題解

高斯消元+期望
首先好像有那麽一句話叫概率正著推期望倒著推==
讓你計算期望
那麽考慮倒著推
設\(f_S\)表示剩余血量為\(S\)時距離結束的期望步數
然後顯然\(f_0=0\)
考慮每一輪,先有\(1\)次回血,再有\(k\)段攻擊
那麽可以計算出每次扣\(t\)滴血的期望是\(atk[t]=C(_{t}^{k})(\frac{1}{m+1})^t(\frac{m}{m+1})^{k-t}\)
然後先預處理出來\(atk[]\)
轉移是成環的,只能高斯消元
因為是倒著推得
所以顯然枚舉所有的後繼即可:\(f_i=1+\sum_{j=1}^{i}{f[j]*atk[i-j]*\frac{m}{m+1} + f[j]*atk[i-j+1]*\frac{1}{m+1}}\)

代碼

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
const int M = 1505 ;
const int mod = 1e9 + 7 ;
using namespace std ;

inline int read() {
    char c = getchar() ; int x = 0 , w = 1 ;
    while(c>'9' || c<'0') { if(c=='-') w = -1 ; c = getchar() ; }
    while(c>='0' && c<='9') { x = x*10+c-'0' ; c = getchar() ; }
    return x*w ;
}

int n , stp , m , k ;
int fac[M] , hit[M] , miss[M] , atk[M] ; 
int B[M][M] , wei[M] ;
inline int Fpw(int Base , int k) {
    int temp = 1 ;
    while(k) {
        if(k & 1) temp = (1LL * temp * Base) % mod ; 
        Base = (1LL * Base * Base) % mod ; k >>= 1 ;
    }
    return temp ;
}
inline int Inv(int x) { return Fpw(x , mod - 2) ; }

inline int Gfac(int l , int r) {
    int temp = 1 ;
    for(int i = l ; i <= r ; i ++) temp = 1LL * temp * i % mod ;
    return temp ;
}
inline void Pre_Solve() {
    for(int i = 1 ; i <= n ; i ++) 
        for(int j = 1 ; j <= n + 1 ; j ++) 
            B[i][j] = 0 ;
    fac[0] = 1 ; 
    for(int i = 1 ; i <= n ; i ++)
        fac[i] = 1LL * fac[i - 1] * i % mod ;
    hit[0] = miss[0] = 1 ;
    hit[1] = Inv(m + 1) ; miss[1] = 1LL * m * Inv(m + 1) % mod ;
    for(int i = 0 ; i <= min(n , k) ; i ++)
        atk[i] = 1LL * Gfac(k - i + 1 , k) * Inv(fac[i]) % mod * Fpw(hit[1] , i) % mod * Fpw(miss[1] , k - i) % mod ;
}
inline bool gauss() {
    for(int i = 1 ; i <= n ; i ++) {
        wei[i] = min(i + 1 , n) ;
        for(int j = 1 ; j <= n + 1 ; j ++)
            B[i][j] = (B[i][j] % mod + mod) % mod ;
    }
    for(int i = 1 ; i <= n ; i ++) {
        if(B[i][i] == 0) return false ;
        int tinv = Inv(B[i][i]) ;
        for(int j = i + 1 ; j <= n ; j ++) {
            if(B[j][i] == 0) continue ;
            int tp = 1LL * B[j][i] * tinv % mod ;
            B[j][i] = ((B[j][i] - 1LL * tp * B[i][i] % mod) % mod + mod) % mod ;
            if(i + 1 <= n) B[j][i + 1] = ((B[j][i + 1] - 1LL * tp * B[i][i + 1] % mod) % mod + mod) % mod ;
            B[j][n + 1] = ((B[j][n + 1] - 1LL * tp * B[i][n + 1] % mod) % mod + mod) % mod ;
        }
    }
    for(int i = n ; i >= 1 ; i --) {
        for(int j = i + 1 ; j <= n ; j ++)
            B[i][n + 1] = ((B[i][n + 1] - 1LL * B[i][j] * B[j][n + 1] % mod) % mod + mod) % mod ;
        B[i][n + 1] = (1LL * B[i][n + 1] * Inv(B[i][i]) % mod + mod) % mod ;
        if(i == stp) break ;
    }
    return true ;
}

int main() {
    int T = read() ;
    while(T --) {
        n = read() ; stp = read() ; m = read() ; k = read() ; 
        if(k == 0) { printf("-1\n") ; continue ; }
        else if(k == 1 && m == 0) { printf("-1\n") ; continue ; }
        Pre_Solve() ;
        for(int i = 1 ; i <= n ; i ++) {
            B[i][i] = (B[i][i] + 1) % mod ; B[i][n + 1] = (B[i][n + 1] + 1) % mod ;
            if(i < n) for(int j = 0 ; j <= i + 1 ; j ++) {
                if(i - j <= k) B[i][j] = ((B[i][j] - 1LL * atk[i - j] * miss[1] % mod) % mod + mod) % mod ;
                if(i - j + 1 <= k) B[i][j] = ((B[i][j] - 1LL * atk[i - j + 1] * hit[1] % mod) % mod + mod) % mod ;
            }
            else for(int j = 0 ; j <= i ; j ++) {
                if(i - j <= k) B[i][j] = ((B[i][j] - 1LL * atk[i - j] % mod) % mod + mod) % mod ;
            }
        }
        if(!gauss()) printf("-1\n") ;
        else printf("%d\n",(B[stp][n + 1] % mod + mod) % mod) ;
    }
    return 0 ;
}
 

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