CF621E Wet Shark and Blocks（矩阵加速DP）

题意：

n位数，有b组，从每一组中选取一位数，依次组成一个b位数，计算有多少种选法可以满足结果%x=k

题解：

推导出转移方程后用矩阵加速，这里构造矩阵的思路学习一位巨佬的写法，还是要多练练矩阵！

/*
 *author: zlc
 *zucc_acm_lab
 *just do it
 */
#include<bits/stdc++.h> 
using namespace std;
typedef long long ll;
const double pi=acos(-1.0);
const double eps=1e-6;
const int mod=1e9+7;
const int inf=1e9;
const int maxn=2e5+100;
inline ll read () {ll x=0,f=1;char ch=getchar();while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}while (ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}
ll qpow (ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll n,x,k,b;
struct matrix {
    ll m[105][105];
}ans,base;
void init () {
    memset(ans.m,0,sizeof(ans.m));
    memset(base.m,0,sizeof(base.m));
    for (int i=0;i<x;i++) ans.m[i][i]=1;
}
matrix mul (matrix a,matrix b) {
    matrix wjm;
    memset(wjm.m,0,sizeof(wjm.m));
    for (int i=0;i<x;i++)
        for (int j=0;j<x;j++) {
            wjm.m[i][j]=0;
            for (int k=0;k<x;k++)
                wjm.m[i][j]=(wjm.m[i][j]+a.m[i][k]*b.m[k][j]%mod)%mod;
        }
    return wjm;
}
void qpow (ll p) {
    while (p) {
        if (p&1) ans=mul(ans,base);
        base=mul(base,base);
        p>>=1;
    }
}
int cnt[maxn];
int main () {
    n=read(),b=read(),k=read(),x=read();
    for (int i=0;i<n;i++) {
        ll tt;
        tt=read();
        cnt[tt]++;
    }
    init();
    for (int i=0;i<x;i++)
        for (int j=0;j<=9;j++) {
            base.m[i][(i*10+j)%x]=(base.m[i][(i*10+j)%x]+cnt[j])%mod;//dp[i][j]表示从余数i到余数j的方法数 
        }
    qpow(b);
    printf("%lld
",ans.m[0][k]); 
}