牛客练习赛39 C 流星雨 (概率dp)

题意：

现在一共有n天，第i天如果有流星雨的话，会有 ${\mathrm{wi颗流星雨。}}^{}$

第1天有流星雨的概率是 ${\mathrm{p1。}}^{}$

${\mathrm{如果第i-1 (i\ge 2)天有流星雨，第i天有流星雨的可能性是pi+P，否则是pi。}}^{}$

求n天后，流星雨颗数的期望。

分数以逆元形式输出

 

思路：

直接在逆元情况下做

第i天有流星雨的概率为t[i]=t[i-1]*(p[i-1]+P)+(1-t[i-1])*p[i]

注意减法的时候要加mod

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
ll fp(ll a,ll n){
    ll ans = 1;
    while(n){
        if(n&1)ans*=a;
        n>>=1;
        a*=a;
        a%=mod;
        ans%=mod;
    }
    return ans;
}
ll f[maxn],t[maxn],w[maxn];
ll p;
int main() {
    ll n, a, b;
    scanf("%lld %lld %lld", &n, &a, &b);
    p = a*fp(b,mod-2)%mod;
    for(int i = 1; i <=n; i++){
        scanf("%lld", &w[i]);
    }
    for(int i = 1; i <= n; i++){
        scanf("%lld %lld", &a, &b);
        f[i] = a*fp(b,mod-2)%mod;
    }
    ll tmp = 1;
    ll ans = f[1]*w[1]%mod;
    t[1]=f[1];
    for(int i = 2; i <= n; i++){
        t[i] += t[i-1]*(f[i]+p)%mod+(1-t[i-1]+mod)*f[i]%mod;
        t[i]%=mod;
        ans+=t[i]*w[i]%mod;
        ans%=mod;
    }
    printf("%lld",ans);
    return 0;
}