CF 316E3(Summer Homework-广意Fib数列在p,q=1时的性质-Fib线段树)

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本文解析了一道复杂的算法题目，介绍了如何通过构建特殊的数据结构和运用斐波那契数列特性来高效处理一系列更新与查询操作。

By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:

You are given a sequence of integers a₁, a₂, ..., a_n. Your task is to perform on it m consecutive operations of the following type:

For given numbers x_i and v_i assign value v_i to element a_{x_i}.
For given numbers l_i and r_i you've got to calculate sum , where f₀ = f₁ = 1 and at i ≥ 2: f_i = f_i - 1 + f_i - 2.
For a group of three numbers l_i r_i d_i you should increase value a_x by d_i for all x (l_i ≤ x ≤ r_i).

Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 2·10⁵) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁵). Then follow m lines, each describes an operation. Each line starts with an integer t_i (1 ≤ t_i ≤ 3) — the operation type:

if t_i = 1, then next follow two integers x_i v_i (1 ≤ x_i ≤ n, 0 ≤ v_i ≤ 10⁵);
if t_i = 2, then next follow two integers l_i r_i (1 ≤ l_i ≤ r_i ≤ n);
if t_i = 3, then next follow three integers l_i r_i d_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ d_i ≤ 10⁵).

The input limits for scoring 30 points are (subproblem E1):

It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.

The input limits for scoring 70 points are (subproblems E1+E2):

It is guaranteed that there will be queries of the 1-st and 2-nd type only.

The input limits for scoring 100 points are (subproblems E1+E2+E3):

No extra limitations.

Output

For each query print the calculated sum modulo 1000000000 (10⁹).

Sample test(s)

input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5

output
12
32
8
50

input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5

output
12
45

艰难滴完成了E题……感觉不会再爱了

好吧，题目那个解法由于智商太低没看懂……

说一下我最后乱推的结果

设S(0)=F1a1+F2a2+..+Fnan

+)S(1)=F2a2+F3a3+..+F3a3

= (F1+F2)a2+(F2+F3)a3+..+(Fn-1+Fn)an

= F3 a3+ F4 a3+..+ Fn+1 an

于是我XXXXX……

只要存S0和S1就行了 Sx可以用关于S0和S1的表达式表示：Sx=Fn-2*S0 +Fn-1*S1 (系数为什么是Fib数？因为Sx=Sx-1+Sx-2的不明公式,可用数学归纳法证)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MAXN (200000+10)
#define F (1000000000)
#define Lson (x<<1)
#define Rson ((x<<1)+1)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
//void inv(long long a,long long b){return exgcd(a,}
int n,m;
long long f[MAXN],sf[MAXN]={0},f2[MAXN],sf2[MAXN];
struct arr_t
{
    long long a[MAXN*4],s0[MAXN*4],s1[MAXN*4],b[MAXN*4];
    arr_t(){MEM(a)MEM(s0)MEM(s1)MEM(b)}
    long long S(int x,int n) //求的是S(x)
    {
        if (n==1) return s0[x];
        if (n==2) return s1[x];
        return add(mul(f[n-2],s0[x]),mul(f[n-1],s1[x]));
    }
    void update(int x,int l,int r)
    {
        int n=r-l+1,m=(l+r)>>1,ls=m-l+1,rs=n-ls;
        s0[x]=(s0[Lson]+S(Rson,ls+1))%F;
        s1[x]=(s1[Lson]+S(Rson,ls+2))%F;
    }
    void change(int x,int n,int c)
    {
        s0[x]=add(s0[x],mul(sf[n],c));
        s1[x]=add(s1[x],mul(sf2[n],c));
    }
    void pushdown(int x,int l,int r)
    {
        if (l==r) return;
        else if (b[x])
        {
            int n=r-l+1,m=(l+r)>>1,ls=m-l+1,rs=n-ls;
            s0[Lson]=add(s0[Lson],mul(sf[ls],b[x]));
            s0[Rson]=add(s0[Rson],mul(sf[rs],b[x]));
            s1[Lson]=add(s1[Lson],mul(sf2[ls],b[x]));
            s1[Rson]=add(s1[Rson],mul(sf2[rs],b[x]));
            b[Lson]+=b[x];b[Rson]+=b[x];b[x]=0;
        }
    }
    void ins(int x,int l,int r,int L,int R,int c)
    {
        pushdown(x,l,r);
        int n=r-l+1,m=(l+r)>>1;
        if (L<=l&&r<=R) {b[x]+=c;change(x,n,c);return;}
        if (L<=m) ins(Lson,l,m,L,R,c);
        if (m<R) ins(Rson,m+1,r,L,R,c);
        update(x,l,r);
    }
    void modify(int x,int l,int r,int L,int c)
    {
        pushdown(x,l,r);
        int n=r-l+1,m=(l+r)>>1;
        if (l==r) {s0[x]=s1[x]=c;return;}
        if (L<=m) modify(Lson,l,m,L,c);
        else modify(Rson,m+1,r,L,c);
        update(x,l,r);
    }

    void build(int x,int l,int r)
    {
        if (l==r) {scanf("%d",&a[x]);s0[x]=s1[x]=a[x]; return ;}
        int n=r-l+1,m=(l+r)>>1;
        build(Lson,l,m);
        build(Rson,m+1,r);
        update(x,l,r);
    }
    long long qur(int x,int l,int r,int L,int R)
    {
        pushdown(x,l,r);
        int n=r-l+1,m=(l+r)>>1;
        if (L<=l&&r<=R) {return S(x,l-(L-1));}
        long long ans=0;
        if (L<=m) ans=add(ans,qur(Lson,l,m,L,R));
        if (m<R) ans=add(ans,qur(Rson,m+1,r,L,R));
        return ans;
    }
}T;
int main()
{
	//freopen("homework.in","r",stdin);
	//freopen(".out","w",stdout);
    scanf("%d%d",&n,&m);
    f[1]=f[2]=sf[1]=1,sf[2]=2;Fork(i,3,n) f[i]=(f[i-1]+f[i-2])%F,sf[i]=(sf[i-1]+f[i])%F;
    f2[1]=sf2[1]=1,f2[2]=2;sf2[2]=3;Fork(i,3,n) f2[i]=(f2[i-1]+f2[i-2])%F,sf2[i]=(sf2[i-1]+f2[i])%F;

    T.build(1,1,n);
    For(i,m)
    {
        int p,x,v,l,r,d;scanf("%d",&p);
        if (p==1)
        {
            scanf("%d%d",&x,&v);T.modify(1,1,n,x,v);
        }
        else if (p==2)
        {
            scanf("%d%d",&l,&r);cout<<T.qur(1,1,n,l,r)<<endl;
        }
        else
        {
            scanf("%d%d%d",&l,&r,&d);T.ins(1,1,n,l,r,d);

        }
    }


	return 0;
}