BZOJ 3231: [Sdoi2008]递归数列 (JZYZOJ 1353) 矩阵快速幂

http://www.lydsy.com/JudgeOnline/problem.php?id=3231

 

和斐波那契一个道理在最后加一个求和即可

#include<cstdio>
#include<cstring>
#include<iostream>
//using namespace std;
const int maxn=10010;
const double eps=1e-8;
long long modn;
long long n,l,r;
long long b[20]={};
struct mat{
    long long e[20][20];
    mat(){ memset(e,0,sizeof(e)); }
};
mat a;
mat Mul(mat x,mat y){
    mat z;
    for(int i=1;i<=n+1;i++){
        for(int j=1;j<=n+1;j++){
            for(int k=1;k<=n+1;k++){
                z.e[i][j]+=x.e[i][k]*y.e[k][j];
                z.e[i][j]%=modn;
            }
        }
    }
    return z;
}
mat Pow(mat x,long long k){
    mat z;
    for(int i=1;i<=n+1;i++){
        z.e[i][i]=1;
    }
    while(k>0){
        if(k&1){
            z=Mul(z,x);
        }
        x=Mul(x,x);
        k/=2;
    }/*for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            std::cout<<z.e[i][j]<<' ';
        }
        std::cout<<std::endl;
    }*/
    return z;
}
long long doit(long long x){
    if(x<n){
        return b[x+1];
    }
    mat z=Pow(a,x-n+1);
    long long ans=0,s=0,d=0;
    for(int i=1;i<=n+1;i++){
        d+=z.e[n][i]*b[i];
        s+=z.e[n+1][i]*b[i];
        d%=modn;s%=modn;
    }
    ans=(s+d)%modn;
    ans%=modn;
    return ans;
}
int main(){
    scanf("%lld",&n);
    n+=1;
    for(int i=2;i<=n;i++){
        scanf("%lld",&b[i]);
        b[n+1]+=b[i];
    }
    b[n+1]-=b[n];
    for(int i=n;i>1;i--){
        scanf("%lld",&a.e[n][i]);
    }
    long long l,r;
    scanf("%lld%lld%lld",&l,&r,&modn);
    for(int i=2;i<=n;i++){
        b[i]%=modn;
        a.e[i-1][i]=1;a.e[n][i]%=modn;
    }
    a.e[n+1][n+1]=1,a.e[n+1][n]=1;
    /*for(int i=1;i<=n+1;i++){
        for(int j=1;j<=n+1;j++){
            std::cout<<a.e[i][j]<<' ';
        }
        std::cout<<std::endl;
    }*/
    long long ans=(doit(r)-doit(l-1)+modn)%modn;
    printf("%lld
",ans);
    return 0;
}