BZOJ5340: [Ctsc2018]假面【概率+期望】【思维】

LINK


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\(n^3\)

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\(f_{i-1,j}=\frac{f_{i,j}-(1-p_{now})*f_{i-1,j-1}}{p_{now}}\)

\(p_{now}=0\)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


#include<bits/stdc++.h>

using 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;
}
posted @ 2018-12-06 20:37 Dream_maker_yk 阅读(...) 评论(...) 编辑 收藏