BNUOJ 34985 Elegant String 2014 Beijing Invitational competition question E Dynamic Planning Matrix Quick power

Elegant StringTime Limit: 1000 msMemory Limit: 65536KB64-bit integer IO format: % lld Java class name: Main

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 ). inputInput 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. outputFor 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 Input21 17 6 Sample OutputCase #1: 2 Case #2: 818503Source2014 ACM-ICPC Beijing Invitational Programming Contest question

When I was playing in Beijing, I read the wrong question... The meaning is that a string with n length only uses (0, 1, 2,..., k) Numbers (k + 1. If none of the substrings in this string appear (0, 1, 2 ,..., k. Ask how many elegant strings are in all strings whose lengths are n (k + 1) digits.

For example, a string ("112345678910") is an elegant string, but a string ("963852741023") is not an elegant string, because the latter has a substring ("9638527410 ") is an arrangement.

The correct idea is to change the direction of thinking! Try to construct a string instead of directly using the exclusion principle.

Set n and k to 6 and 3 respectively.

Suppose this string is an elegant string, then all the substrings of this string with a length of 4 (= k + 1) must not appear (0, 1, 2, 3).

Let me assume that I have constructed a substring ("023"). If I want this string to be an elegant one, the next character will not be "1 ". Because once "1" is obtained, there is an arrangement...

Then I thought about it. If the substring is ("02"), there are two cases. The first one is to select a dangerous string with a length of three from 1 and 3. Why is it a dangerous string? Because the next character must not be 1 to ensure that the entire string is an elegant string. Of course, the second case is to select 0 and 2, which is equivalent to the fact that this sub-string is secure (almost secure.

So it is a planning problem.

When dp [I] [j] is used to indicate the number of j-bit security when the string is increased to the I-bit.

Transfer equation:

Therefore, it is the calculation of the Rapid power of the matrix.

Now that we are all at this step, the matrix is very easy to write: <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> VcD4KPHA + pgltzybzcm9 "http://www.2cto.com/uploadfile/Collfiles/20140523/20140523091449228.gif" alt = "\">

The final result is

Res is the answer.

Sample Code

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     #include using namespace std;typedef long long int64;const int MAXN = 11;const int MAXM = 11;const int Mod = 20140518;struct Matrax{    int n,m;    int64 mat[MAXN][MAXM];    Matrax():n(-1),m(-1){}    Matrax(int _n,int _m):n(_n),m(_m){        memset(mat,0,sizeof(mat));    }    void Unit(int _s){        n=_s; m=_s;        for (int i = 0; i < n; i++){            for (int j = 0; j < n; j++){                mat[i][j] = (i == j)? 1: 0;            }        }    }    void print(){        printf("n = %d, m =  %d\n", n, m);        for (int i = 0; i < n; i++){            for (int j = 0; j < m; j++)                printf("%8d", mat[i][j]);            printf("\n");        }    }};Matrax add_mod(const Matrax& a,const Matrax& b,const int64 mod){    Matrax ans(a.n,a.m);    for (int i = 0; i < a.n; i++){        for (int j = 0; j < a.m; j++){            ans.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % mod;        }    }    return ans;}Matrax mul(const Matrax& a,const Matrax& b){    Matrax ans(a.n, b.m);    for (int i = 0; i < a.n; i++){        for (int j = 0; j < b.m; j++){            int64 tmp = 0;            for (int k = 0; k < a.m; k++){                tmp += a.mat[i][k] * b.mat[k][j];            }            ans.mat[i][j] = tmp;        }    }    return ans;}Matrax mul_mod(const Matrax& a, const Matrax& b, const int mod){    Matrax ans(a.n, b.m);    for (int i = 0; i < a.n; i++){        for (int j = 0; j < b.m; j++){            int64 tmp = 0;            for (int k = 0; k < a.m; k++){                tmp += (a.mat[i][k] * b.mat[k][j]) % mod;            }            ans.mat[i][j] = tmp % mod;        }    }    return ans;}Matrax pow_mod(const Matrax& a, int64 k, const int mod){    Matrax p(a.n,a.m), ans(a.n,a.m);    p = a; ans.Unit(a.n);    if (k==0) return ans;    else if (k==1) return a;    else {        while (k){            if (k & 1){                ans=mul_mod(ans, p, mod);                k--;            }            else {                k /= 2;                p = mul_mod(p, p, mod);            }        }        return ans;    }}int64 n;int  k, t, tt;void solve(){    cin >> n >> k;    Matrax ans(k, 1);    //tmp = cef ^ (n - 1);    //ans = tmp * beg;    //res = ans.mat[0][0];    Matrax cef(k, k);    for (int i = 0; i < k; i++)        for (int j = 0; j <= i; j++)            cef.mat[i][j] = 1;    for (int i = 0; i < k - 1; i++)        cef.mat[i][i + 1] = k - i;    //cef.print();    Matrax beg(k, 1);    for (int i = 0; i < k; i++)        beg.mat[i][0] = k + 1;    Matrax tmp(k, k);    tmp = pow_mod(cef, n - 1, Mod);    //tmp.print();    ans = mul_mod(tmp, beg, Mod);    int res = ans.mat[0][0];    printf("Case #%d: %d\n", ++tt, res);}int main(){    cin >> t;    while (t--) solve();    return 0;}