湖南大学ACM程序设计新生杯大赛（同步赛）H-Yuanyuan Long and His Ballons【公式】_湖南大学acm程序设计新生杯大赛(同步赛)-优快云博客

本文链接：https://blog.youkuaiyun.com/Irish_Moonshine/article/details/78886379

探讨一种特定条件下的组合计数问题：给定数量的气球和颜色种类，气球排列成环并要求相邻气球颜色不同，求所有可能的染色方案总数，并通过代码实现解决方案。

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时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
Yuanyuan Long is a dragon like this picture?

I don’t know, maybe you can ask him. But I’m sure Yuanyuan Long likes ballons, he has a lot of ballons.
One day, he gets n white ballons and k kinds of pigment, and he thought a problem:
1. Number these ballons b1, b2, … , bi, …, to bn.
2. Link these ballons to a circle in order, and color these ballons.
3. He need to make sure that any neighbor ballons have different colors.
He wants to know how many solutions to solve this problem. Can you get the answer to him? The answer maybe very large, get the answer MOD 100000007.
For Example: with 3 ballons and 3 kinds of pigment

Your answer is 3*2*1 MOD 100000007 = 6.
The answer maybe large, you can use integer type long long.
输入描述:
The first line is the cases T. ( T <=
100)
For next T lines, each line contains n and
k. (2<= n <= 10000, 2<= k
<=100)
输出描述:
For each test case, output the answer on
each line.
示例1
输入

3
3 3
4 2
5 3
输出

6
2
30

题意：一个环上涂颜色，相邻的色不能相同。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long int
using namespace std;
const int maxn = 1e6 + 10;
const ll mod = 1e8 + 7;
int a[maxn];
int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        int n, k;
        scanf("%d%d", &n, &k);
        int tmp = k - 1;
        ll ans = 1;
        for (int i = 1; i <= n; i++) {
            tmp *= -1;
            (ans = 1ll * ans*(k - 1)) %= mod;
        }
        (ans += 1ll*tmp)%=mod;
        printf("%lld\n", ans);
    }
    return 0;
}