2017 ACM-ICPC 亚洲区（西安赛区）网络赛 Coin（组合数）

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 $k$ times, 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 $T$, 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)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

22 1 13 1 2

样例输出

500000004555555560

这题的答案公式很好推，但是怎么化简不好构造；利用了组合数中的二项式展开技巧，因为正面朝上的次数为偶数 所以需要把奇数项消除 构造二项式 令第二项为负数即可消除

思路：

#include <iostream>#include <bits/stdc++.h>using namespace std;typedef long long LL;const LL mod = 1e9+7;LL quick(LL x, LL n){    LL r=1;    while(n)    {        if(n&1) r=r*x%mod;        n>>=1;        x=x*x%mod;    }    return r%mod;}int main(){    //cout<<quick(2,mod-2)<<endl;    int t;    scanf("%d", &t);    while(t--)    {        LL p, q, k;        scanf("%lld %lld %lld", &p, &q, &k);        LL x=quick(p-2*q,k)*quick(quick(p,k)%mod,mod-2)%mod;        x=(x+1)*quick(2,mod-2)%mod;        printf("%lld\n",x);    }    return 0;}