2017 ACM-ICPC西安网赛B-Coin

B-Coin

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

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

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

Input Format

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

Then Each line has 33 integer p,q,k(1\le 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

题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

 //2017-10-24
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define ll long long

 using namespace std;

 const int MOD = ;

 ll quickPow(ll a, ll n){
     ll ans = ;
     while(n){
         if(n&)
             ans = (a*ans)%MOD;
         a = (a*a)%MOD;
         n >>= ;
     }
     return ans;
 }

 int main()
 {
     ll p, q, k, T;
     cin>>T;
     while(T--){
         cin>>p>>q>>k;
         ll X = quickPow(p-*q, k);
         ll Y = quickPow(p, k);
         cout<<(((+X*quickPow(Y, MOD-))%MOD) * quickPow(, MOD-))%MOD<<endl;
     }

     return ;
 }