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

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