2017 ACM-ICPC 亚洲区（西安赛区）网络赛： B. Coin

Bob has a not even coin, every time he tosses the coin, the probability that the coin’s front face up is qp \frac{q}{p} ( qp \frac{q}{p} ≤12 \le \frac{1}{2})
The question is, when Bob tosses the coin kk times, what’s the probability that the frequency of the coin facing up is even number.

If the answer is XY \frac{X}{Y} , 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).

Input Format

First line an integer T, indicates the number of test cases ( T≤100 T\le100).

Then Each line has 3 integer p,q,k( 1≤p,q,k≤107 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

思路:一个高中的组合数学题。。。。。

ans=∑i=0i∗2≤k(qp)2i(1−qp)(k−2i)C2ik

ans=\sum_{i=0}^{i*2\le k}(\frac{q}{p})^{2i} (1-\frac{q}{p})^{(k-2i)}C_{k}^{2i}

ans=1pk∑i=0i∗2≤kq2i(p−q)(k−2i)C2ik

ans=\frac{1}{p^{k}}\sum_{i=0}^{i*2\le k}{q}^{2i} (p-q)^{(k-2i)}C_{k}^{2i}

ans=1pk((p−q)+q)k+((p−q)−q)k2

ans=\frac{1}{p^{k}}\frac{((p-q)+q)^{k}+((p-q)-q)^{k}}{2}

#include<bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
long long POW(long long x,long long n)
{long long res=1;while(n){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;n/=2;}return res;
}
int main()
{int T;cin>>T;while(T--){long long p,q,k;scanf("%lld%lld%lld",&p,&q,&k);long long x=POW(p-2*q,k);long long y=POW(p,k);cout<<(  ((1+x*POW(y,MOD-2))%MOD) * POW(2,MOD-2) )%MOD<<endl;}return 0;
}

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2017 ACM-ICPC 亚洲区（西安赛区）网络赛： B. Coin

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