Coin 2017 西安网络赛 快速幂 + 二项式定理

题目链接

---

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)qp(qp≤12)​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 XY​ ​​, because the answer could be extremely large, you only need to print (X∗Y−1)mod(109+7)(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≤100T≤100T≤100).

Then Each line has 333 integer p,q,k(1≤p,q,k≤107)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.

---

题意：

给你一枚不均匀的硬币， 正面朝上的概率是 q / p；现在扔 k 次， 求正面朝上次数为偶数次的概率。

思路：

计算方式类似于二项分布的概率的计算， 可以得出 ∑ki=0C(k,i)∗(qp)i∗(1−qp)k−i，i是奇数 , 观察式子我们发现分母岁固定的 ： pk , 分子是二项式的偶数项， 至于计算二项式的偶数项就很简单了 即是， qk+(p−2∗q)k2 ,最后就是除法逆元了，

代码：

#include
#include
#include
#include
#include
#include
using namespace std;

typedef long long LL;
const int maxn = 1;
const double eps = 1e-7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;

LL qpow(LL x, LL n){
    x %= mod;
    LL ans = 1;
    while(n){
        if(n&1) ans = ans * x % mod;
        x = x * x % mod;
        n >>= 1;
    }return ans;
}

int main(){
    int t, k, p, q;
    scanf("%d", &t);
    while(t--){
        scanf("%d %d %d", &p, &q, &k);
        LL t = qpow(p, k);
        LL tt = (qpow(p, k) + qpow( p - 2 * q, k)) % mod;
        LL ans = tt * qpow(t * 2, mod -2) % mod;
        printf("%lld\n", (ans % mod +mod) % mod);
    }
    return 0;
}