Choosing number 写出递推公式，然后用矩阵的快速幂。

Description

There are n people standing in a row. And There are m numbers, 1.2...m. Every one should choose a number. But if two persons standing adjacent to each other choose the same number, the number shouldn't equal or less than k. Apart from this rule, there are no more limiting conditions.

And you need to calculate how many ways they can choose the numbers obeying the rule.

Input

There are multiple test cases. Each case contain a line, containing three integer n (2 ≤ n ≤ 10⁸), m (2 ≤ m ≤ 30000), k(0 ≤ k ≤ m).

Output

One line for each case. The number of ways module 1000000007.

Sample Input

4 4 1

Sample Output

216
***********************************************************************************************************************************************************
矩阵的快速幂dp[n][1]即取>=k的数时的方案总数，dp[n][2]即取<k时的方案总数
dp[n][1]=dp[n-1][1]*(m-k)+dp[n-1][2]*(m-k);
dp[n][2]=dp[n-1]*k+dp[n-1][2]*(k-1);
列出矩阵
： m-k   m-k     dp[n-1][1]   dp[n][1]
   k     k-1     dp[n-1][2]   dp[n][2]
可得由矩阵的快速幂快速求解
***********************************************************************************************************************************************************

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstdio>
 6 #include<algorithm>
 7 #define LL long long
 8 #define MOD 1000000007
 9 using namespace std;
10 LL n,m,k;
11 LL save[5][5];
12 LL mod(LL m,LL n)
13 {
14     LL sum=1;
15     while(n)
16     {
17         if(n&1)
18            sum=(sum*m)%MOD;
19         m=(m*m)%MOD;
20         n=n>>1;
21     }
22     return sum;
23 }
24 int main()
25 {
26     while(scanf("%lld%lld%lld",&n,&m,&k)!=EOF)
27     {
28         if(k==0)
29         {
30             printf("%lld\n",mod(m,n)%MOD);
31             continue;
32         }
33         if(n==1)
34         {
35             printf("%lld\n",m);
36             continue;
37         }
38         save[1][1]=m-k;
39         save[1][2]=m-k;
40         save[2][1]=k;
41         save[2][2]=k-1;
42         LL ans1=m-k;
43         LL ans2=k;
44         n--;
45         while(n)
46         {
47             if(n&1)
48             {
49                 LL temp1=(((ans1*save[1][1])%MOD)+((ans2*save[1][2])%MOD))%MOD;
50                 LL temp2=(((ans1*save[2][1])%MOD)+((ans2*save[2][2])%MOD))%MOD;
51                 ans1=temp1;
52                 ans2=temp2;
53             }
54             n=n>>1;
55             LL a1=(((save[1][1]*save[1][1])%MOD)+((save[1][2]*save[2][1])%MOD))%MOD;
56             LL a2=(((save[1][2]*save[2][2])%MOD)+((save[1][1]*save[1][2])%MOD))%MOD;
57             LL b1=(((save[2][2]*save[2][1])%MOD)+((save[2][1]*save[1][1])%MOD))%MOD;
58             LL b2=(((save[2][2]*save[2][2])%MOD)+((save[2][1]*save[1][2])%MOD))%MOD;
59             save[1][1]=a1; save[1][2]=a2;
60             save[2][1]=b1; save[2][2]=b2;
61 
62         }
63         printf("%lld\n",(ans1+ans2)%MOD);
64     }
65     return 0;
66 }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3446454.html