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  • 【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)

    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;
    }
    
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/10006053.html
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