Bnu oj 34985 elegant string (matrix + dp)

Elegant string
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 (random SIVE ). 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" Case # X:", And X is the case number. Then output function (n, k) mod 20140518 in this case. sample input

21 17 6

Sample output

Case #1: 2Case #2: 818503

Source 2014 ACM-ICPC Beijing Invitational Programming Contest

Question: 0 ~ The numbers k construct a sequence with a length of N, so that any subsequence is not arranged by K and the number of seeds is required.
Idea: DP [I] [J] indicates that the length is I, followed by J different types of numbers, and there are two cases for filling in the following: 1. it is different from the last J. The K + 1-j filling method is used to obtain the status DP [I + 1] [J + 1]. 2. same as the previous J, obtain DP [I + 1] [1], DP [I + 1] [2]... DP [I + 1] [J]. Since N is very large and K is very small, we can use a matrix to accelerate it. By constructing a recursive relationship of one row, we can use a matrix to quickly power the matrix.
Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#define maxn 205#define MAXN 200005#define INF 0x3f3f3f3f#define mod 20140518#define eps 1e-6const double pi=acos(-1.0);typedef long long ll;using namespace std;ll n,k,ans;struct Matrix{    int row,col;    ll v[15][15];    Matrix operator*(Matrix &tt)    {        int i,j,k;        Matrix temp;        temp.row=row;        temp.col=tt.col;        for(i=1; i<=row; i++)            for(j=1; j<=tt.col; j++)            {                temp.v[i][j]=0;                for(k=1; k<=col; k++)                {                    temp.v[i][j]+=v[i][k]*tt.v[k][j];                    temp.v[i][j]%=mod;                }            }        return temp;    }};Matrix pow_mod(Matrix x,ll i) // x^i{    Matrix tmp;    tmp.row=x.row; tmp.col=x.col;    memset(tmp.v,0,sizeof(tmp.v));    for(int j=1;j<=x.row;j++) tmp.v[j][j]=1;    while(i)    {        if(i&1) tmp=tmp*x;        x=x*x;        i>>=1;    }    return tmp;}void solve(){    int i,j;    Matrix A,res,x;    x.row=1; x.col=k;    res.row=1; res.col=k;    memset(x.v,0,sizeof(x.v));    x.v[1][1]=k+1;    A.row=A.col=k;    memset(A.v,0,sizeof(A.v));    for(i=1;i<=k;i++)    {        A.v[i-1][i]=k-i+2;        for(j=i;j<=k;j++)        {            A.v[j][i]=1;        }    }    res=pow_mod(A,n-1);    res=x*res;    ans=0;    for(i=1;i<=k;i++)    {        ans+=res.v[1][i];        ans%=mod;    }}int main(){    int i,j,test,ca=0;    scanf("%d",&test);    while(test--)    {        scanf("%lld%lld",&n,&k);        solve();        printf("Case #%d: %lld\n",++ca,ans);    }    return 0;}

Bnu oj 34985 elegant string (matrix + dp)