P2461 [SDOI2008]递归数列 矩阵乘法+构造

还好$QwQ$

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思路：矩阵快速幂

提交：1次

题解：

如图：

注意$n,m$如果小于$k$就不要快速幂了，直接算就行、。。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define ull unsigned long long
#define ll long long
#define R register ll
using namespace std;
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
namespace Fread {
static char B[1<<15],*S=B,*D=B;
#ifndef JACK
#define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
#endif
inline ll g() {
    R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
    if(ch==EOF) return EOF; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=36||ch>=127);}
inline void gs(char* s) {
    register char ch; while(isempty(ch=getchar()));
    do *s++=ch; while(!isempty(ch=getchar()));
}
} using Fread::g; using Fread::gs;
namespace Luitaryi {
const int N=20;
int k,mod,b[N],c[N];
ll n,m,sumn,summ,sum[N];
ll ans[N],s[N],a[N][N],mem[N][N];
inline void mul(ll a[][N],ll b[][N]) {
    R tmp[N][N]; memset(tmp,0,sizeof(tmp));
    for(R i=1;i<=k+1;++i) for(R l=1;l<=k+1;++l) for(R j=1;j<=k+1;++j)
        tmp[i][j]=(tmp[i][j]+a[i][l]*b[l][j])%mod;
    memcpy(a,tmp,sizeof(tmp));
}
inline void qpow(ll p) {
    R ret[N][N]; memset(ret,0,sizeof(ret));
    for(R i=1;i<=k+1;++i) ret[i][i]=1;
    for(;p;p>>=1,mul(a,a)) if(p&1) mul(ret,a);
    memcpy(a,ret,sizeof(a));
}
inline void main() {
    k=g(); for(R i=1;i<=k;i++) b[i]=g();
    for(R i=1;i<=k;++i) c[i]=g();
    m=g()-k-1,n=g()-k,mod=g();
    const int M=mod;
    for(R i=1;i<=k;++i) sum[i]=(b[i]+sum[i-1])%M;
    for(R i=1;i<=k;++i) s[i]=b[i]%M; s[k+1]=sum[k]%M;
    for(R i=1;i<k;++i) a[i+1][i]=1;
    for(R i=1;i<=k;++i) a[i][k]=a[i][k+1]=c[k-i+1]%M; a[k+1][k+1]=1;
    if(n<=0) return (void)printf("%lld
",((sum[k+n-1]-sum[k+m-1])%M+M)%M);
    memcpy(mem,a,sizeof(a));
    qpow(n); for(R i=1;i<=k+1;++i) for(R j=1;j<=k+1;++j) ans[j]=(ans[j]+s[i]*a[i][j])%M; sumn=ans[k+1];
    if(m<=0) return (void)printf("%lld
",((sumn-sum[k+m])%M+M)%M);
    memset(ans,0,sizeof(ans)); memcpy(a,mem,sizeof(a));
    qpow(m); for(R i=1;i<=k+1;++i) for(R j=1;j<=k+1;++j) ans[j]=(ans[j]+s[i]*a[i][j])%M; summ=ans[k+1];
    printf("%lld
",((sumn-summ)%M+M)%M);
}
}
signed main() {
    Luitaryi::main();
    return 0;
}

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2019.07.21