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=12;
const ll mod=20140518;

ll n,k;
struct matrix
{
    ll an[maxn][maxn];
}A,B;
matrix multiply(matrix x,matrix y)
{
    matrix sum;
    memset(sum.an,0,sizeof(sum.an));
    for (int i=1 ;i<=k ;i++)
    {
        for (int j=1 ;j<=k ;j++)
        {
            for (int k2=1 ;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=1 ;i<=k ;i++)
    {
        for (int j=1 ;j<=k ;j++)
        temp.an[i][j]= i==j ;
    }
    while (K)
    {
        if (K&1) temp=multiply(temp,q);
        q=multiply(q,q);
        K >>= 1;
    }
    return temp;
}
int main()
{
    int t,ncase=1;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%lld%lld",&n,&k);
        if (n==1)
        {
            printf("Case #%d: %d
",ncase++,k+1);continue;
        }
        matrix q;
        for (int i=1 ;i<=k ;i++)
        {
            for (int j=1 ;j<=k ;j++)
            {
                if (j>=i) q.an[i][j]=(ll)1;
                else if (j==i-1) q.an[i][j]=(ll)(k+2-i);
                else q.an[i][j]=(ll)0;
            }
        }
        q=power(n-1,q);
//        for (int i=1 ;i<=k ;i++)
//        {
//            for (int j=1 ;j<=k ;j++)
//            cout<<q.an[i][j]<<" ";
//            cout<<endl;
//        }
        ll ans=0;
        for (int i=1 ;i<=k ;i++)
        {
            ans=(ans+(ll)q.an[i][1]*(ll)(k+1))%mod;
        }
        printf("Case #%d: %lld
",ncase++,ans);
    }
    return 0;
}

 

 

 

 

后续：感谢提出宝贵的意见。。。。