【BZOJ4872】【六省联考2017】分手是祝愿（概率dp）

Description

click me

Solution

设fifi为从还剩ii个位置没有操作到剩i−1i−1个位置没有操作的期望操作次数。

那么有：fi=in+n−in(1+fi+fi+1)fi=in+n−in(1+fi+fi+1)

解方程得：fi=1+(n−i)(1+fi+1)ifi=1+(n−i)(1+fi+1)i

Source

/**********************************
* Au: Hany01
* Prob: BZOJ4872 & 六省联考2017 分手是祝愿
* Date: Feb 7th, 2018
* Email: hany01@foxmail.com
**********************************/

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>

using namespace std;

typedef long long LL;
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define rep(i, k) for (register int i = 0, i##_end_ = (k); i < i##_end_; ++ i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (100003)
#define debug(...) fprintf(stderr, __VA_ARGS__)

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
register int _ = 0, __ = 1; register char c_ = getchar();
for ( ; c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File()
{
#ifdef hany01
freopen("bzoj4872.in", "r", stdin);
freopen("bzoj4872.out", "w", stdout);
#endif
}

const int maxn = 100005;

int n, k, pend, stt[maxn];
LL Ans, inv[maxn], fac[maxn], ifac[maxn], f[maxn];

inline LL Pow(LL a, LL b) {
LL Ans = 1;
for ( ; b; a = a * a % Mod, b >>= 1) if (b & 1) (Ans *= a) %= Mod;
return Ans;
}

inline void Get_inv()
{
fac[0] = 1;
For(i, 1, n) fac[i] = fac[i - 1] * i % Mod;
ifac
= Pow(fac
, Mod - 2);
Fordown(i, n, 1) ifac[i - 1] = ifac[i] * i % Mod;
For(i, 1, n) inv[i] = fac[i - 1] * ifac[i] % Mod;
}

vector<int> d[maxn];
inline void Init()
{
n = read(), k = read();
For(i, 1, n) stt[i] = read();
For(i, 1, n) for (register int j = 2; j * i <= n; ++ j) d[j * i].pb(i);
Fordown(i, n, 1) if (stt[i]) {
rep(j, SZ(d[i])) stt[d[i][j]] ^= 1;
++ pend;
}
Get_inv();
}

inline void mod(LL &Ans) { if (Ans >= Mod) Ans -= Mod; }

inline LL Solve()
{
if (pend <= k) return pend;
f
= 1;
Fordown(i, n - 1, k + 1) f[i] = (1 + (n - i) * (1ll + f[i + 1]) % Mod * inv[i]) % Mod;
LL Ans = 0;
For(i, k + 1, pend) mod(Ans += f[i]);
mod(Ans += k);
return Ans;
}

int main()
{
File();
Init();
Ans = Solve();
For(i, 1, n) (Ans *= i) %= Mod;
printf("%lld\n", Ans);
return 0;
}