Codeforces 380D Sereja and Cinema （看题解)

Sereja and Cinema

首先我们可以发现除了第一个人， 其他人都会坐在已入坐人的旁边。

难点在于计算方案数。。 我们可以从外往里把确定的人用组合数算上去，然后缩小范围。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#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 ull unsigned long long

using namespace std;

const int N = 1e5 + 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 power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a % mod;
        a = 1LL * a * a % mod; b >>= 1;
    }
    return ans;
}

int inv[N], Finv[N], F[N];
void init() {
    inv[1] = F[0] = Finv[0] = 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;
}
int comb(int n, int m) {
    if(n < m || n < 0) return 0;
    return 1LL * F[n] * Finv[m] % mod * Finv[n - m] % mod;
}

int n, a[N], prefix[N];

int solve(int L, int R) {
    if(prefix[L - 1] == prefix[R]) return power(2, R - L);
    int p, q;
    for(p = L; p <= R; p++) if(a[p] != 0) break;
    for(q = R; q >= L; q--) if(a[q] != 0) break;
    if(p == q && a[p] == 1) return comb(R - L, p - L);
    int ans = 0;
    if(a[p] >= a[q]) {
        int L2 = p, R2 = L2 + a[p] - 1;
        if(R2 >= q && R2 <= R) add(ans, 1LL * solve(L2 + 1, R2) * comb(R - L - R2 + L2, L2 - L) % mod);
    }
    if(a[q] >= a[p]) {
        int R2 = q, L2 = R2 - a[q] + 1;
        if(L2 >= L && L2 <= p) add(ans, 1LL * solve(L2, R2 - 1) * comb(R - L - R2 + L2, L2 - L) % mod);
    }
    return ans;
}

int main() {
    init();
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(int i = 1; i <= n; i++) prefix[i] = prefix[i - 1] + (a[i] != 0);
    printf("%d
", solve(1, n));
    return 0;
}

/*
*/