北邀 E Elegant String

E. Elegant String

Time Limit: 1000ms

Case Time Limit: 1000ms

Memory Limit: 65536KB

 

64-bit integer IO format: %lld      Java class name: Main

Submit Status PID: 34985

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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.

 

 

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. 

 

 

Sample Input

2
1 1
7 6

 

Sample Output

Case #1: 2
Case #2: 818503

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;

const LL p = 20140518;
struct Matrix
{
    LL mat[11][11];
    void init()
    {
        memset(mat,0,sizeof(mat));
    }
    void first(int n)
    {
        int i,j;
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
                if(i==j)mat[i][j]=1;
        else mat[i][j]=0;
    }
}M_hxl,M_tom;
Matrix multiply(Matrix cur,Matrix ans,int n)
{
    Matrix now;
    now.init();
    int i,j,k;
    for(i=1;i<=n;i++)
    {
        for(k=1;k<=n;k++)
        {
            for(j=1;j<=n;j++)
            {
                now.mat[i][j]=now.mat[i][j]+cur.mat[i][k]*ans.mat[k][j];
                now.mat[i][j]%=p;
            }
        }
    }
    return now;
}
Matrix pow_mod(Matrix ans,LL n,LL m)
{
    Matrix cur;
    cur.first(m);
    while(n)
    {
        if(n&1) cur=multiply(cur,ans,m);
        n=n>>1;
        ans=multiply(ans,ans,m);
    }
    return cur;
}
LL solve(LL n,LL k)
{
    M_hxl.init();
    int i,j;
    for(i=1;i<=k-1;i++)M_hxl.mat[1][i]=1;
    for(i=2;i<=k-1;i++)
    {
        for(j=1;j<=k-1;j++)
        {
            if(i-j>1)continue;
            if(i-j==1) M_hxl.mat[i][j]=k-i+1;
            else M_hxl.mat[i][j]=1;
        }
    }/**ok**/
    M_tom = pow_mod(M_hxl,n,k-1);
    LL sum=(M_tom.mat[1][1]*k)%p;
    return sum;

}
int main()
{
    int T,t;
    LL n,k;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%lld%lld",&n,&k);
        k++;
        LL ans=solve(n,k);
        printf("Case #%d: %lld
",t,ans);
    }
    return 0;
}