Codeforces Round #341 (Div. 2) E. Wet Shark and Blocks (Matrix optimized DP)

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Test instructions: The number of N as a block, there is a B block, from which a number of options, each block can only select one number, and finally constitute a number modulo x equals K method number.

Idea: It is easy to think of such a DP equation: using DP[I][J] to represent the present I-bit, the remainder is J. Then Dp[i + 1][(J * + K)% x] = dp[i][j] * Cnt[k],k refers to which of the enumerations to put 1~9.

Because B is particularly large, obviously to matrix optimization, know the DP equation, in fact, the structure of the matrix is particularly simple. According to the rules of matrix multiplication, in fact this matrix is a[j][(J*10+a[k])%x]++, where A[k] is the K from the 1~n of each number.

In fact, it is very regular, the key is to construct a good DP equation.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long Long ll;const double PI = ACOs ( -1.0); const double EPS = 1e-6;const int INF = 10000 00000;const int mod = 1000000000 + 7;const int maxn = 50000 + 10;int t,n,q,u,bb,k,m,x,y,a[maxn];typedef vector<int>    Vec;typedef vector<vec> Mat;mat Mul (Mat &a, Mat &b) {Mat C (a.size (), VEC (A.size ())); for (int i=0;i<x;i++) {for (int. k=0;k<x;k++) {for (int j=0;j<x;j++) {C[i][j] = (            (ll) C[i][j] + (LL) a[i][k] * b[k][j])% MoD; }}} return C;} Mat Pow (Mat A, ll k) {Mat B (a.size (), VEC (a.size()));    for (int i=0;i<x;i++) {B[i][i] = 1;        } while (k > 0) {if (K & 1) b = Mul (b, a);        A = Mul (A, a);    K >>= 1; } return B;        int main () {while (~scanf ("%d%d%d%d", &n,&bb,&k,&x)) {Mat A (x+3, VEC (x+3));        for (int i=1;i<=n;i++) {scanf ("%d", &a[i]);            } for (int i=0;i<x;i++) {for (int j=1;j<=n;j++) {a[i][(I * + a[j])% x]++;        }} A = Pow (A, BB);    printf ("%d\n", A[0][k]); } return 0;}

Codeforces Round #341 (Div. 2) E. Wet Shark and Blocks (Matrix optimized DP)