【BZOJ4362】isn

【BZOJ4362】isn

题面

bzoj

题解

设(f[i][j])表示当前在(i)，长度为(j)的最长不降子序列有多少个

这个可以用树状数组(n^2logn)求出

设(g[i])为长度为(i)的不降子序列的和

则(g[i]=sum_{j=1}^nf[j][i])

最后的答案乍一看是((n-i)!sum_{i=1}^ng[i])

但是因为我们取到非降就(break)

所以需要容斥一下

不难想到

[ans_i=(n-i)!sum_{i=1}^ng[i]-(n-i-1)!g[i+1]*(i+1)\ Ans=sum_{i=1}^nans_i ]

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
using namespace std; 
inline int gi() {
    register int data = 0, w = 1; 
    register char ch = 0; 
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
}
const int MAX_N = 2e3 + 5; 
const int Mod = 1e9 + 7;
inline void pls(int &x, int y) { x += y; if (x >= Mod) x -= Mod; } 
inline int dec(int x, int y) { x -= y; if (x < 0) x += Mod; return x; } 
int N, a[MAX_N], f[MAX_N][MAX_N], c[MAX_N][MAX_N], g[MAX_N], fac[MAX_N]; 
int X[MAX_N], cnt; 
inline int lb(int x) { return x & -x; } 
void add(int *bit, int x, int v) { while (x <= cnt) pls(bit[x], v), x += lb(x); } 
int sum(int *bit, int x) { int res = 0; while (x > 0) pls(res, bit[x]), x -= lb(x); return res; } 
 
int main () {
    N = gi(); for (int i = 1; i <= N; i++) X[++cnt] = a[i] = gi(); 
    sort(&X[1], &X[cnt + 1]); cnt = unique(&X[1], &X[cnt + 1]) - X - 1; 
    for (int i = 1; i <= N; i++) a[i] = lower_bound(&X[1], &X[cnt + 1], a[i]) - X;
    add(c[0], 1, 1); 
    for (int i = 1; i <= N; i++) { 
        for (int j = N; j >= 1; j--) { 
            f[i][j] = sum(c[j - 1], a[i]); 
            add(c[j], a[i], f[i][j]); 
        } 
    } 
    for (int i = 1; i <= N; i++) 
        for (int j = i; j <= N; j++) pls(g[i], f[j][i]); 
    fac[0] = 1;
    for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod; 
    int ans = 0; 
    for (int i = 1; i <= N; i++)
        pls(ans, dec(1ll * fac[N - i] * g[i] % Mod, 1ll * fac[N - i - 1] * g[i + 1] % Mod * (i + 1) % Mod));
    printf("%d
", ans); 
    return 0; 
}