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})​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 \frac{X}{Y}​Y​​X​​, because the answer could be extremely large, you only need to print (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\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:  扔 k次硬币   给你 一次硬币向上的 概率为p/q   让你求  k次硬币后 向上的次数为偶数的概率:

这个地方用到高中组合知识:

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

代码:

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};

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;
    //cout<<5*inv1(8)<<endl;
    cin>>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;
}

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