CF GYM 100548 容斥原理+快速幂+逆元

Recently, Mr. Big recieved n owers from his fans. He wants to recolor those owers with m colors. The
owers are put in a line. It is not allowed to color any adjacent owers with the same color. Flowers i and
i + 1 are said to be adjacent for every i, 1  i < n. Mr. Big also wants the total number of different
colors of the n owers being exactly k.
Two ways are considered different if and only if there is at least one ower being colored with different
colors.
Input
The rst line of the input gives the number of test cases, T. T test cases follow. T is about 300 and in
most cases k is relatively small.
For each test case, there will be one line, which contains three integers n, m, k (1  n;m  109,
1  k  106, k  n;m).
Output
For each test case, output one line containing `Case #x: y', where x is the test case number (starting
from 1) and y is the number of ways of different coloring methods modulo 109 + 7.
Sample Input
2
3 2 2
3 2 1
Sample Output
Case #1: 2
Case #2: 0

题意：

n个格子排成一行，有m种颜色，问用恰好k种颜色进行染色，使得相邻格子颜色不同的方案数。 
k≤106n,m≤109

题解：

首先，我们可以从m个颜色中取出k个，即 C k m 。

接着容易想到

k ∗ ( k − 1 ) n − 1

, 这个是使用不超过k种颜色的所有方案。

但我们要求的是恰好使用k种颜色。

假设选出的k种颜色标号为1，2，3，…, k，那么记 A i 为 不使用

颜色i的方案数，求的就是

| S | − | A 1 ⋃ A 2 ⋃ ⋯ ⋃ A n |

。

也就是反过来考虑，我们不考虑用了哪些颜色，我们考虑哪些颜色没用！减去所有有没使用颜色的方案的并集，剩下的方案就是使用了所有k种颜色的方案。上式中的 | S| 即 k ∗ ( k − 1 ) n − 1 ，后者就可以用容斥原理来求了。注意到我们只是给颜色标了个号，所以后面每一项的应为 C i k ∗ ( k − i ) ∗ ( k − i − 1 ) n − 1 的形式，即选出i个不使用的颜色，用剩余颜色去涂的方案数。