Description
有 (n(nle 10^5)) 个点，在 (i) 有 (p[i]) 的概率走到 (i+1) ，(1-p[i]) 的概率走到 (i-1) 。有 (Q(Qle10^5)) 次操作。操作有两种：
- 单点修改概率。
- 询问从 (L) 走到 (R+1) ，且不经过小于 (L) 的点的概率。
Solution
- (f[i]) ：从 (i) 到 (R+1) 不经过小于 (i) 的点的概率。
[f[L-1]=0,f[R+1] =1
]
[f[i] = p[i] imes f[i+1]+(1-p[i]) imes f[i-1]
]
[f[i]-f[i-1]=p[i] imes(f[i+1]-f[i-1])
]
- (g[i]=f[i]-f[i-1])
[g[i]=p[i] imes (g[i+1]+g[i])
]
[g[i+1]=cfrac{1-p[i]}{p[i]} imes g[i]
]
- (u[i]=cfrac{1-p[i]}{p[i]})
则有
[sum_{i=L}^{R+1}g[i]=f[R+1]-f[L-1]=1
]
[g[L] imes(1+u[L]+u[L] imes u[L+1]+cdots +u[L] imes u[L+1] imes cdots imes u[R])=1
]
[f[L]=g[L]=cfrac{1}{1+u[L]+u[L] imes u[L+1]+cdots +u[L] imes u[L+1] imes cdots imes u[R]}
]
- (A_{L,R}=u[L] imes u[L+1] imes cdots imes u[R])
- (B_{L,R}=u[L]+u[L] imes u[L+1]+cdots +u[L] imes u[L+1] imes cdots imes u[R])
[B_{L,R}=B_{L,mid}+A_{L,mid} imes B_{mid+1,R}
]
用线段树维护 (A) 和 (B) 即可。
#include<bits/stdc++.h>
using namespace std;
template <class T> inline void read(T &x) {
x = 0; static char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar());
for (; ch >= '0' && ch <= '9'; ch = getchar()) (x *= 10) += ch - '0';
}
#define N 100001
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define ll long long
double A[N << 2], B[N << 2];
#define ls rt << 1
#define rs ls | 1
#define mid (l + r >> 1)
void update(int rt, int l, int r, int pos, double val) {
if (l == r) { A[rt] = B[rt] = (1.0 - val) / val; return; }
if (pos <= mid) update(ls, l, mid, pos, val);
else update(rs, mid + 1, r, pos, val);
A[rt] = A[ls] * A[rs], B[rt] = B[ls] + A[ls] * B[rs];
}
#define pdd pair<double, double>
pdd query(int rt, int l, int r, int L, int R) {
if (L <= l && r <= R) return pdd(A[rt], B[rt]);
if (R <= mid) return query(ls, l, mid, L, R);
if (L > mid) return query(rs, mid + 1, r, L, R);
if (L <= mid && R > mid) {
pdd ansl = query(ls, l, mid, L, R), ansr = query(rs, mid + 1, r, L, R);
return pdd(ansl.first * ansr.first, ansl.second + ansl.first * ansr.second);
}
}
int main() {
int n, Q; read(n), read(Q);
rep(i, 1, n) {
double a, b; read(a), read(b);
update(1, 1, n, i, a / b);
}
while (Q--) {
int op; read(op);
if (op == 1) {
int pos; double a, b; read(pos), read(a), read(b);
update(1, 1, n, pos, a / b);
}
else {
int l, r; read(l), read(r);
pdd ans = query(1, 1, n, l, r);
printf("%.12lf
", ans.second <= 2e15 ? 1.0 / (1.0 + ans.second) : 0);
}
}
return 0;
}