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。。。。
 
posted @ 2014-12-19 12:45  huangxf  阅读(218)  评论(0)  编辑  收藏  举报