luoguP3750 [六省联考2017]分手是祝愿 概率期望DP + 贪心

...........真的神状态了，没办法去想的状态...................

考试的时候选择$50$分贪心+$15$分状压吧，别的点就放弃算了........

令$f[i]$表示从最小步数为$i$时走到最小步数为$i - 1$的状态的期望步数

（所以题目中的$k$实际上是个提示...........................）

那么当$i > k$时，有$f[i] = frac{i}{n} + frac{n - i}{n} * (1 + f[i] + f[i + 1])$

移项后转移就是递推式了

当$i leqslant k$时，有$f[i] = f[i + 1] + 1$

怎么求解初始状态的最小步数呢？

可以发现，我们一定是从$n$慢慢点到$1$最优

那么，$1$个点会不会被点就跟它的倍数有多少个$1$有关

倒叙枚举$i$，再枚举$i$的倍数看看就好了.....

复杂度$O(n log n)$

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

extern inline char gc() {
    static char RR[23456], *S = RR + 23333, *T = RR + 23333;
    if(S == T) fread(RR, 1, 23333, stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
    return p * w;
}

#define ri register int
#define sid 200500

const int mod = 100003;
int n, k, nj = 1, mis, ans;
int inv[sid], f[sid], v[sid];

int main() {
    n = read(); k = read();
    for(ri i = 1; i <= n; i ++) v[i] = read();

    for(ri i = n; i >= 1; i --)
    for(ri j = i + i; j <= n; j += i) v[i] ^= v[j];
    for(ri i = 1; i <= n; i ++) mis += v[i];

    inv[1] = 1;
    for(ri i = 2; i <= n; i ++)
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    for(ri i = 1; i <= n; i ++) nj = 1ll * nj * i % mod;

    for(ri i = n; i > k; i --)
    f[i] = (n + 1ll * (n - i) * f[i + 1] % mod) * inv[i] % mod;
    for(ri i = k; i; i --) f[i] = 1;
    
    for(ri i = 1; i <= mis; i ++) (ans += f[i]) %= mod;
    printf("%d
", 1ll * ans * nj % mod); 
    return 0;
}