【SRM 717 DIV2 C】DerangementsDiv2

Problem Statement

You are given two ints: n and m.

Let D be the number of permutations of the set {1,2,…,n+m} such that the first m values are not fixed points of the permutation. Formally, we are interested in permutations p such that for each j between 1 and m, inclusive, we have p(j) != j.

Compute and return D modulo 1,000,000,007.

Definition

Class:

DerangementsDiv2

Method:

count

Parameters:

int, int

Returns:

int

Method signature:

int count(int n, int m)

(be sure your method is public)

Limits

Time limit (s):

2.000

Memory limit (MB):

512

Stack limit (MB):

512

Constraints

n will be between 0 and 50, inclusive.

m will be between 1 and 50, inclusive.

Examples

0)

0

2

Returns: 1

Here we are looking for permutations of {1, 2} such that p(1) != 1 and p(2) != 2. There is only one such permutation: the permutation (2, 1). In other words, the permutation p such that p(1) = 2 and p(2) = 1.

1)

2

1

Returns: 4

Here we are counting permutations of {1, 2, 3} such that p(1) != 1. There are four such permutations: (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). Here, (a, b, c) denotes a permutation p for which p(1) = a, p(2) = b, and p(3) = c.

2)

1

2

Returns: 3

This time we want permutations of {1, 2, 3} such that p(1) != 1 and p(2) != 2. The three such permutations are (2, 1, 3), (2, 3, 1), and (3, 1, 2).

3)

3

5

Returns: 21234

20

27

Returns: 88437461

Watch out for integer overflow.

【题目链接】:

【题意】

给你两个整数n和m;

然后让你求1..n+m的一些满足以下要求的排列p的个数:

要求i从1..m满足p[i]!=i;

【题解】

容斥原理搞;

设ci表示1..m中有i个位置满足pi==i的方案数;

ci=C(m,i)*(n+m-i)!

则答案就为(n+m)!-c1∪c2∪c3…..∪cm

减号右边那个东西,用容斥原理搞

为了不重复计数;

先加上每一个位置都不同的方案,然后减去有两个位置不同的方案,然后加上有3个位置不同的方案,然后减去有4个位置不同的方案…

【Number Of WA】

0

【反思】

取模过程中会出现负数的话,要注意加上MOD数;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 50+5;
const LL MOD = (int) 1e9 + 7;
//head
LL c[N][N],fac[N+N];
class DerangementsDiv2
{
public:
    int count(int n, int m)
    {
        rep1(i, 1, 50)
            c[i][i] = c[i][0] = 1;
        rep1(i, 1, 50)
            rep1(j, 1, i - 1)
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
        fac[0] = 1;
        rep1(i, 1, 100)
            fac[i] = (fac[i - 1] * i) % MOD;
        LL ans = fac[n + m],temp = 0,p = 1;
        rep1(i, 1, m) {
            temp += (p*c[m][i]%MOD + MOD) % MOD*fac[n + m - i] % MOD;
            p = -p;
        }
        ans = ((ans - temp)%MOD + MOD) % MOD;
        return (int) ans;
    }
};