codeforces 719E E. Sasha and Array(线段树)

最新推荐文章于 2024-08-31 12:42:46 发布

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最新推荐文章于 2024-08-31 12:42:46 发布

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原文链接：http://www.cnblogs.com/zhangchengc919/p/5911293.html

本文介绍了一个关于数组操作的问题，包含两种操作：一是增加指定区间内所有整数的值；二是查询区间内整数之和对应的斐波那契数列值。通过线段树维护矩阵来高效解决此问题。

题目链接：

E. Sasha and Array

Sasha has an array of integers a1, a2, ..., an. You have to perform m queries. There might be queries of two types:

1 l r x — increase all integers on the segment from l to r by values x;
2 l r — find $\text{[math]}$ , where f(x) is the x-th Fibonacci number. As this number may be large, you only have to find it modulo109 + 7.

In this problem we define Fibonacci numbers as follows: f(1) = 1, f(2) = 1, f(x) = f(x - 1) + f(x - 2) for all x > 2.

Sasha is a very talented boy and he managed to perform all queries in five seconds. Will you be able to write the program that performs as well as Sasha?

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of elements in the array and the number of queries respectively.

The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Then follow m lines with queries descriptions. Each of them contains integers tpi, li, ri and may be xi (1 ≤ tpi ≤ 2, 1 ≤ li ≤ ri ≤ n,1 ≤ xi ≤ 109). Here tpi = 1 corresponds to the queries of the first type and tpi corresponds to the queries of the second type.

It's guaranteed that the input will contains at least one query of the second type.

Output

For each query of the second type print the answer modulo 109 + 7.

Examples

input

output

5
7
9

题意：

两个操作,1是把这个区间里的数都加x,2是求这个区间的和函数和,函数是斐波那契数列;

思路：

显然是一个线段树的题,不过维护的是矩阵,具体的可以看题解,写的太挫,跑了2000+ms;

AC代码：

#include <bits/stdc++.h>
#define lson o<<1
#define rson o<<1|1
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
const LL mod=1e9+7;
LL a[maxn];
struct matrix
{
    LL a[2][2];
};
matrix add(matrix A,matrix B)
{
    matrix C;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            C.a[i][j]=A.a[i][j]+B.a[i][j];
            if(C.a[i][j]>=mod)C.a[i][j]-=mod;
        }
    }
    return C;
}
matrix mul(matrix A,matrix B)
{
    matrix C;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            C.a[i][j]=0;
            for(int k=0;k<2;k++)
            {
                C.a[i][j]+=A.a[i][k]*B.a[k][j];
                C.a[i][j]%=mod;
            }
        }
    }
    return C;
}
matrix pow_mod(LL x)
{
    matrix s,base;
    s.a[0][0]=s.a[1][1]=1;s.a[0][1]=s.a[1][0]=0;
    base.a[0][0]=base.a[0][1]=base.a[1][0]=1;base.a[1][1]=0;
    while(x)
    {
        if(x&1)s=mul(s,base);
        base=mul(base,base);
        x>>=1;
    }
    return s;
}

struct Tree
{
    int l,r,mark;
    matrix sum,fs;
}tr[4*maxn];

inline void pushup(int o)
{
    tr[o].sum=add(tr[lson].sum,tr[rson].sum);
}
inline void pushdown(int o)
{
    if(tr[o].mark)
    {
        tr[o].mark=0;tr[lson].mark=1;tr[rson].mark=1;
        tr[lson].sum=mul(tr[lson].sum,tr[o].fs);tr[rson].sum=mul(tr[rson].sum,tr[o].fs);
        tr[lson].fs=mul(tr[lson].fs,tr[o].fs);tr[rson].fs=mul(tr[rson].fs,tr[o].fs);
        tr[o].fs.a[0][0]=tr[o].fs.a[1][1]=1;tr[o].fs.a[1][0]=tr[o].fs.a[0][1]=0;
    }
}
void build(int o,int L ,int R)
{
    tr[o].l=L;tr[o].r=R;tr[o].mark=0;
    tr[o].fs.a[0][0]=tr[o].fs.a[1][1]=1;tr[o].fs.a[0][1]=tr[o].fs.a[1][0]=0;
    if(L>=R)
    {
        tr[o].sum=pow_mod(a[L]);
        return ;
    }
    int mid=(L+R)>>1;
    build(lson,L,mid);
    build(rson,mid+1,R);
    pushup(o);
}

LL query(int o,int L,int R)
{
    //cout<<o<<" "<<L<<" "<<R<<endl;
    if(L<=tr[o].l&&R>=tr[o].r)return tr[o].sum.a[0][0];
    int mid=(tr[o].l+tr[o].r)>>1;
    pushdown(o);
    LL ans=0;
    if(L<=mid)ans+=query(lson,L,R);
    if(R>mid)ans+=query(rson,L,R);
    pushup(o);
    return ans%mod;
}

void update(int o,int L,int R,matrix num)
{
    if(L<=tr[o].l&&R>=tr[o].r)
    {
        tr[o].fs=mul(tr[o].fs,num);
        tr[o].mark=1;
        tr[o].sum=mul(tr[o].sum,num);
        return ;
    }
    pushdown(o);
    int mid=(tr[o].l+tr[o].r)>>1;
    if(L<=mid)update(lson,L,R,num);
    if(R>mid)update(rson,L,R,num);
    pushup(o);
}
int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%I64d",&a[i]),a[i]--;
    build(1,1,n);
    int op,u,v;
    LL temp;
    while(m--)
    {
        scanf("%d",&op);
        if(op==1)
        {
            scanf("%d%d%I64d",&u,&v,&temp);
            matrix num=pow_mod(temp);
            update(1,u,v,num);
        }
        else 
        {
            scanf("%d%d",&u,&v);
            printf("%I64d\n",query(1,u,v));
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/zhangchengc919/p/5911293.html