Niu Ke—Happy Brushing

Niu Ke—Happy Brushing

Topic link: Happy painting

Title

There are n grids and m colors. It is necessary to find out the matching number of two adjacent grids with the same color.

Problem-solving ideas

To understand how many coating methods there are, it is obviously $m^n$ , then as long as you find all the coating methods with different adjacent colors, you can show that there are two adjacent grids with the same color.
How to express the color of any adjacent grid? It is very simple that you can choose any color for the first grid, and if the color of the second grid is different from that of the previous grid, there are only (m-1) colors to choose from, and so on, a total of n *$(m-1{)}^{n-1}$ coating method.
Then you can use fast power to calculate$m^n$,$(m-1{)}^{n-1}$ .
Upload code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
ll fastpow(ll a , ll n){
	ll res = 1, base = a % mod;//这里一定要取mod因为m可能会大于mod
	while(n){
		if(n & 1)
			res = (res * base) % mod;
		base = (base * base) % mod;
		n >>= 1;
	}
	return res;
}
int main(){
	ll n, m, x, y, ans;
	cin >> n >> m;
	x = fastpow(m, n) % mod;
	y = (m % mod * fastpow(m - 1, n - 1)) % mod;
	if(x - y > 0)//因为取mod了，所以有可能x < y ，小于的话加一个mod就ok
		ans = (x - y) % mod;
	else
		ans = (x - y + mod) % mod;
	cout << ans << endl;
	return 0;
}