2017 ACM-ICPC 亚洲区（西安赛区）网络赛 B

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 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}​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 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≤10​7​​)
indicates the i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer.http://write.blog.csdn.net/postedit

样例输入

2
2 1 1
3 1 2

样例输出

500000004

555555560

B

正确通过

2017-09-16 16:15:53

1ms

236kB

c++

推一下概率。

设一次向上的概率为x,向下的概率为y，x + y == 1

设A为向上n（n <=k,为偶数）次，则易知P（A） = C（k, n）x^n * y ^(k - n)

所以有  sigma（i = 0， k）= P（偶数次） + P（奇数次） = C（k, i）x^i * y ^(k - i) = （x + y）^ k = 1

为了得到偶数的概率，可以用-x替换上式的x

有sigma（i = 0， k） = C（k, i）（-x）^i * y ^(k - i) = （-x + y）^ k = （1 - 2 * x）^k

 两式相加 = 1 + （1 - 2 * x）^k = 2 *P（偶数次）

 所以 ans = (1 + （1 - 2 * x）^k) / 2

                = (1 + (1 - 2 * q / p) ^ k) / 2

题目要求对1e9 + 7取模.除法用费马小定理求逆元（

因为 a ^ (p - 1) = 1 (mod p)

所以 a * a ^ (p - 2) = 1 (mod p)

所以此题a^(1e9 + 5)就是a的逆元

）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<cstdlib>
#include<string>
#include<set>
#include<stack>
#define mod 1000000007

using namespace std;

long long p, q, k;
long long  PowerMod(long long  a, long long  b)
{
long long c = mod;
long long  ans = 1;
a = a % c;
while(b)
{
if(b & 1)
ans = (ans * a) % c;
b >>= 1;
a = (a * a) % c;
}
return ans;
}

int main()
{
int t;
cin >> t;
while(t--)
{
scanf("%lld %lld %lld", &p, &q, &k);
long long ans = PowerMod(2, mod - 2);
long long tmp = PowerMod(p, mod - 2);
long long x = 1 + PowerMod(1 - 2 * q * tmp % mod, k);
ans *= x;
ans %= mod;
cout << (ans + mod) % mod<< endl;
}
return 0;
}