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?xn?1?x)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;
}

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