牛客网多校赛第9场 E-Music Game【概率期望】【逆元】

链接：https://www.nowcoder.com/acm/contest/147/E

来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 262144K，其他语言524288K

64bit IO Format: %lld

题目描述

Niuniu likes to play OSU!

We simplify the game OSU to the following problem.

Given n and m, there are n clicks. Each click may success or fail.

For a continuous success sequence with length X, the player can score X^m.

The probability that the i-th click success is p[i]/100.

We want to know the expectation of score.

As the result might be very large (and not integral), you only need to output the result mod 1000000007.

输入描述:

The first line contains two integers, which are n and m.
The second line contains n integers. The i-th integer is p[i].

1 <= n <= 1000
1 <= m <= 1000
0 <= p[i] <= 100

输出描述:

You should output an integer, which is the answer.

示例1

输入

复制

3 4
50 50 50

输出

复制

750000020

说明

000 0
001 1
010 1
011 16
100 1
101 2
110 16
111 81

The exact answer is (0 + 1 + 1 + 16 + 1 + 2 + 16 + 81) / 8 = 59/4.
As 750000020 * 4 mod 1000000007 = 59
You should output 750000020.

备注:

If you don't know how to output a fraction mod 1000000007,
You may have a look at https://en.wikipedia.org/wiki/Modular_multiplicative_inverse

感觉最近需要好好补一下数学方面的题目包括概率期望什么的 平时接触的太少了

看题解说要用什么第二类斯特林数 什么玩意啊 根本没听过 多校赛到处都是别人很熟的而我们听都没听过的东西

但是看了一下别的题解其实也没有那么难吧

顺便学习了一下分数取模的方法

一个数a乘一个数b对p取模 和a乘b的逆元对p取模结果是一样的

求逆元最方便的方法是用扩展欧几里得

https://blog.csdn.net/stray_lambs/article/details/52133141这篇博客讲欧几里得讲的还是蛮细的

但是我还是没有很明白题解里求的inv 是什么道理

下面说说题目的主体思路

首先用快速幂预处理得分

用pos[i][j]表示从 i 到 j 这段区间全部是获胜的概率

pos[i][j]肯定是可以有pos[i][j - 1]推出来的

枚举i ， j 表示从i + 1 到 j - 1都连续胜利利用期望的可加性E(X+Y)=E(X)+E(Y);

这段的期望就是（i失败的概率）*（j失败的概率）*（i-j连续获胜的概率）*（分数）

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#define inf 1e18
using namespace std;

const long long mod = 1e9 + 7;
const int maxn = 1005;
long long m, n, p[maxn];
long long ipowm[maxn], pos[maxn][maxn];
long long qpow(long long x, long long m)
{
    long long ans = 1;
    while(m){
        if(m & 1){
            ans = ans * x % mod;
        }
        x = x * x % mod;
        m = m >> 1;
    }
    return ans;
}

long long exgcd(long long a, long long b, long long &x, long long &y)
{
    if(a == 0 && b == 0)
        return -1;
    if(b == 0){
        x = 1;
        y = 0;
        return a;
    }
    long long d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
//扩展欧几里得求逆元 用于分数取模
long long mod_rev(long long a, long long n)
{
    long long x, y;
    long long d = exgcd(a, n, x, y);
    if(d == 1)
        return (x % n + n) % n;
    else return -1;
}

int main()
{
    long long inv = mod_rev(100ll, mod);//因为这里概率还要除以100，所以第一个参数写100， 第二个参数就是模
    //cout<<inv;
    while(scanf("%lld%lld", &n, &m) != EOF){
            //cout<<1;
        for(int i = 1; i <= n; i++){
            scanf("%lld", &p[i]);
            ipowm[i] = qpow(i, m);
        }
        //cout<<1;
        p[0] = p[n + 1] = 0;
        long long ans = 0;

        for(int i = 1; i <= n; i++){
            pos[i][i] = p[i] * inv % mod;
            for(int j = i + 1; j <= n; j++){
                pos[i][j] = pos[i][j - 1] * p[j] % mod * inv % mod;
            }
        }
        for(int i = 0; i <= n; i++){
            for(int j = i + 2; j <= n + 1;  j++){
                ans += pos[i + 1][j - 1] * ipowm[j - i - 1] % mod
                    * (100 - p[i]) % mod * inv % mod * (100 - p[j]) % mod * inv % mod;
                ans %= mod;
            }
        }

        printf("%lld\n", ans);
    }
    return 0;
}