UVa 11481 Arrange the Numbers (组合数学)

题意：给定 n，m，k，问你在 1 ~ n 的排列中，前 m 个恰好有 k 个不在自己位置的排列有多少个。

析：枚举 m+1 ~ n 中有多少个恰好在自己位置，这个是C(n-m, i)，然后前面选出 k 个，是C(m, k)，剩下 n - k - i 个是都不在自己位置，也就是错排 D[n-k-i]，求一个和就Ok了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define aLL 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-6;

const int maxn = 1000 + 10;

const int maxm = 76543;

const int mod = 1000000007;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int f[maxn], D[maxn], inv[maxn];

int fast_pow(int a, int n){

  int res = 1;

  while(n){

    if(n&1)  res = (LL)res * a % mod;

    n >>= 1;

    a = (LL)a * a % mod;

  }

  return res;

}

int main(){

  D[1] = 0;  D[0] = D[2] = 1;

  f[0] = f[1] = 1;  f[2] = 2;

  for(int i = 3; i <= 1000; ++i){

    f[i] = (LL)f[i-1] * i % mod;

    D[i] = (LL)(i-1) * (D[i-1] + D[i-2]) % mod;

  }

  inv[1000] = fast_pow(f[1000], mod-2);

  for(int i = 999; i >= 0; --i)

    inv[i] = (LL)(i+1) * inv[i+1] % mod;

  int T, k;  cin >> T;

  for(int kase = 1; kase <= T; ++kase){

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

    int ans = 0;

    int cmk = (LL)f[m] * inv[k] % mod * inv[m-k] % mod;

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

      ans = (ans + (LL)cmk * f[n-m] % mod * inv[i] % mod * inv[n-m-i] % mod * D[n-k-i] % mod) % mod;

    printf("Case %d: %d\n", kase, ans);

  }

  return 0;

}