牛客练习赛39 - C 流星雨 概率dp+除法逆元

题目链接

题意：每天都有概率下流星雨，前一天有下的话当天下流星雨的概率为 $\frac{x}{y}+\frac{a}{b}$ ，如果没有，则当天是 $\frac{x}{y}$ ，求 $n$ 天过后下流星雨总数在 $mod$ $10^9+7$ 意义下的期望。

思路：求出每天下流星雨的概率，把每天的期望算出来，求和即是答案，由于中间答案会爆 $long$ $long$，所以用除法逆元，每次计算都取模，同时每次概率化成最简分数。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
#define mod 1000000007
#define ll long long
int n;
ll a, b;
ll dp[2][2], w[100005];
inline ll calc(ll x, ll y)
{
    ll mul = 1;
    while (y)
        if (y & 1)
        {
            --y;
            (mul *= x) %= mod;
        }
        else
        {
            y >>= 1;
            (x *= x) %= mod;
        }
    return mul;
}
inline ll gcdxy(ll x, ll y)
{
    ll t;
    while (y)
    {
        t = x % y;
        x = y;
        y = t;
    }
    return x;
}
inline ll solve(ll x,ll y)
{
    ll gcd = gcdxy(x, y);
    x /= gcd, y /= gcd;
    return x * calc(y, mod - 2) % mod;
}
int main()
{
    ll x, y, ab, xy;
    ll ans = 0;
    scanf("%d%lld%lld", &n, &a, &b);
    ab = solve(a, b);
    for (int i = 1; i <= n; ++i)
        scanf("%lld", &w[i]);
    int j = 1;
    scanf("%lld%lld", &x, &y);
    dp[0][0] = solve(y - x, y), dp[0][1] = solve(x, y);
    (ans += (dp[0][1] * w[1] % mod)) %= mod;
    for (int i = 2; i <= n; ++i)
    {
        scanf("%lld%lld", &x, &y);
        xy = solve(x, y);
        dp[j][1] = ((dp[1 - j][1] * ((xy + ab) % mod)) % mod + dp[1 - j][0] * xy % mod) % mod;
        dp[j][0] = (1 + mod - dp[j][1]) % mod;
        (ans += (dp[j][1] * w[i] % mod)) %= mod;
        j = 1 - j;
    }
    printf("%lld\n", ans);
    return 0;
}