计算客网络赛 Coin 二项式定理+逆元

https://nanti.jisuanke.com/t/17115

Bob has a not even coin, every time he tosses the coin, the probability that the coin's front face up is qp(qp≤12)frac{q}{p}(frac{q}{p} le frac{1}{2})​p​​q​​(​p​​q​​≤​2​​1​​).

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

If the answer is XYfrac{X}{Y}​Y​​X​​, because the answer could be extremely large, you only need to print (X∗Y−1)mod(109+7)(X * Y^{-1}) mod (10^9+7)(X∗Y​−1​​)mod(10​9​​+7).

Input Format

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

Then Each line has 333 integer p,q,k(1≤p,q,k≤107)p,q,k(1le p,q,k le 10^7)p,q,k(1≤p,q,k≤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
咳咳
  当时想到了做法无奈组合数学太垃圾写错了公式最后才发现，要求k次抛中有偶数次正面向上的概率，设每一次向上概率为P，显然答案是SUM{C(k,x)*Px*(1-P)k-x},当时把二项式系数给扔了卧槽。
要求x必须是偶数只要xjb构造一个就好了，令S1=(P+(1-P))k , S2=((-P)+(1-P))k  显然S2中P的幂数为奇数的项都是负的，二式相加在除以二之后就是答案了，再求一下逆元就是答案。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define LL long long
LL mod=1e9+7;
LL inv(LL i) {  if(i==1)return 1;return (mod-mod/i)*inv(mod%i)%mod; }
LL qpow(LL a,LL b){LL r=1;for(;b;b>>=1,a=a*a%mod)if(b&1)r=r*a%mod;return r;}
int main()
{
    int T;
    LL p,q,k,ans;
    cin>>T;
    while(T--){ans=0;
        cin>>p>>q>>k;
        ans=qpow(p,k);
        ans=(ans+qpow(p-2*q,k))%mod;
        LL fm=qpow(inv(p),k);
        ans=ans*fm%mod;
        ans=ans*inv(2)%mod;
        cout<<ans<<endl;
    }
    return 0;
}