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('
')
#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;
}