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})pq(pq≤21).
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\frac{X}{Y}YX, 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(109+7).
Input Format
First line an integer TTT, indicates the number of test cases (T≤100T \le 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≤107) 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;
}