;
auto res = Gauss(a);
int x = (-s[i + 1]) + 1;
r1[i][j] = res[s[i] - x];
r2[i][j] = res[-s[i] - x];
}
}
dp[0] = 1;
FOR (i, 0, 1 << n)
{
int idx = __builtin_popcount(i);
FOR (j, 0, n)
{
if (i & (1 << j))
continue;
db P = r1[idx][j] * dp[i] + r2[idx][j] * (1 - dp[i]);
if (P > dp[i | (1 << j)])
{
dp[i | (1 << j)] = P;
}
}
}
cout << dp[(1 << n) - 1] << '\n';
cerr << db(clock()) / CLOCKS_PER_SEC << '\n';
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 187ms
memory: 5060kb
input:
17 0.49 0.45 0.50 0.47 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.50 0.49 0.50 0.49 0.50 0.49 15 16 17 18 19 20 26 29 86 88 89 93 95 96 97 98 100
output:
0.000006056144143
result:
wrong answer 1st numbers differ - expected: '0.4516422', found: '0.0000061', error = '0.4516362'