uva 11637 - Garbage Remembering Exam(概率)

题目链接：uva 11637 - Garbage Remembering Exam

题目大意：大白数里有很详细的题意。

解题思路：对于有序的序列来说，考虑每个位置为有效性的概率。C(2∗kn−1−x)∗A(2k2k)∗A(n−1−2kn−1−2k)A(n−1n−1)
x为考虑当前位置，然后与该位置距离小于等于k的位置个数。该位置有效的话，则对应的要将原先邻近的2*k个单词放到另外的位置上。

#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;}