codeforces 935D. Fafa and Ancient Alphabet (math)

题意：给你两个字符串，长度均为n,第二第三行输入为A,B字符串，其中0表示这里缺失，现在给你m种选择可填入0之中，问你A比B字典序大的可能性模1e9+7

AC code:

#include
using namespace std;
typedef long long ll;

const int mod = 1000000007;
const int Maxn = 100005;
int n, m, res;
int A[Maxn], B[Maxn];

int qpow(int a, int p)
{
	int res = 1;
	while (p)
    {
		if (p & 1)
            res = ll(res) * a % mod;
		p >>= 1;
        a = ll(a) * a % mod;
	}
	return res;
}

int Inv(int x)
{
    return qpow(x, mod - 2);
}

int main()
{
	scanf("%d %d", &n, &m);
	for (int i = 0; i < n; i++)
		scanf("%d", &A[i]);
	for (int i = 0; i < n; i++)
		scanf("%d", &B[i]);

	int one = Inv(m);
	int two = Inv(ll(m) * m % mod);
	int larger = ll(m) * (m - 1) / 2 % mod;
	int cur = 1;

	for (int i = 0; i < n; i++)
        {
		if (A[i] == 0 && B[i] == 0)
                {
			res = (res + ll(cur) * ll(larger) % mod * two) % mod;
			cur = ll(cur) * m % mod * two % mod;//表示到此处可能相等的概率
		}
		else if (A[i] != 0 && B[i] == 0)
                {
			res = (res + ll(cur) * ll(A[i] - 1) % mod * one) % mod;
			cur = ll(cur) * one % mod;
		}
		else if (A[i] == 0 && B[i] != 0)
                {
			res = (res + ll(cur) * ll(m - B[i]) % mod * one) % mod;
			cur = ll(cur) * one % mod;
		}
		else if (A[i] > B[i])
                {
                    res = (res + cur) % mod;
                    break;
                }
                else if (A[i] < B[i])
                    break;
        }
	printf("%d\n", res);
	return 0;
}