2017 ACM-ICPC 亚洲区（西安赛区）网络赛_Coin

#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define rep0(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)
#define rep_1(i, n) for (int i = n; i > 0; i--)
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define mem(x, y) memset(x, y, sizeof(x))
/**
题目大意
给出抛硬币正面朝上的概率为q/p
求抛k次偶数次正面朝上的概率
思路
Σ C(i, k) * (q / p)^i * (1 - q / p)^(k - i)  (i = 0, 2, 4...)

x = Σ C(i, k) * a^i * b^(k - i)  (i = 0, 1, 2, 3...)
y = Σ C(i, k) * (-a)^i * b^(k - i)  (i = 0, 1, 2, 3...)

(x + y) / 2 即为所求：
(1 + (p - 2 * q)^k / p^k) / 2
=
(p^k + (p - 2 * q)^k) / (2 * p^k)

除法取模
设分子为i 分母为j
i / j % MOD = x

法1: 扩展欧几里得
j * x = i(mod MOD)

法2: i / j % MOD <=> i * j^(-1) (j^(-1) 为j的逆元)
由费马小定理，MOD为质数，且与j互质
j^(MOD - 1) = 1(mod MOD)
j * j^(MOD - 2) = 1(mod MOD)
则j^(MOD - 2)为j的逆元

*/
using namespace std;
typedef long long LL;
const int MOD = 1e9 + 7;

LL exgcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL r = exgcd(b, a % b, x, y);
    LL temp = x;
    x = y;
    y = temp - a / b * y;
    return r;
}

LL multi(LL a, LL b)
{
    LL ans = 1;
    while (b)
    {
        if (b & 1)
            ans = (ans * a) % MOD;
            
        a = (a * a) % MOD;
        b >>= 1;
    }
    return ans;
}

int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int t, p, q, k;
    LL ans;

    scanf("%d", &t);
    while (t--)
    {
        scanf("%d %d %d", &p, &q, &k);

        LL a, b = MOD, x, y, p_k = multi(p, k);

        a = (p_k + multi(p - 2 * q, k)) % MOD;
        b = 2 * p_k % MOD;
        ans = a * multi(b, MOD - 2) % MOD;

        /**
        a = 2 * p_k % MOD;
        exgcd(a, b, x, y);
        LL temp = (p_k + multi(p - 2 * q, k)) % MOD;
        ans = ((x % MOD + MOD) % MOD * temp) % MOD;
        */

        printf("%lld\n", ans);


    }

    return 0;
}