【牛客小白月赛16】I. 石头剪刀布

题意：
在一维直线上，\(A\)和\(B\)玩游戏，\(A\)刚开始在\(1\)，他和\(B\)进行剪刀石头布：

• 如果赢了，就往前走一格，赢得概率的\(a\)，(如果在\(n\)就不走)
• 如果平局，就不动，平局的概率为\(b\)
• 如果输了，就往后退一格，输的概率为\(1 - a - b\)(如果在\(1\)就不退)
  问\(A\)走到\(n\)的期望局数。

思路：
考虑\(f[i]\)表示从\(i\)走到\(n\)的期望步数，那么有：

\[\begin{eqnarray*} f[i] = \left\{ \begin{array}{cccc} Af[i + 1] + (1 - A)f[i] + 1 && i = 1\\ Af[i + 1] + Bf[i] + (1 - A - B)f[i - 1] && 1 < i < n \\ 0 && i = n \end{array} \right. \end{eqnarray*} \]

高斯消元即可。

代码：

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 110
int n; ll A, B;
const ll p = 1e9 + 7;
ll a[N][N], x[N];
ll gcd(ll a, ll b) {
	return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}
ll inv(ll a, ll p) {
	if (a == 1) return 1;
	return inv(p % a, p) * (p - p / a) % p;
}
int Gauss(int equ, int var) {
	int max_r, col, k;
	for (k = 0, col = 0; k < equ && col < var; ++k, ++col) {
		max_r = k;
		for (int i = k + 1; i < equ; ++i) {
			if (abs(a[i][col]) > abs(a[max_r][col])) {
				max_r = i;
			}
		}
		if (a[max_r][col] == 0) {
			--k;
			continue;
		}
		if (max_r != k) {
			for (int j = col; j < var + 1; ++j) {
				swap(a[k][j], a[max_r][j]);
			}
		}
		for (int i = k + 1; i < equ; ++i) {
			if (a[i][col] != 0) {
				ll LCM = lcm(abs(a[i][col]), abs(a[k][col]));
				ll ta = LCM / abs(a[i][col]);
				ll tb = LCM / abs(a[k][col]);
				if (a[i][col] * a[k][col] < 0) tb = -tb;
				for (int j = col; j < var + 1; ++j) {
					a[i][j] = ((a[i][j] * ta - a[k][j] * tb) % p + p) % p;
				}
			}
		}
	}
	for (int i = k; i < equ; ++i) {
		if (a[i][col] != 0) {
			return -1;
		}
	}
	if (k < var) return var - k;
	for (int i = var - 1; i >= 0; --i) {
		ll tmp = a[i][var];
		for (int j = i + 1; j < var; ++j) {
			if (a[i][j] != 0) {
				tmp -= a[i][j] * x[j];
				tmp = (tmp % p + p) % p;
			}
		}
		x[i] = (tmp * inv(a[i][i], p)) % p;
	}
	return 0; 
}


int main() {
	while (scanf("%d%lld%lld", &n, &A, &B) != EOF) {
		memset(a, 0, sizeof a);
		a[0][0] = p - A; a[0][1] = A;
		a[0][n] = -1;
		for (int i = 2; i < n; ++i) {
			a[i - 1][i - 2] = (1 - A - B + 2ll * p) % p;
			a[i - 1][i - 1] = (B - 1 + p) % p;
			a[i - 1][i] = A;
			a[i - 1][n] = -1;
		}		
		a[n - 1][n - 1] = 1;
		Gauss(n, n);
		printf("%lld\n", x[0]);
	}
	return 0;
}