Hdu 4695 线性递推模版(m^2logn)

#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long ll;
#define LL long long
const int mod=1e9+7;
const int MOD=1e9+7;
const int MAXN=205;
ll n;
int u, d;
int a[100], b[100];
ll dp[205];
ll A[205];
ll C[205];

// given first m a[i] and coef (0 based)
// calc a[n] % MOD in O(m*m*log(n))
// a[n] = sum(c[m - i] * a[n - i]) i = 1....m
// a[m] = sum(c[i] * a[i]), i = 0.....m-1

LL linear_recurrence(LL n, int m, LL a[], LL c[], int p){ //n->a[i], m -> c[i]
    LL v[MAXN] = {1 % MOD}, u[MAXN << 1], msk = !!n;
    for(LL i = n; i > 1; i >>= 1) msk <<= 1;
    for(LL x = 0; msk; msk >>= 1, x <<= 1){
        fill_n(u, m << 1, 0);
        int b = !!(n & msk); x |= b;
        if(x < m) u[x] = 1 % p;
        else{
            for(int i = 0; i < m; i++){
                for(int j = 0, t = i + b; j < m; ++j, ++t)
                    u[t] = (u[t] + v[i] * v[j]) % MOD;
            }
            for(int i = (m << 1) - 1; i >= m; --i){
                for(int j = 0, t = i - m; j < m; ++j, ++t){
                    u[t] = (u[t] + c[j] * u[i]) % MOD;
                }
            }
        }
        copy(u, u+m, v);
    }
    LL ans = 0;
    for(int i = 0; i < m; i++){
        ans = (ans + v[i] * a[i]) % MOD;
    }
    return ans;
}

bool vis[205];
int main()
{
    while(~scanf("%lld", &n)){
        int ma=0;
        memset(dp, 0, sizeof(dp));
        memset(A, 0, sizeof(A));
        memset(C, 0, sizeof(C));
        memset(vis, false, sizeof(vis));
        scanf("%d", &u);
        for(int i=1;i<=u;i++){scanf("%d", &a[i]);}
        scanf("%d", &d);
        for(int i=1;i<=d;i++){scanf("%d", &b[i]);ma=max(b[i], ma);vis[b[i]]=true;}
        dp[0]=1;
        for(int i=1;i<=ma;i++){
            for(int j=1;j<=u;j++){
                if(i<a[j])continue;
                dp[i]=(dp[i-a[j]]+dp[i])%mod;
            }
        }
        vis[0]=true;
        for(int i=0;i<=ma;i++)if(!vis[i])dp[i]=0;
        for(int i=0;i<ma;i++){C[i]=dp[ma-i];}
        A[0]=1;
        for(int i=1;i<ma;i++){
            for(int j=1;j<=i;j++){
                A[i]=(A[i]+A[i-j]*dp[j])%mod;
            }
        }
        ll ans=linear_recurrence(n, ma, A, C, mod);
        printf("%lld\n", ans);
    }
}

UPD：杜教BM模板无敌！此贴终结

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(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;}
// head

int _;
ll n;
namespace linear_seq {
    const int N=10010;
    ll res[N],base[N],_c[N],_md[N];

    vector<int> Md;
    void mul(ll *a,ll *b,int k) {
        rep(i,0,k+k) _c[i]=0;
        rep(i,0,k) if (a[i]) rep(j,0,k) _c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
        for (int i=k+k-1;i>=k;i--) if (_c[i])
            rep(j,0,SZ(Md)) _c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
        rep(i,0,k) a[i]=_c[i];
    }
    int solve(ll n,VI a,VI b) { // a 系数 b 初值 b[n+1]=a[0]*b[n]+...
        //        printf("%d\n",SZ(b));
        ll ans=0,pnt=0;
        int k=SZ(a);
        assert(SZ(a)==SZ(b));
        rep(i,0,k) _md[k-1-i]=-a[i];_md[k]=1;
        Md.clear();
        rep(i,0,k) if (_md[i]!=0) Md.push_back(i);
        rep(i,0,k) res[i]=base[i]=0;
        res[0]=1;
        while ((1ll<<pnt)<=n) pnt++;
        for (int p=pnt;p>=0;p--) {
            mul(res,res,k);
            if ((n>>p)&1) {
                for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
                rep(j,0,SZ(Md)) res[Md[j]]=(res[Md[j]]-res[k]*_md[Md[j]])%mod;
            }
        }
        rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
    VI BM(VI s) {
        VI C(1,1),B(1,1);
        int L=0,m=1,b=1;
        rep(n,0,SZ(s)) {
            ll d=0;
            rep(i,0,L+1) d=(d+(ll)C[i]*s[n-i])%mod;
            if (d==0) ++m;
            else if (2*L<=n) {
                VI T=C;
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                L=n+1-L; B=T; b=d; m=1;
            } else {
                ll c=mod-d*powmod(b,mod-2)%mod;
                while (SZ(C)<SZ(B)+m) C.pb(0);
                rep(i,0,SZ(B)) C[i+m]=(C[i+m]+c*B[i])%mod;
                ++m;
            }
        }
        return C;
    }
    int gao(VI a,ll n) {
        VI c=BM(a);
        c.erase(c.begin());
        rep(i,0,SZ(c)) c[i]=(mod-c[i])%mod;
        return solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
    }
};

int a[55],an;
int b[55],bn;
int dp[403];
int dp2[403];

int main() {
    while (~scanf("%lld",&n)) {
        memset(dp,0,sizeof(dp));
        memset(dp2,0,sizeof(dp2));
        scanf("%d",&an);
        for(int i=1;i<=an;i++){
            scanf("%d",&a[i]);
        }
        dp[0]=1;
        for(int i=1;i<=400;i++){
            for(int j=1;j<=an;j++){
                if(i>=a[j]){
                    dp[i]=(dp[i]+dp[i-a[j]])%mod;
                }
            }
        }
        scanf("%d",&bn);
        for(int i=1;i<=bn;i++){
            scanf("%d",&b[i]);
        }
        dp2[0]=1;
        for(int i=1;i<=400;i++){
            for(int j=1;j<=bn;j++){
                if(i>=b[j]){
                    dp2[i]=(dp2[i]+1ll*dp2[i-b[j]]*dp[b[j]])%mod;
                }
            }
        }
        vector<int>v;
        for(int i=0;i<=400;i++)v.push_back(dp2[i]);
        //VI{1,2,4,7,13,24}
        printf("%d\n",linear_seq::gao(v,n));
    }
}