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

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #define inf 0x7fffffff
 using namespace std;
 typedef long long ll;
 const int maxn=;
 const ll mod=;

 ll n,k;
 struct matrix
 {
     ll an[maxn][maxn];
 }A,B;
 matrix multiply(matrix x,matrix y)
 {
     matrix sum;
     memset(sum.an,,sizeof(sum.an));
     for (int i= ;i<=k ;i++)
     {
         for (int j= ;j<=k ;j++)
         {
             for (int k2= ;k2<=k ;k2++) {
                 sum.an[i][j]=(sum.an[i][j]+x.an[i][k2]*y.an[k2][j]%mod);
                 if (sum.an[i][j]>=mod) sum.an[i][j] -= mod;
             }
         }
     }
     return sum;
 }
 matrix power(ll K,matrix q)
 {
     matrix temp;
     for (int i= ;i<=k ;i++)
     {
         for (int j= ;j<=k ;j++)
         temp.an[i][j]= i==j ;
     }
     while (K)
     {
         if (K&) temp=multiply(temp,q);
         q=multiply(q,q);
         K >>= ;
     }
     return temp;
 }
 int main()
 {
     int t,ncase=;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%lld%lld",&n,&k);
         if (n==)
         {
             printf("Case #%d: %d\n",ncase++,k+);continue;
         }
         matrix q;
         for (int i= ;i<=k ;i++)
         {
             for (int j= ;j<=k ;j++)
             {
                 if (j>=i) q.an[i][j]=(ll);
                 else if (j==i-) q.an[i][j]=(ll)(k+-i);
                 else q.an[i][j]=(ll);
             }
         }
         q=power(n-,q);
 //        for (int i=1 ;i<=k ;i++)
 //        {
 //            for (int j=1 ;j<=k ;j++)
 //            cout<<q.an[i][j]<<" ";
 //            cout<<endl;
 //        }
         ll ans=;
         for (int i= ;i<=k ;i++)
         {
             ans=(ans+(ll)q.an[i][]*(ll)(k+))%mod;
         }
         printf("Case #%d: %lld\n",ncase++,ans);
     }
     return ;
 }

 

 

 

 

后续：感谢提出宝贵的意见。。。。