【LOJ】#2513. 「BJOI2018」治疗之雨

题解

具体就是列一个未知数方程\(dp[i]\)表示有\(i\)滴血的时候期望多少轮

\(dp[i] = 1 + \sum_{j = 1}^{i + 1} a_{i,j}dp[j]\)

\(dp[n] = 1 + \sum_{j = 1}^{n} a_{i,j}dp[j]\)

\(a_{i,j}\)表示从\(i\)滴血到\(j\)滴血的概率

可以高斯消元？

但是发现这个似乎和递推形式只差一点点

\(a_{i,i + 1} dp[i + 1] = -1 - \sum_{j = 1}^{i - 1} a_{i,j}dp[j] + (1 - a_{i,i})dp[i]\)

但是我们不知道\(dp[1]\)

我们可以把\(dp[1]\)设成\(X\)，然后用前\(n - 1\)个式子推出来\(dp[n] = A_1x + B_1\)

用第\(n\)个式子再推出来\(dp[n] = A_2x + B_2\)就可以解出来\(x\)了

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 20005
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int C[1505],N,M,P,K;
pii dp[1505];
int g[1505],f[1505],d[1505],inv[1505],ad[1505];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
const pii operator + (const pii &a,const pii &b) {
    return mp(inc(a.fi,b.fi),inc(a.se,b.se));
}
const pii operator * (const pii &a,const int &d) {
    return mp(mul(a.fi,d),mul(a.se,d));
}
void update(pii &a,pii b) {
    a = a + b;
}
void Solve() {
    read(N);read(P);read(M);read(K);
    if(K == 0) {puts("-1");return;}
    if(M == 0) {
	if(K == 1) {
	    puts("-1");
	}
	else {
	    int cnt = 0;
	    if(P == N) {P = max(0,P - K);++cnt;}
	    if(P) cnt += (P - 1) / (K - 1) + 1;
	    out(cnt);enter;
	}
	return;
    }

    int T = min(N,K);
    int InvM = fpow(M,MOD - 2),InvM1 = fpow(M + 1,MOD - 2);
    C[0] = 1;C[1] = K;
    g[0] = 1;f[0] = fpow(mul(InvM1,M),K);int t = mul(InvM,M + 1);
    g[1] = InvM1,f[1] = mul(f[0],t);
    for(int i = 2 ; i <= T ; ++i) {
	C[i] = mul(C[i - 1],mul(inv[i],inc(K,MOD - i + 1)));
	g[i] = mul(g[i - 1],g[1]);
	f[i] = mul(f[i - 1],t);
    }
    for(int i = 0 ; i <= T ; ++i) {
	t = mul(C[i],mul(g[i],f[i]));
	d[i] = mul(t,mul(M,InvM1));
	ad[i] = mul(t,InvM1);
    }
    for(int i = T + 1 ; i <= N ; ++i) d[i] = ad[i] = 0;
    dp[1] = mp(1,0);
    for(int i = 2 ; i <= N ; ++i) {
	dp[i] = mp(0,MOD - 1);
	update(dp[i],dp[i - 1] * inc(1,MOD - inc(d[0],ad[1])));
	for(int j = 1 ; j < i - 1; ++j) {
	    update(dp[i],dp[j] * (MOD - inc(d[i - 1 - j],ad[i - j])));
	}
	dp[i] = dp[i] * fpow(ad[0],MOD - 2);
    }
    pii another = mp(0,1);
    for(int i = 1 ; i < N ; ++i) {
	if(N - i <= T) {
	    t = mul(C[N - i],mul(g[N - i],f[N - i]));
	    update(another,dp[i] * t);
	}
    }
    another = another * fpow(inc(1,MOD - f[0]),MOD - 2);
    if(another.fi == dp[N].fi) {puts("-1");return;}
    int x = mul(inc(dp[N].se,MOD - another.se),fpow(inc(another.fi,MOD - dp[N].fi),MOD - 2));
    out(inc(mul(dp[P].fi,x),dp[P].se));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    inv[1] = 1;
    for(int i = 2 ; i <= 1500 ; ++i) {
	inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    }
    int T;read(T);
    while(T--) Solve();
    return 0;
}