Sha 11637-garbage remembering exam (probability)

Link to the question: Ultraviolet A 11637-garbage remembering exam

The answer to the question is very detailed in big white.

Solution: For an ordered sequence, consider the probability that each location is valid. C (2? Kn? 1? X )? A (2k2k )? A (n? 1? XN? 1? X) a (n? 1n? 1)
X is the number of positions at which the distance from the current location is less than or equal to K. If this position is valid, the corresponding 2 * k adjacent words should be placed in another position.

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int maxn = 1e5;int N, K;long double frc[maxn+5];double solve () {    if (N == 1)        return 0;    if (N - 2 * K - 1 <= 0)        return N;    double ret = 0;    for (int i = 1; i <= N; i++) {        int x = min(K, i - 1)  + min(K, N - i);        if (N >= x + 2 * K + 1)            ret += exp(frc[N-1-2*K] + frc[N-1-x] - frc[N-1] - frc[N - 1 - x - 2 * K]);    }    return N - ret;}int main () {    int cas = 1;    frc[0] = 0;    for (int i = 1; i <= maxn; i++)        frc[i] = frc[i-1] + log((long double)i);    while (scanf("%d%d", &N, &K) == 2 && N + K) {        printf("Case %d: %.4lf\n", cas++, solve());    }    return 0;}