Happy Doodle (combination mathematics)

Link: https://ac.nowcoder.com/acm/problem/14718
Source: Niuke

Title description

There are m colors and n grids, it is necessary to calculate the number of solutions of the same color in two adjacent grids

answer

First of all, I found that it is not easy to calculate directly, so consider the
total number of plans: $m^n$
consider that the first position has$m$ kinds of placement methods, the second position in order to keep the color of adjacent grids different, only$m-1$ method, and so on,nnThere are onlym − 1 m-1 in$n$ positions$m-1$ method of placement The
number of schemes with different adjacent grid colors:$m\times (m-1{)}^{(n-1)}$

Knowledge integration

Fast power

When my $a,When\; b\; is$ greater than my modulus, for$a$ modulo and$b$ Euler power

Code

#include<iostream>
using namespace std;

#define mod 1000000007
typedef long long ll;
const int N = 1e6 + 10;

ll qpow(ll a, ll b, ll p) {
    
    
	//当我的a, b大于我的p时
	//取模并欧拉降幂
	a %= p, b %= p - 1;
	ll res = 1;
	while (b) {
    
    
		if (b & 1) {
    
    
			res = (res * a) % p;
		}
		a = (a * a) % p;
		b >>= 1;
	}
	return res;
}

int main() {
    
    

	ll n, m;
	cin >> n >> m;

	ll ans = qpow(m, n, mod);
	//注意数据范围和取模技巧
	ans = (ans - m % mod * qpow(m - 1, n - 1, mod) % mod);
	//把答案映射到正整数
	ans = (ans % mod + mod) % mod;

	cout << ans << endl;

	return 0;
}