2017 ACM-ICPC 亚洲区（西安赛区）网络赛_Coin

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#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ji / 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;}