Codeforces 285E Positions in Permutations dp + 容斥原理

先dp出选 k 个的方案数，这个很简单, dp[ i ][ j ][ u ][ v ]表示到 i 为止选了 j 个， i - 1的选取情况是 u， i 的选取情况是 v 的方案数。

然后最后容斥一下，容斥系数是 C(i, k)。(虽然我不会证明，我感觉就是这个，逃

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, k;
int dp[N][N][2][2];
int way[N];
int F[N], Finv[N], inv[N];
int C(int n, int m) {
    if(n < m || n < 0) return 0;
    return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;
}
int main() {
    F[0] = Finv[0] = inv[1] = 1;
    for(int i = 2; i < N; i++) inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
    for(int i = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % mod;
    for(int i = 1; i < N; i++) Finv[i] = 1LL * Finv[i - 1] * inv[i] % mod;

    scanf("%d%d", &n, &k);
    dp[0][0][1][0] = 1;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j <= i; j++) {
            for(int u = 0; u < 2; u++) {
                for(int v = 0; v < 2; v++) {
                    if(!dp[i][j][u][v]) continue;
                    add(dp[i + 1][j][v][0], dp[i][j][u][v]);
                    if(!u) {
                        add(dp[i + 1][j + 1][v][0], dp[i][j][u][v]);
                        if(i + 1 < n) add(dp[i + 1][j + 1][v][1], dp[i][j][u][v]);
                    } else {
                        if(i + 1 < n) add(dp[i + 1][j + 1][v][1], dp[i][j][u][v]);
                    }
                }
            }
        }
    }
    for(int j = 0; j <= n; j++)
        for(int u = 0; u < 2; u++)
            for(int v = 0; v < 2; v++)
                add(way[j], dp[n][j][u][v]);
    int ans = 0;
    for(int i = k, op = 1; i <= n; i++, op = -op) {
        ans = (ans + 1LL * op * C(i, k) * way[i] % mod * F[n - i] % mod + mod) % mod;
    }
    printf("%d\n", ans);
    return 0;
}

/*
*/

Codeforces 285E Positions in Permutations dp + 容斥原理

猜你喜欢