P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫

合集 - 动态规划dp(55)

1.P8784 [蓝桥杯 2022 省 B] 积木画2024-02-17 2.P9847 [ICPC2021 Nanjing R] Crystalfly2024-02-18 3.G. Vlad and Trouble at MIT2024-02-20 4.P2899 [USACO08JAN] Cell Phone Network G2024-02-20 5.P5322 [BJOI2019] 排兵布阵2024-03-03 6.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-03 7.P1156 垃圾陷阱2024-03-03 8.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-04 9.P2466 [SDOI2008] Sue 的小球2024-03-04 10.P5020 [NOIP2018 提高组] 货币系统2024-03-04 11.P1757 通天之分组背包2024-03-04 12.P2135 方块消除2024-03-06 13.D - String Bags2024-03-10 14.P4395 [BOI2003] Gem 气垫车2024-03-10 15.A - A Multiply2024-03-24 16.P2642 双子序列最大和2024-03-25 17.P1121 环状最大两段子段和2024-03-25 18.P2422 良好的感觉2024-03-25 19.P2854 [USACO06DEC] Cow Roller Coaster S2024-03-26 20.C. Theofanis' Nightmare2024-03-27 21.D. Birthday Gift2024-03-27 22.P1284 三角形牧场2024-03-28 23.P1353 [USACO08JAN] Running S2024-03-28 24.P8736 [蓝桥杯 2020 国 B] 游园安排2024-03-30 25.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-01 26.P7859 [COCI2015-2016#2] GEPPETTO2024-04-05 27.P1435 [IOI2000] 回文字串2024-05-08 28.P2340 [USACO03FALL] Cow Exhibition G2024-05-13 29.P1854 花店橱窗布置2024-05-13 30.P3147 [USACO16OPEN] 262144 P2024-05-13 31.Game on Tree2024-05-14 32.P2938 [USACO09FEB] Stock Market G2024-05-18 33.P9691 [GDCPC2023] Base Station Construction2024-05-18 34.P8624 [蓝桥杯 2015 省 AB] 垒骰子2024-05-20 35.E. Money Buys Happiness2024-05-22 36.P8675 [蓝桥杯 2018 国 B] 搭积木2024-05-22

37.P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫2024-05-22

38.P8806 [蓝桥杯 2022 国 B] 搬砖2024-05-23 39.P5662 [CSP-J2019] 纪念品2024-05-24 40.B. Mashmokh and ACM2024-06-08 41.C. Minimizing the Sum2024-06-12 42.H. Don't Blame Me2024-06-13 43.D. Bicolorings2024-06-14 44.D. Armchairs2024-06-18 45.C. Chef Monocarp2024-06-18 46.D. Invertible Bracket Sequences2024-06-20 47.P1351 [NOIP2014 提高组] 联合权值2024-06-26 48.D. World is Mine2024-07-02 49.F. Feed Cats2024-07-07 50.P3522 [POI2011] TEM-Temperature2024-07-21 51.P3572 [POI2014] PTA-Little Bird2024-07-22 52.P3089 [USACO13NOV] Pogo-Cow S2024-07-22 53.E - Maximum Glutton2024-07-28 54.Projects2024-07-28 55.Elevator Rides2024-07-28

原题链接

题解

设

$E_{i}$ 为树根到高度为 $i$ 的点的期望用时
$P_{i}$ 为 $i - 1$ 到 $i$ 时，掉下去概率

则有
$E_{i} = E_{i - 1} + 1 + (1 - P_{i}) \cdot 0 + P_{i} \cdot E_{i}$

式子解读：
当我花一单位时间到i点时，有p的概率掉下去重新来，有1-p的概率保持住了i点

所以
$E_{i} = \frac{E_{i - 1} + 1}{1 - p}$

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod=998244353;
ll ny(ll x)
{
    ll res=1;
    ll len=mod-2;
    while(len)
    {
        if(len&1)res=res*x%mod;
        x=x*x%mod;
        len>>=1;
    }
    return res;
}
int main()
{
    ll n;
    cin>>n;

    ll e=0;
    for(ll i=1;i<=n;i++)
    {
        ll x,y;
        cin>>x>>y;
        e=(e+1LL)*ny(1LL-x*ny(y)%mod+mod)%mod;
    }

    cout<<e;
    return 0;
}