Codeforces Round 718C（维护矩阵的线段树）
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
, where f(x) is the x-th Fibonacci number. As this number may be large, you only have to find it modulo 109 + 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?
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.
For each query of the second type print the answer modulo 109 + 7.
5 4 1 1 2 1 1 2 1 5 1 2 4 2 2 2 4 2 1 5
5 7 9
Initially, array a is equal to 1, 1, 2, 1, 1.
The answer for the first query of the second type is f(1) + f(1) + f(2) + f(1) + f(1) = 1 + 1 + 1 + 1 + 1 = 5.
After the query 1 2 4 2 array a is equal to 1, 3, 4, 3, 1.
The answer for the second query of the second type is f(3) + f(4) + f(3) = 2 + 3 + 2 = 7.
The answer for the third query of the second type is f(1) + f(3) + f(4) + f(3) + f(1) = 1 + 2 + 3 + 2 + 1 = 9.
题目大意：
有一棵线段树，查询以每个端点为下标值的斐波那契值的和。
解题思路：
一般情况下，求任意斐波那契数我们使用的是矩阵快速幂，这题是一个对于斐波那契数的线段树，所以我们在线段树上维护矩阵即可。
AC代码（用了Q巨的模版）：
#include
#include
#include
#include
#include
using namespace std;
#define MOD 1000000007
const int MAXN=100000+3;
int a[MAXN];
struct Matrix
{
int a[2][2];//矩阵大小根据需求修改
Matrix()
{
memset(a,0,sizeof(a));
}
void init()
{
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
a[i][j]=(i==j);
}
Matrix operator + (const Matrix &B)const
{
Matrix C;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
C.a[i][j]=(a[i][j]+B.a[i][j])%MOD;
return C;
}
Matrix operator * (const Matrix &B)const
{
Matrix C;
for(int i=0;i<2;i++)
for(int k=0;k<2;k++)
for(int j=0;j<2;j++)
C.a[i][j]=(C.a[i][j]+1LL*a[i][k]*B.a[k][j])%MOD;
return C;
}
Matrix operator ^ (const int &t)const
{
Matrix A=(*this),res;
res.init();
int p=t;
while(p)
{
if(p&1)res=res*A;
A=A*A;
p>>=1;
}
return res;
}
};
Matrix f,s[MAXN<<2],lz[MAXN<<2];
void pushUp(int n)
{
s[n]=s[n<<1]+s[n<<1|1];
}
void pushDown(int n)
{
s[n<<1]=lz[n]*s[n<<1];
s[n<<1|1]=lz[n]*s[n<<1|1];
lz[n<<1]=lz[n]*lz[n<<1];
lz[n<<1|1]=lz[n]*lz[n<<1|1];
lz[n].init();
}
void build(int l,int r,int n)//使用时输入(1,N,1)，N为总长度
{
lz[n].init();
if(l==r)
{
s[n]=f^(a[l]-1);
return;
}
int m=(l+r)/2;
build(l,m,n<<1);
build(m+1,r,n<<1|1);
pushUp(n);
}
void update(int ql,int qr,int l,int r,Matrix v,int n)//使用时输入(l,r.1,N,x,1)
{
if(l==ql && r==qr)
{
s[n]=v*s[n];
lz[n]=v*lz[n];
return;
}
pushDown(n);
int m=(l+r)/2;
if(qr<=m)
update(ql,qr,l,m,v,n<<1);
else if(ql>m)
update(ql,qr,m+1,r,v,n<<1|1);
else
{
update(ql,m,l,m,v,n<<1);
update(m+1,qr,m+1,r,v,n<<1|1);
}
pushUp(n);
}
int query(int ql,int qr,int l,int r,int n)
{
if(l==ql && r==qr)
return s[n].a[0][0];
pushDown(n);
int m=(l+r)/2;
if(qr<=m)
return query(ql,qr,l,m,n<<1);
if(ql>m)
return query(ql,qr,m+1,r,n<<1|1);
return (query(ql,m,l,m,n<<1)+query(m+1,qr,m+1,r,n<<1|1))%MOD;
}
int N,M;
int main()
{
f.a[0][0]=1; f.a[0][1]=1;
f.a[1][0]=1; f.a[1][1]=0;
scanf("%d%d",&N,&M);
for(int i=1;i<=N;++i)
scanf("%d",&a[i]);
build(1,N,1);
while(M--)
{
int type,l,r;
scanf("%d%d%d",&type,&l,&r);
if(type==1)
{
int x;
scanf("%d",&x);
update(l,r,1,N,f^x,1);
}
else printf("%d\n",query(l,r,1,N,1));
}
return 0;
}