1360: Good Serial Inc.（不知道是什么类型的题）

1360: Good Serial Inc.

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Description

Good Serial Inc. (GSI) produces integer sequences with length N, in which every element is an integer in range [1, M].
They call a sequence S is good if the sequence has the following property: for every substring of S with length M, i.e., S[i→i+m-1], all the elements in the substring are the same or all the elements are distinct(different from each other).
The company GIS is designed to produce good sequences. But how many different good sequences are there? Since the answer will be very large, just output the result after module 987654321.

Input

There are several cases. For each case, there is a line with two integers N, and M ( 1 ≤ N ≤ 1000000, 1 ≤ M ≤ 1000000 ).
The input ends up with two negative numbers, which should not be processed as a case.

Output

Print the number of different good sequences module 987654321 in a line for each case.

Sample Input

4 4
3 5
-1 -1

Sample Output

28
125

Hint

Source

做法：
1.当m==1的时候，序列全部由相同的1组成
2.m==2的时候，对序列的每个位置都可以有两种情况，所以是m的n次方

注意m==2要放在n>=m的情况中讨论
3.n>=m m长度的子串全部相同的情况：m种，全1到m
    m长度的子串全部不同的情况：m的全排列 所以是m的阶乘
    所以是二种情况的总和
4.n

#include
#include
#include
#include
#include
#include
using namespace std;
#define mod 987654321
typedef long long LL;
LL f(int n,int m)
{
    LL ans=1;

    if(m==1)
        return 1;
    if(n>=m)
    {
        if(m==2)
        {
            for(int i=1;i<=n;i++)
                ans=ans*m%mod;//m的n次方
        }
        else
        {
            for(int i=1;i<=m;i++)
                ans=(ans*i)%mod;//m的阶乘
            ans+=m;
        }
        return ans;
    }
    else if(n<m)
    {
        for(int i=1;i<=n;i++)
            ans=ans*m%mod;//m的n次方
        return ans;
    }
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        if(n<0&&m<0)
            break;
        printf("%lld\n",f(n,m));
    }
    return 0;
}
/*
做法：
1.当m==1的时候，序列全部由相同的1组成
2.m==2的时候，对序列的每个位置都可以有两种情况，所以是m的n次方
3.n>=m m长度的子串全部相同的情况：m种，全1到m
    m长度的子串全部不同的情况：m的全排列 所以是m的阶乘
    所以是二种情况的总和
4.n*/