bnuoj 34985 Elegant String DP+矩阵快速幂

题目链接：http://acm.bnu.edu.cn/bnuoj/problem_show.php?pid=34985

We define a kind of strings as elegant string: among all the substrings of an elegant string, none of them is a permutation of "0, 1,…, k".

Let function(n, k) be the number of elegant strings of length n which only contains digits from 0 to k (inclusive). Please calculate function(n, k).

INPUT

Input starts with an integer T (T ≤ 400), denoting the number of test cases.

Each case contains two integers, n and k. n (1 ≤ n ≤ 10¹⁸) represents the length of the strings, and k (1 ≤ k ≤ 9) represents the biggest digit in the string

2

1 1

7 6

OUTPUT

For each case, first output the case number as "Case #x: ", and x is the case number. Then output function(n, k) mod 20140518 in this case.

Case #1: 2

Case #2: 818503

source:2014 ACM-ICPC Beijing Invitational Programming Contest

 

题意：首先定义一个串elegant string：在这个串里的任何子串都没有包括0～k的一个全排列，现在给出n和k，要求求出长度为n的elegant string的个数。

解法：DP+矩阵快速幂。我们先用DP推出递推公式，然后用矩阵快速幂解决这个公式。具体思想如下：

定义DP数组：dp[12][12]（dp[i][j]表示长度为 i 时字符串末尾连续有 j 个不同数字，例：1233末尾只有一个数字3，2234末尾有3个：2，3，4）。

以第二组数据为例：n=7   k=6

 

初始化：dp[1][1]=k+1，dp[1][2,,,k]=0;

dp[i+1][1]=dp[i][1]+dp[i][2]+dp[i][3]+dp[i][4]+dp[i][5]+dp[i][6]

dp[i+1][2]=6*dp[i][1]+dp[i][2]+dp[i][3]+dp[i][4]+dp[i][5]+dp[i][6]

dp[i+1][3]=5*dp[i][2]+dp[i][3]+dp[i][4]+dp[i][5]+dp[i][6]

dp[i+1][4]=4*dp[i][3]+dp[i][4]+dp[i][5]+dp[i][6]

dp[i+1][5]=3*dp[i][4]+dp[i][5]+dp[i][6]

dp[i+1][6]=2*dp[i][5]+dp[i][6]

把递推式改为如下矩阵：

 

然后我们就可以用快速幂来解决了

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 #include<cmath>

 6 #include<algorithm>

 7 #define inf 0x7fffffff

 8 using namespace std;

 9 typedef long long ll;

10 const int maxn=12;

11 const ll mod=20140518;

12 

13 ll n,k;

14 struct matrix

15 {

16     ll an[maxn][maxn];

17 }A,B;

18 matrix multiply(matrix x,matrix y)

19 {

20     matrix sum;

21     memset(sum.an,0,sizeof(sum.an));

22     for (int i=1 ;i<=k ;i++)

23     {

24         for (int j=1 ;j<=k ;j++)

25         {

26             for (int k2=1 ;k2<=k ;k2++) {

27                 sum.an[i][j]=(sum.an[i][j]+x.an[i][k2]*y.an[k2][j]%mod);

28                 if (sum.an[i][j]>=mod) sum.an[i][j] -= mod;

29             }

30         }

31     }

32     return sum;

33 }

34 matrix power(ll K,matrix q)

35 {

36     matrix temp;

37     for (int i=1 ;i<=k ;i++)

38     {

39         for (int j=1 ;j<=k ;j++)

40         temp.an[i][j]= i==j ;

41     }

42     while (K)

43     {

44         if (K&1) temp=multiply(temp,q);

45         q=multiply(q,q);

46         K >>= 1;

47     }

48     return temp;

49 }

50 int main()

51 {

52     int t,ncase=1;

53     scanf("%d",&t);

54     while (t--)

55     {

56         scanf("%lld%lld",&n,&k);

57         if (n==1)

58         {

59             printf("Case #%d: %d\n",ncase++,k+1);continue;

60         }

61         matrix q;

62         for (int i=1 ;i<=k ;i++)

63         {

64             for (int j=1 ;j<=k ;j++)

65             {

66                 if (j>=i) q.an[i][j]=(ll)1;

67                 else if (j==i-1) q.an[i][j]=(ll)(k+2-i);

68                 else q.an[i][j]=(ll)0;

69             }

70         }

71         q=power(n-1,q);

72 //        for (int i=1 ;i<=k ;i++)

73 //        {

74 //            for (int j=1 ;j<=k ;j++)

75 //            cout<<q.an[i][j]<<" ";

76 //            cout<<endl;

77 //        }

78         ll ans=0;

79         for (int i=1 ;i<=k ;i++)

80         {

81             ans=(ans+(ll)q.an[i][1]*(ll)(k+1))%mod;

82         }

83         printf("Case #%d: %lld\n",ncase++,ans);

84     }

85     return 0;

86 }

 

 

 

 

 

后续：感谢提出宝贵的意见。。。。