2017 ACM-ICPC 亚洲区（西安赛区）网络赛 Coin（组合数）

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Bob has a not even coin, every time he tosses the coin, the probability that the coin's front face up is $frac{q}{p}(frac{q}{p} le frac{1}{2})$ .

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 $frac{X}{Y}$ , because the answer could be extremely large, you only need to print $Y^{-1}) mod (10^9+7)$ .

Input Format

First line an integer $T$ , indicates the number of test cases ( $T \leq 100$ ).

Then Each line has $3$ integer $10^7)$ 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

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样例输出

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这题的答案公式很好推，但是怎么化简不好构造；利用了组合数中的二项式展开技巧，因为正面朝上的次数为偶数所以需要把奇数项消除构造二项式令第二项为负数即可消除

$思路：$

复制代码

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#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod = 1e9+7;
LL quick(LL x, LL n)
{
LL r=1;
while(n)
{
if(n&1) r=r*x%mod;
n>>=1;
x=x*x%mod;
}
return r%mod;
}
int main()
{
//cout<<quick(2,mod-2)<<endl;
int t;
scanf("%d", &t);
while(t--)
{
LL p, q, k;
scanf("%lld %lld %lld", &p, &q, &k);
LL x=quick(p-2*q,k)*quick(quick(p,k)%mod,mod-2)%mod;
x=(x+1)*quick(2,mod-2)%mod;
printf("%lldn",x);
}
return 0;
}