2017 ACM-ICPC 亚洲区（西安赛区）网络赛 B.Coin(基本概率+二项式展开)

Posted 2022-04-16 ITAK

传送门

Bob has a not even coin, every time he tosses the coin, the probability that the coin’s front face up is $\\fracqp(\\fracqp \\le \\frac12)$.

The question is, when Bob tosses the coin $k$ times, what’s the probability that the frequency of the coin facing up is even number.

If the answer is $\\fracXY$, because the answer could be extremely large, you only need to print $(X * Y^-1) \\mod (10^9+7)$.

Input Format

First line an integer $T$, indicates the number of test cases ($T \\le 100$).

Then Each line has $3$ integer $p,q,k(1\\le p,q,k \\le 10^7)$ indicates the i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer.

样例输入

2 

2 1 1 

3 1 2

样例输出

500000004 

555555560

题目大意：

Bob有一个不公平的硬币，正反的概率是不一样的，现在给出这个硬币朝上的概率为 $\\frac pq$，现在求抛 $k$ 次之后正面朝上的概率是多少，输出的是 $X*Y^-1$，就是求分母关于$MOD$ 的逆元。

解题思路：
根据概率所学的知识，很容易想到的是分母为 $fenmu=p^k$,分子为：$fenzi=\\sum_i=0^kC_k^i*q^i*(p-q)^k-i(i为偶数)$
根据二项式定理有：
现在有 $(q+p-q)^k=\\sum_i=0^kC_k^i*q^i*(p-q)^k-i——(1)$
设有： $(q-(p-q))^k=\\sum_i=0^kC_k^i*q^i*(-1)^k-i*(p-q)^k-i——(2)$
那么我们发现：
当 $k$ 是奇数的时候 $(1)-(2)=fenzi$
当 $k$ 是偶数的时候 $(1)+(2)=fenzi$
显然分子可求，最后在乘以分母关于 $mod$ 的逆元就OK了。

代码：

#include <iostream>
#include <string.h>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
const double PI = acos(-1);
const double eps = 1e-8;
const LL MOD = 1e9+7;
LL Pow(LL a, LL b)
    LL ans = 1;
    while(b)
        if(b & 1) ans = ans * a % MOD;
        b>>=1;
        a = a * a % MOD;
    
    return ans;

void Exgcd(LL a, LL b, LL &x, LL &y)
    if(b == 0)
        x = 1;
        y = 0;
        return ;
    
    LL x1, y1;
    Exgcd(b, a%b, x1, y1);
    x = y1;
    y = x1-(a/b)*y1;

int main()
    //freopen("C:/Users/yaonie/Desktop/in.txt", "r", stdin);
    //freopen("C:/Users/yaonie/Desktop/out.txt", "w", stdout);
    int T; scanf("%d", &T);
    while(T--)
        LL p, q, k; scanf("%lld%lld%lld",&p,&q,&k);
        LL tmp1 = Pow(2*q-p, k);
        LL tmp2 = Pow(p, k);
        int flag = 1;
        if(k & 1) flag = -1;
        LL fenzi = (tmp2+flag*tmp1)%MOD*500000004LL%MOD;
        LL x, y;
        Exgcd(p, MOD, x, y);
        x = (x%MOD+MOD)%MOD;
        LL ans = Pow(x, k)*fenzi%MOD;
        printf("%lld\\n",ans);
    
    return 0;