斐波拉契数列前n项和的求法
一、递推关系构建系数矩阵–矩阵快速幂
⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪S[n]=S[n−1]+fac[n]fac[n+1]=fac[n]+fac[n−1]fac[n]=fac[n] { S [ n ] = S [ n − 1 ] + f a c [ n ] f a c [ n + 1 ] = f a c [ n ] + f a c [ n − 1 ] f a c [ n ] = f a c [ n ]
系数矩阵如下：
⎡⎣⎢⎢⎢⎢⎢⎢S[n]fac[n+1]fac[n]⎤⎦⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢100111010⎤⎦⎥⎥⎥⎥⎥⎥∗⎡⎣⎢⎢⎢⎢⎢⎢S[n−1]fac[n]fac[n−1]⎤⎦⎥⎥⎥⎥⎥⎥ [ S [ n ] f a c [ n + 1 ] f a c [ n ] ] = [ 1 1 0 0 1 1 0 1 0 ] ∗ [ S [ n − 1 ] f a c [ n ] f a c [ n − 1 ] ]
递推得到：
⎡⎣⎢⎢⎢⎢⎢⎢S[n]fac[n+1]fac[n]⎤⎦⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢100111010⎤⎦⎥⎥⎥⎥⎥⎥n∗⎡⎣⎢⎢⎢⎢⎢⎢S[0]fac[1]fac[0]⎤⎦⎥⎥⎥⎥⎥⎥ [ S [ n ] f a c [ n + 1 ] f a c [ n ] ] = [ 1 1 0 0 1 1 0 1 0 ] n ∗ [ S [ 0 ] f a c [ 1 ] f a c [ 0 ] ]
⎡⎣⎢⎢⎢⎢⎢⎢S[0]fac[1]fac[0]⎤⎦⎥⎥⎥⎥⎥⎥=⎡⎣⎢⎢⎢⎢⎢⎢111⎤⎦⎥⎥⎥⎥⎥⎥ [ S [ 0 ] f a c [ 1 ] f a c [ 0 ] ] = [ 1 1 1 ]
所以我们需要求出系数矩阵的n次方，S[n] = 系数矩阵第一行的和
牛客OI赛制测试赛3 D
#include
}
} 二、线性递推式—杜教BM模板
⎧⎩⎨S[n]=S[n−1]+fac[n]fac[n]=fac[n−1]+fac[n−2] { S [ n ] = S [ n − 1 ] + f a c [ n ] f a c [ n ] = f a c [ n − 1 ] + f a c [ n − 2 ]
得到S[n] 为线性递推式！
输入前0~8项数：1,2,4,7,12,20,33,54,88；(n可以为0）
输入n，即可得到答案！
牛客OI赛制测试赛3 D
前面的1不能删去，因为n可以为0！
#include
#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=998244353;
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];
vectorfor (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)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)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 main() {
scanf("%lld",&n);
printf("%d\n",linear_seq::gao(VI{1,2,4,7,12,20,33,54,88},n));
} 三、斐波拉契n项和公式
S[n]=fac[n+2]−1 S [ n ] = f a c [ n + 2 ] − 1
证明：S[n]=fac0+fac1+fac2+fac3+……+facn=1+fac0+fac1+fac2+fac3+……+facn−1=fac1+fac0+fac1+fac2+fac3+……+facn−1=(fac1+fac0)+fac1+fac2+fac3+……+facn−1=(fac2+fac1)+fac2+fac3+……+facn−1=(fac3+fac2)+fac3+……+facn−1=(fac4+fac3)+……+facn−1……=facn+1+facn−1=facn+2−1(1)(2)(3)(4)(5)(6)(7)(8)(9)(10) (1) 证 明 ： S [ n ] = f a c 0 + f a c 1 + f a c 2 + f a c 3 + … … + f a c n (2) = 1 + f a c 0 + f a c 1 + f a c 2 + f a c 3 + … … + f a c n − 1 (3) = f a c 1 + f a c 0 + f a c 1 + f a c 2 + f a c 3 + … … + f a c n − 1 (4) = ( f a c 1 + f a c 0 ) + f a c 1 + f a c 2 + f a c 3 + … … + f a c n − 1 (5) = ( f a c 2 + f a c 1 ) + f a c 2 + f a c 3 + … … + f a c n − 1 (6) = ( f a c 3 + f a c 2 ) + f a c 3 + … … + f a c n − 1 (7) = ( f a c 4 + f a c 3 ) + … … + f a c n − 1 (8) … … (9) = f a c n + 1 + f a c n − 1 (10) = f a c n + 2 − 1
求第n+2项斐波拉契！
⎧⎩⎨fac[n]=fac[n−1]+fac[n−2]fac[n−1]=fac[n−1] { f a c [ n ] = f a c [ n − 1 ] + f a c [ n − 2 ] f a c [ n − 1 ] = f a c [ n − 1 ]
系数矩阵如下：
⎡⎣⎢fac[n]fac[n−1]⎤⎦⎥=⎡⎣⎢1110⎤⎦⎥∗⎡⎣⎢fac[n−1]fac[n−2]⎤⎦⎥ [ f a c [ n ] f a c [ n − 1 ] ] = [ 1 1 1 0 ] ∗ [ f a c [ n − 1 ] f a c [ n − 2 ] ]
递推得到：
⎡⎣⎢fac[n]fac[n−1]⎤⎦⎥=⎡⎣⎢1110⎤⎦⎥n−1∗⎡⎣⎢fac[1]fac[0]⎤⎦⎥ [ f a c [ n ] f a c [ n − 1 ] ] = [ 1 1 1 0 ] n − 1 ∗ [ f a c [ 1 ] f a c [ 0 ] ]
⎡⎣⎢fac[1]fac[0]⎤⎦⎥=⎡⎣⎢11⎤⎦⎥ [ f a c [ 1 ] f a c [ 0 ] ] = [ 1 1 ]
即fac[n] = 系数矩阵n-1方的第一行之和！
#include
}
}