UVa12024 - Hats(错排问题)

最新推荐文章于 2022-08-04 15:17:36 发布

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最新推荐文章于 2022-08-04 15:17:36 发布

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分类专栏：算法设计与分析 OJ

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本文讨论了John Hatman解决皇家伦敦剧院观众拿错帽子问题的方法，通过错排递推关系式和排列的递推式计算错排个数与排列总数的比例，并提供了具体的编程实现。

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Background

John Hatman, the honest cloakroom attendant of the RoyalTheatre of London, would like to know the solution to the followingproblem.

Whenthe show finishes, all spectators in the theatre are in a hurry to see the Final of the UEFAChampionship. So, they run to the cloakroom to take their hats back.

Some of them take a wrong hat. But, how likely is thateveryone take a wrong hat?

The first lineof the input contains an integer, t,indicating the number of test cases. For each test case, one lineappears, that contains anumber n, 2<=n<=12,representing the number of people and hats.

TheOutput

For each test case, the output should contain asingle line with the number representing the number of favourable cases(i.e., the number of cases where all people take a wrong hat),followed by a bar, "/", and followed by a number representing thetotal number of possible cases.

SampleInput

SampleOutput

1/2
2/6
9/24

题意：求错排个数与排列的个数之比

思路：错排递推关系式为f(n) = (n-1)(f(n - 2) + f(n - 1)),排列的递推式为d(n) = n * d(n -1)

#include <cstdio>

using namespace std;

const int MAXN = 13;

typedef long long LL;

LL f[MAXN], fact[MAXN];
int n;

void init()
{
    f[0] = 0; f[1] = 0; f[2] = 1;
    fact[1] = 1; fact[2] = 2;
    
    for (int i = 3; i < MAXN; i++) {
        f[i] = (i - 1) * (f[i - 2] + f[i - 1]);
        fact[i] = i * fact[i - 1];
    }
}

void input()
{
    scanf("%d", &n);
}

void solve()
{
    printf("%d/%d\n", f[n], fact[n]);
}

int main(int argc, char **argv) 
{
#ifndef ONLINE_JUDGE
    freopen("d:\\OJ\\uva_in.txt", "r", stdin);
#endif
    
    init();
    
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        input();
        solve();
    }
    
    return 0;
}