2017 ACM-ICPC 亚洲区（西安赛区）网络赛 B Coin （概率计算）

传送门: https://nanti.jisuanke.com/t/17115

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})pq(pq≤21).

The question is, when Bob tosses the coin kk times, what's the probability that the frequency of the coin facing up is even number.

If the answer is \frac{X}{Y}YX, because the answer could be extremely large, you only need to print (X * Y^{-1}) \mod (10^9+7)(X∗Y−1)mod(109+7).

Input Format

First line an integer TT, indicates the number of test cases (T \le 100T≤100).

Then Each line has 33 integer 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

题目的意思很简单，就是让你扔硬币，给你p,q.向上的概率为q/p;然后问

你向上次数为偶数的概率。

这个地方用到高中组合知识:(组合数学，概率方面的问题）

设向上概率概率是 a 向下为b 则 a+b =1;

k次扔取时向上为偶数的为: C(k,0) *a^0 * b^k + C(k,2) *a^2 *b^(k-

2) +C(k,4)*a^4 *b^(k-4) ......... + C(k,k)*a^k*b^0

(a+b)^k 展开为 : 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

(a-b) ^k 展开为: C(k,0) *a^0 *(-b)^k +C(k,1)^a* (-b)^(k-1)

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

所以应为: ((a+b)^k +(a-b)^k ) /2

题意要求逆元:

求逆元方法;:(三种)

ll inv_exgcd(ll a,ll n){lld,x,y;ex_gcd(a,n,d,x,y);return d==1?

(x+n)%n:-1;}

ll inv1(ll b){returnb==1?1:(MOD-MOD/b)*inv1(MOD%

b)%MOD;}

ll inv2(ll b){return qpow(b,MOD-2);}

求逆元应用费马小定理的那个即为:

ll inv2(ll b){return qpow(b,MOD-2);}

若用带/ 法的求逆元则会出现 /0 的情况 WA

#include<stdio.h>
#include<iostream>
#include <algorithm>
#include<string.h>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#define LL long long
#define INf 0x3f3f3f3f
const int MOD=1e9+7;
using namespace std;
LL qpow(LL x,LL n)
{LL res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;
}
LL inv2(LL b)
{return qpow(b,MOD-2);
}
int main()
{int t;  scanf("%d",&t);while(t--)  {  LL p,q,k;  scanf("%lld %lld %lld",&p,&q,&k);  LL a=qpow(p,k);  LL sum=(a+qpow((p-2*q),k))%MOD;;  sum=sum*(inv2(2*a)%MOD);  printf("%lld\n",sum%MOD);  }  return 0;
}