题目链接

BZOJ5340

题解

我们能很容易维护每个人当前各种血量的概率
设\(p[u][i]\)表示\(u\)号人血量为\(i\)的概率
每次攻击的时候，讨论一下击中不击中即可转移
是\(O(Qm^2)\)的

现在考虑一下结界
如果我们设\(f[u][i]\)表示除了\(u\)还存活\(i\)个人的概率
那么
\[ans[u] = (1 - p[u][0]) \sum\limits_{i = 0}^{k - 1} \frac{f[u][i]}{i + 1}\]
所以我们只需计算\(f[u][i]\)
\(f[u][i]\)同样可以通过枚举剩余每个人存活与否进行转移，是\(O(n^3)\)的，复杂度过高
我们考虑计算\(g[i]\)表示剩余\(i\)人的概率
枚举\(u\)
\[g'[i] = g[i]p[u][0] + g[i - 1](1 - p[u][0])\]
即可\(O(n^2)\)计算\(g[i]\)
如果我们拿\(f[u][i]\)来计算\(g[i]\)的话
\[g[i] = f[u][i]p[u][0] + f[u][i - 1](1 - p[u][0])\]
那么
\[f[u][i] = \frac{g[i] - f[u][i - 1](1 - p[u][0])}{p[u][0]}\]
也可以\(O(n^2)\)递推

这样我们就做完了

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 205,maxm = 105,INF = 1000000000,P = 998244353;
inline int read(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}return out * flag;
}
LL p[maxn][maxm],f[maxn][maxn],g[maxn][maxn],n,m[maxn],id[maxn];
LL INV[maxn];
inline LL qpow(LL a,LL b){LL ans = 1;for (; b; b >>= 1,a = 1ll * a * a % P)if (b & 1) ans = 1ll * ans * a % P;return ans;
}
inline LL inv(int x){if (x <= n) return INV[x];return qpow(x,P - 2);
}
int main(){n = read();REP(i,n) m[i] = read(),p[i][m[i]] = 1;INV[0] = 1; INV[1] = 1;for (int i = 2; i <= n; i++) INV[i] = 1ll * (P - P / i) * INV[P % i] % P;LL Q = read(),opt,u,v,pp,x,k;while (Q--){opt = read();if (!opt){x = read(); u = read(); v = read(); pp = u * inv(v) % P;p[x][0] = (p[x][0] + pp * p[x][1] % P) % P;for (int i = 1; i <= m[x]; i++)p[x][i] = ((p[x][i] * (1 - pp) % P + p[x][i + 1] * pp % P) % P + P) % P;}else {k = read();cls(g); g[0][0] = 1;for (int i = 1; i <= k; i++){u = id[i] = read();g[i][0] = g[i - 1][0] * p[u][0] % P;for (int j = 1; j <= i; j++){g[i][j] = ((g[i - 1][j] * p[u][0] % P + g[i - 1][j - 1] * (1 - p[u][0]) % P) % P + P) % P;}}for (int i = 1; i <= k; i++){u = id[i];LL ans = 0,Inv = inv(p[u][0]);if (!p[u][0]){for (int j = 0; j < k; j++)f[u][j] = g[k][j + 1];}else {f[u][0] = 1ll * g[k][0] * Inv % P;for (int j = 1; j < k; j++){f[u][j] = (1ll * (g[k][j] - 1ll * f[u][j - 1] * (1 - p[u][0]) % P) % P * Inv % P + P) % P;}}for (int j = 0; j < k; j++){ans = (ans + 1ll * f[u][j] * inv(j + 1) % P) % P;}ans = (1ll * ans * (1ll - p[u][0]) % P + P) % P;printf("%lld",ans);if (i < k) putchar(' ');else puts("");}}}for (int i = 1; i <= n; i++){LL ans = 0;for (int j = 1; j <= m[i]; j++)ans = (ans + 1ll * j * p[i][j] % P) % P;printf("%lld",ans);if (i < n) putchar(' ');else puts("");}return 0;
}