解法

概要

DP。とりあえず$O(Nk^2)$を書いて適当にすると$O(Nk)$に落ちる。

実装

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define ALL(x) begin(x), end(x)
using namespace std;

double solve(int n, int k, vector<int> const & t) {
double acc = 0;
vector<double> cur(k), prv;
cur[0] = 1;
REP (i, n) {
cur.swap(prv);
cur.assign(k, 0);

/*
// O(k^2)
REP (arrive, k) {
REP3 (depart, arrive, k) {
int wait = depart - arrive;
double p = prv[arrive] * (1.0 / k);
acc += p * wait;
cur[(depart + t[i]) % k] += p;
}
REP (depart, k) {
int wait = k + depart - arrive;
double p = prv[arrive] * ((double)arrive / k) * (1.0 / k);
acc += p * wait;
cur[(depart + t[i]) % k] += p;
}
}
*/

// O(k)
REP (arrive, k) {  // arrive <= depart
double p = prv[arrive] * (1.0 / k);
double sum_wait = (k - arrive) * ((arrive + k - 1) / 2.0 - arrive);
acc += p * sum_wait;
if (arrive + t[i] % k < k) {
cur[0] += p;
cur[arrive + t[i] % k] += p;
} else {
cur[(arrive + t[i]) % k] += p;
}
cur[t[i] % k] -= p;
}
REP (j, k - 1) {
cur[j + 1] += cur[j];  // imos's method
}
double prob = 0;
REP (arrive, k) {  // depart < arrive
double average_wait = k + ((k - 1) / 2.0) - arrive;
double p = prv[arrive] * ((double)arrive / k) * (1.0 / k);
acc += p * average_wait * k;
prob += p;
}
REP (depart, k) {
cur[depart] += prob;
}

}
return accumulate(ALL(t), 0ll) + acc;
}

int main() {
int n, k; cin >> n >> k;
vector<int> t(n);
REP (i, n) cin >> t[i];
cout << setprecision(16) << solve(n, k, t) << endl;
return 0;
}