Bob has a not even coin（就是一個不均勻的硬幣，朝上的概率不一定是1/2）, every time he tosses the coin, the probability that the coin‘s front face up is q/p(q/p<=1/2).

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（就是硬幣朝上次數為偶數次的概率和，0次也算）.

If the answer is X/Y??, 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 TT, indicates the number of test cases (T≤100).

Then Each line has 33 integer p,q,k(1<= p,q,k ,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.

樣例輸入

2
2 1 1
3 1 2

樣例輸出

500000004
555555560
 

思路:

題目說的有點晦澀。

一個組合數學題。。。。。

　

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int MOD=1e9+7;
 7 typedef long long ll;
 8 ll p,q,k;
 9 
10 ll qmod(ll x,ll n)
11 {
12     ll res=1;
13     while(n){
14         if
 
(n&1) res=(res*x)%MOD;
15         x=(x*x)%MOD;
16         n/=2;
17     }
18     return res;
19 }
20 
21 int main()
22 {
23     int T;
24     cin>>T;
25     while(T--)
26     {
27         cin>>p>>q>>k;
28         ll x=qmod(p,k);
29         ll y=qmod(p-2*q,k);
30         ll ans=((x+y)%MOD*qmod(2*x,MOD-2))%MOD;
31         cout<<ans<<endl;
32     }
33     return 0;
34 }

2017 ACM-ICPC 亞洲區（西安賽區）網絡賽： B. Coin 【概率題】