題目連結：
http://codeforces.com/problemset/problem/621/E
E. Wet Shark and Blocks
time limit per test2 seconds
memory limit per test256 megabytes
#### 問題描述
> There are b blocks of digits. Each one consisting of the same n digits, which are given to you in the input. Wet Shark must choose exactly one digit from each block and concatenate all of those digits together to form one large integer. For example, if he chooses digit 1 from the first block and digit 2 from the second block, he gets the integer 12.
>
> Wet Shark then takes this number modulo x. Please, tell him how many ways he can choose one digit from each block so that he gets exactly k as the final result. As this number may be too large, print it modulo 109 + 7.
> Note, that the number of ways to choose some digit in the block is equal to the number of it's occurrences. For example, there are 3 ways to choose digit 5 from block 3 5 6 7 8 9 5 1 1 1 1 5.
#### 輸入
> The first line of the input contains four space-separated integers, n, b, k and x (2 ≤ n ≤ 50 000, 1 ≤ b ≤ 109, 0 ≤ k < x ≤ 100, x ≥ 2) — the number of digits in one block, the number of blocks, interesting remainder modulo x and modulo x itself.
> The next line contains n space separated integers ai (1 ≤ ai ≤ 9), that give the digits contained in each block.
#### 輸出
> Print the number of ways to pick exactly one digit from each blocks, such that the resulting integer equals k modulo x.
####樣例輸入
> 12 1 5 10
> 3 5 6 7 8 9 5 1 1 1 1 5
樣例輸出
3
題意
給你n個數ai(ai>=1&&ai<=9)，你每次要在其中選一個數，可以重複選，你現在要取b次，将選出來的數按選擇的順序組成一個b位的整數，現在問要使最後的結果%x==k，總共有多少種選法。
題解
dp[i][j]表示選出來的前i個數拼成的數%x==j的一共有多少種，則容易得到狀态轉移表達式：dp[i][(k10+j)%10]+=dp[i-1][k]cntv[j](cntv[j]表示n個數中等于j的有多少個）。
b有10^9，明顯是需要矩陣加速一下！！！
代碼
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
const int maxn=111;
const int mod=1e9+7;
struct Matrix {
LL mat[maxn][maxn];
Matrix() { memset(mat, 0, sizeof(mat)); }
friend Matrix operator *(const Matrix& A, const Matrix& B);
friend Matrix operator +(const Matrix &A,const Matrix &B);
friend Matrix pow(Matrix A, int n);
};
Matrix I;
Matrix operator +(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
ret.mat[i][j] = (A.mat[i][j] + B.mat[i][j])%mod;
}
}
return ret;
}
Matrix operator *(const Matrix& A, const Matrix& B) {
Matrix ret;
for (int i = 0; i < maxn; i++) {
for (int j = 0; j < maxn; j++) {
for (int k = 0; k < maxn; k++) {
ret.mat[i][j] = (ret.mat[i][j]+A.mat[i][k] * B.mat[k][j]) % mod;
}
}
}
return ret;
}
Matrix pow(Matrix A, int n) {
Matrix ret=I;
while (n) {
if (n & 1) ret = ret*A;
A = A*A;
n /= 2;
}
return ret;
}
int n,m,k,mo;
LL cntv[11];
void solve(){
///狀态轉移矩陣
Matrix A;
for(int j=0;j<mo;j++){
for(int dig=1;dig<=9;dig++){
int i=(j*10+dig)%mo;
A.mat[i][j]+=cntv[dig];
}
}
///初始向量
Matrix vec;
for(int dig=1;dig<=9;dig++){
vec.mat[dig%mo][0]+=cntv[dig];
}
vec=pow(A,m-1)*vec;
prf("%I64d\n",vec.mat[k][0]);
}
void init(){
///機關矩陣
for(int i=0;i<maxn;i++) I.mat[i][i]=1;
clr(cntv,0);
}
int main() {
init();
scf("%d%d%d%d",&n,&m,&k,&mo);
for(int i=1;i<=n;i++){
int x; scf("%d",&x);
cntv[x]++;
}
solve();
return 0;
}
//end-----------------------------------------------------------------------