LightOJ - 1095 - Arrange the Numbers（错排）

时间：2024-11-20 16:05:02

链接：

https://vjudge.net/problem/LightOJ-1095

题意：

Consider this sequence {1, 2, 3 ... N}, as an initial sequence of first N natural numbers. You can rearrange this sequence in many ways. There will be a total of N! arrangements. You have to calculate the number of arrangement of first N natural numbers, where in first M positions; exactly K numbers are in their initial position.

For Example, N = 5, M = 3, K = 2

You should count this arrangement {1, 4, 3, 2, 5}, here in first 3 positions 1 is in 1st position and 3 in 3rd position. So exactly 2 of its first 3 are in there initial position.

But you should not count {1, 2, 3, 4, 5}.

思路：

错排：

F[n] = (n-1)*(F[n-1]+F[n-2]),F[i]为长度i的错排种类。

递推：

第一步，首元素插到剩下n-1个元素位置k。

第二步，可以选择将k插到首位置，此时就剩下n-2个，即F[n-2]

可以不插到首位置，将原本k位置元素删除，k插到首元素位置，看成k位置，则为F[n-1]

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<utility>

using namespace std;

typedef long long LL;

const int INF = 1e9;

const int MAXN = 1e3+10;

const int MOD = 1e9+7;

int n, k, m;

LL P[MAXN], F[MAXN];

LL PowMod(LL a, LL b)

{

    LL res = 1;

    while(b)

    {

        if (b&1)

            res = (1LL*res*a)%MOD;

        a = (1LL*a*a)%MOD;

        b >>= 1;

    }

    return res;

}

LL Com(LL a, LL b)

{

    if (b == 0 || a == b)

        return 1;

    return 1LL*P[a]*PowMod(P[b]*P[a-b]%MOD, MOD-2)%MOD;

}

void Init()

{

    P[1] = 1;

    for (int i = 2;i < MAXN;i++)

        P[i] = 1LL*i*P[i-1]%MOD;

    F[1] = 0, F[0] = F[2] = 1;

    for (int i = 3;i < MAXN;i++)

        F[i] = 1LL*(i-1)*((F[i-1]+F[i-2])%MOD)%MOD;

}

int main()

{

    Init();

    int cnt = 0;

    int t;

    scanf("%d", &t);

    while(t--)

    {

        printf("Case %d:", ++cnt);

        scanf("%d%d%d", &n, &m, &k);

        LL res = 0LL;

        for (int i = 0;i <= n-m;i++)

            res = ((res + 1LL*Com(n-m, i)*F[n-k-i]%MOD)%MOD+MOD)%MOD;

        res = 1LL*res*Com(m, k)%MOD;

        printf(" %lld\n", res);

    }

    return 0;

}

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