计蒜客 2017 ACM-ICPC 亚洲区(西安赛区)网络赛 B coin(求乘法逆元)

Bob has a not even coin, every time he tosses the coin, the probability that the coin's front face up isqp(qp≤12)\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 XY\frac{X}{Y}​Y​​X​​, 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)(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≤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.

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

题意:

计算扔k次硬币，正面朝上的次数为偶数的概率。再将答案mod(1e9+7)

解析:

正面朝上概率为a，反面朝上概率为b

a+b=1；

ans=C(k,0)*a^0*b^k+C(k,2)*a^2*b^(k-2)+...

(a+b)^k= C(k,0)*a^0 *b^k+ C(k,1)*a^1*b^(k-1) +C(k,2)*a^2*b^(k-2) .......+C(k,k)*a^k*b^0
(b-a)^k= C(k,0) *(-a)^0 *b^k +C(k,1)*(-a)* b^(k-1) .........+C(k,k)*(-a)^k*b^0

可得ans=((a+b)^k+(b-a)^k)/2   再将a=p/q,b=(1-p/q)带入

得(（p^k）+(p-2q)^k)/(2*p^k)

除法的模运算就是求乘法逆元

x*(x^-1)=1(mod p) 乘法逆元性质

d=(x/y)(mod p) => y*d=x(mod p) => y*y^-1*d=x*y^-1(mod p) => d=x*y^-1(mod p)

求乘法逆元的三种方法

http://blog.csdn.net/rain722/article/details/53170288

#include
#include
#include
using namespace std;

const long long int N = 1e9+7;

long long p,q,k;
long long zi,mu;

long long quickmulti(long long a,long long b)
{
	long long ans=1;
	while(b)
	{
		if(b&1) ans = ((ans%N)*(a%N)) % N;
		a = ((a%N)*(a%N)) % N;
		b=b>>1;
	}
	return ans;
}


long long inv2(long long b)
{
	return quickmulti(b,N-2);
}

int main()
{
	int t;
	long long x,y,d,x0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld%lld",&p,&q,&k);
		zi=(quickmulti(p,k)+quickmulti(p-2*q,k))%N;
		mu=(2*quickmulti(p,k))%N;
		long long  ans=(zi*inv2(mu))%N;  
		printf("%lld\n",ans);

	}
	return 0;
}