思路
首先考虑减血，直接一个dp做过去，这个部分分不难拿
然后是\(op=1\)的部分
首先因为要知道每个人被打的概率，所以需要算出这个人活着的时候有多少个人活着时概率是什么
那么用\(g_{i,j}\)表示第i个人还活着的时候还有其他的j个人活着的概率
这个东西暴力DP是\(n^3\)的
那么可以考虑优化，用\(f_{i,j}\)表示前i个人有j个人活着的概率
有转移：\(f_{i,j}=f_{i-1,j-1}*(1-p_i)+f_{i-1,j}*p_i\)，其中\(p_i\)表示第i个人已经死了的概率
j等于0特判一下就好了
因为我们用任意i的顺序做f的DP最后的\(f_{n}\)那一行都不会变
所以可以考虑用\(f_n\)逆推回g，因为\(g_{i,j}=f_{n-1,j}\)，我们默认这个时候正在算的i是最后一个
那么根据上面的转移式可以得到\(f_{i-1,j}=\frac{f_{i,j}-(1-p_{now})*f_{i-1,j-1}}{p_{now}}\)
当\(p_{now}=0\)的时候我们发现\(f_{i,j}=f_{i-1,j-1}\)，也就是说最后一个人无论如何是不会死的，那么\(f_{i-1,j}=f_{i,j+1}\)
而当j=0的时候我们又需要特判了，首先入如果\(p_{now}=0\)，\(g_{now,0} = f_{n,1}\)，否则\(g_{now,0}=\frac{f_{n,0}}{p_{now}}\)
剩下的很简单
然后就愉快结束了
#includeusing namespace std;typedef long long ll;namespace io {const int BUFSIZE = 1 << 20;char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;char obuf[BUFSIZE], *os = obuf, *ot = obuf + BUFSIZE - 1;char read_char() { if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); return *is++;}int read_int() { int x = 0, f = 1; char c = read_char(); while (!isdigit(c)) { if (c == '-') f = -1; c = read_char(); } while (isdigit(c)) x = x * 10 + c - '0', c = read_char(); return x * f;}ll read_ll() { ll x = 0, f = 1; char c = read_char(); while (!isdigit(c)) { if (c == '-') f = -1; c = read_char(); } while (isdigit(c)) x = x * 10 + c - '0', c = read_char(); return x * f;}void read_string(char* s) { char c = read_char(); while (isspace(c)) c = read_char(); while (!isspace(c)) *s++ = c, c = read_char(); *s = 0;}void flush() { fwrite(obuf, 1, os - obuf, stdout); os = obuf;}void print_char(char c) { *os++ = c; if (os == ot) flush();}void print_int(int x) { static char q[20]; if (!x) print_char('0'); else { if (x < 0) print_char('-'), x = -x; int top = 0; while (x) q[top++] = x % 10 + '0', x /= 10; while (top--) print_char(q[top]); }}void print_ll(ll x) { static char q[20]; if (!x) print_char('0'); else { if (x < 0) print_char('-'), x = -x; int top = 0; while (x) q[top++] = x % 10 + '0', x /= 10; while (top--) print_char(q[top]); }}struct flusher_t { ~flusher_t() { flush(); }} flusher;};using namespace io;const int Mod = 998244353;const int N = 210;int add(int a, int b) { return (a += b) >= Mod ? a - Mod : a;}int sub(int a, int b) { return (a -= b) < 0 ? a + Mod : a;}int mul(int a, int b) { return 1ll * a * b % Mod;}int fast_pow(int a, int b) { int res = 1; while (b) { if (b & 1) res = mul(res, a); b >>= 1; a = mul(a, a); } return res;}int n, q, m[N], inv[N];int c[N], res[N];int p[N][N], f[N][N], g[N][N];//p[i][j] 第i个人 还有j点血的概率//f[i][j] 前i个人还有j个人活下来的概率//g[i][j] 除了第i个人 还有j个人活下来的概率 int main() { n = read_int(); inv[0] = 1; for (int i = 1; i <= n; ++i) { m[i] = read_int(); p[i][m[i]] = 1; inv[i] = fast_pow(i, Mod - 2); } q = read_int(); while (q--) { int op = read_int(); if (!op) { int x = read_int(), p1 = read_int(), p2 = read_int(); int k = mul(fast_pow(p2, Mod - 2), p1); p[x][0] = add(p[x][0], mul(p[x][1], k)); for (int i = 1; i <= m[x]; ++i) { p[x][i] = add(mul(p[x][i + 1], k), mul(p[x][i], sub(1, k))); } } else { int num = read_int(); for (int i = 1; i <= num; ++i) { c[i] = read_int(); } f[0][0] = 1; for (int i = 1; i <= num; ++i) { f[i][0] = mul(f[i - 1][0], p[c[i]][0]); for (int j = 1; j <= i; ++j) { f[i][j] = add(mul(f[i - 1][j - 1], sub(1, p[c[i]][0])), mul(f[i - 1][j], p[c[i]][0])); } } for (int i = 1; i <= num; ++i) { int invp = fast_pow(p[c[i]][0], Mod - 2); if (!p[c[i]][0]) g[i][0] = f[num][1]; else g[i][0] = mul(f[num][0], invp); //** for (int j = 1; j < num; ++j) { if (!p[c[i]][0]) g[i][j] = f[num][j + 1]; else g[i][j] = mul(invp, sub(f[num][j], mul(g[i][j - 1], sub(1, p[c[i]][0])))); } } for (int i = 1; i <= num; ++i) { res[i] = 0; for (int j = 0; j < num; ++j) { res[i] = add(res[i], mul(inv[j + 1], g[i][j])); } res[i] = mul(res[i], sub(1, p[c[i]][0])); print_int(res[i]), print_char(' '); } print_char('\n'); } } for (int i = 1; i <= n; ++i) { int cur = 0; for (int j = 1; j <= m[i]; ++j) { cur = add(cur, mul(p[i][j], j)); } print_int(cur), print_char(' '); } return 0;}