Description
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?
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.
Sample Input
5 4 1 1 2 1 1 2 1 5 1 2 4 2 2 2 4 2 1 5
5 7 9
Hint
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.
#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
const int maxn = 1e5+5;
struct node
{
long long a[2][2];
void reset()//矩阵清零
{
memset(a,0,sizeof(a));
}
void one()//矩阵初始化
{
reset();
a[0][0]=a[1][1]=1;
}
};
node add(node A,node B)//矩阵相加
{
node k;k.reset();
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
k.a[i][j]=(A.a[i][j]+B.a[i][j])%mod;
return k;
}
node mul(node A,node B)//矩阵相乘
{
node k;memset(k.a,0,sizeof(k.a));
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int t=0;t<2;t++)
k.a[i][j]=(k.a[i][j]+A.a[i][t]*B.a[t][j])%mod;
return k;
}
node qpow(int p)//二分快速矩阵
{
node A;
A.a[0][0]=0,A.a[1][0]=1,A.a[0][1]=1,A.a[1][1]=1;
node K;
K.one();
while(p)
{
if(p%2)//如果是奇数
K=mul(K,A);
A=mul(A,A);
p/=2;
}
return K;
}
typedef node SgTreeDataType;
struct treenode
{
int L , R , flag;
SgTreeDataType sum , lazy;
void update(SgTreeDataType v)//这儿变换成矩阵运算
{
sum=mul(sum,v); //sum更新
lazy=mul(lazy,v); //lazy更新
flag=1; //标志置1
}
};
treenode tree[maxn*4];
int a[maxn];
inline void push_down(int o)
{
if(tree[o].flag)
{
tree[2*o].update(tree[o].lazy);
tree[2*o+1].update(tree[o].lazy);
tree[o].flag = 0;//标志置0
tree[o].lazy.one();//初始化lazy
}
}
inline void push_up(int o)
{
tree[o].sum = add(tree[o*2].sum,tree[o*2+1].sum);//更新父节点
}
node tmp;
inline void build_tree(int L , int R , int o)
{
tree[o].L = L , tree[o].R = R,tree[o].sum.reset(),tree[o].lazy.one(),tree[o].flag=0;
if(L==R)
{
tree[o].sum=qpow(a[L]);//初始化
}
if (R > L)
{
int mid = (L+R) >> 1;
build_tree(L,mid,o*2);
build_tree(mid+1,R,o*2+1);
push_up(o);
}
}
inline void update(int QL,int QR,SgTreeDataType v,int o)
{
int L = tree[o].L , R = tree[o].R;
if (QL <= L && R <= QR) tree[o].update(v);
else
{
push_down(o);
int mid = (L+R)>>1;
if (QL <= mid) update(QL,QR,v,o*2);
if (QR > mid) update(QL,QR,v,o*2+1);
push_up(o);
}
}
inline SgTreeDataType query(int QL,int QR,int o)
{
int L = tree[o].L , R = tree[o].R;
if (QL <= L && R <= QR) return tree[o].sum;
else
{
push_down(o);//important 向下传递lazy
int mid = (L+R)>>1;
SgTreeDataType res; res.reset();
if (QL <= mid) res=add(res,query(QL,QR,2*o));
if (QR > mid) res=add(res,query(QL,QR,2*o+1));
push_up(o);//important 更新父节点
return res;
}
}
int n,q;
int main()
{
tmp.a[0][0]=0,tmp.a[1][0]=1,tmp.a[0][1]=1,tmp.a[1][1]=1;
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
build_tree(1,n,1);
for(int i=1;i<=q;i++)
{
int op;
scanf("%d",&op);
if(op==2)
{
int a,b;scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,1).a[1][0]);
}
else{
int a,b,c;scanf("%d%d%d",&a,&b,&c);
update(a,b,qpow(c),1);
}
}
return 0;
}