【CodeForces 719E】【线段树+矩阵快速幂】 Sasha and Array

传送门：E. Sasha and Array

描述：

E. Sasha and Array

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

1. 1 l r x — increase all integers on the segment from l to r by values x;
2. 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 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

5 4
1 1 2 1 1
2 1 5
1 2 4 2
2 2 4
2 1 5

output

5
7
9

Note

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.

题意:

给出有n个元素的数列ai(1<=i<=n)以及m次操作,操作分为两种:①将区间[l,r]的数加x;②询问∑f(ai)(l<=i<=r),其中f(x)是斐波那契数列的第x个数

题解:

斐波那契数列可以由2*2矩阵A={1 1,1 0}相乘得到,因此可以想到用线段树存储矩阵,每次更新加x就可以用乘以A^x代替

这题写的有点气，初始化没弄对，调半天_(:зゝ∠)_

代码：

#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << "  " ；
#define pl(x) cout << #x << "= " << x << endl;
#define lson l, m, rt<<1  
#define rson m+1, r, rt<<1|1 
#define ll __int64
using  namespace  std;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
const ll M=1e9+7;
const int maxn=1e5+10;

struct Matrix{
    ll v[2][2];
};

void init(Matrix& ad){
  ad.v[0][0]=1; ad.v[0][1]=1;
  ad.v[1][0]=1; ad.v[1][1]=0;
}

Matrix mtAdd(Matrix A, Matrix B){        // 求矩阵 A + B
  int i, j;
  Matrix C;
  for(i = 0; i < 2; i ++)
    for(j = 0; j < 2; j ++){
      C.v[i][j] = (A.v[i][j] + B.v[i][j]) % M;
    }
  return C;
}

Matrix mtMul(Matrix A, Matrix B){        // 求矩阵 A * B
  int i, j, k;
  Matrix C;
  for(i = 0; i < 2; i ++)
    for(j = 0; j < 2; j ++){
      C.v[i][j] = 0;
      for(k = 0; k < 2; k ++){
        C.v[i][j] = (A.v[i][k] * B.v[k][j] + C.v[i][j]) % M;
      }
    }
  return C;
}

Matrix mtPow(Matrix A, int k){           // 求矩阵 A ^ k
  if(k == 0) {
    memset(A.v, 0, sizeof(A.v));
    for(int i = 0; i < 2; i ++){
      A.v[i][i] = 1;
    }
    return A;
  }
  if(k == 1) return A;
  Matrix C = mtPow(A, k / 2);
  if(k % 2 == 0) {
    return mtMul(C, C);
  }
  else {
    return mtMul(mtMul(C, C), A);
  }
}

Matrix sum[maxn<<2],add[maxn<<2],A;

void PushUp(int rt){
  sum[rt]=mtAdd(sum[rt<<1], sum[rt<<1|1]);
}

void PushDown(int rt){
  add[rt<<1]=mtMul(add[rt<<1], add[rt]);
  add[rt<<1|1]=mtMul(add[rt<<1|1], add[rt]);
  sum[rt<<1]=mtMul(sum[rt<<1], add[rt]);
  sum[rt<<1|1]=mtMul(sum[rt<<1|1], add[rt]);
  add[rt]=mtPow(A, 0);
}

void build(int l, int r,int rt){
  add[rt]=mtPow(A, 0);//初始化为单位矩阵
  sum[rt]=mtPow(A, 0); 
  if(l==r){
    int x;read(x);
    sum[rt]=mtPow(A, x-1); 
    return;
  } 
  int m=(l+r)>>1;
  build(lson);  
  build(rson);  
  PushUp(rt); 
}

void update(int L,int R,Matrix c,int l,int r,int rt){
  if(L<=l && R>=r){  
    sum[rt]=mtMul(sum[rt], c);  
    add[rt]=mtMul(add[rt], c); 
    return;  
  } 
  PushDown(rt);
  int m=(l+r)>>1;
  if(L<=m)update(L, R, c, lson);
  if(R>m)update(L, R, c, rson);
  PushUp(rt);
}

ll query(int L,int R,int l, int r,int rt){
  if(L<=l && R>=r){
    return sum[rt].v[0][0];
  }
  PushDown(rt);
  int m=(l+r)>>1;
  ll ret=0;
  if(L<=m)ret=(ret+query(L, R, lson))%M;
  if(R>m)ret=(ret+query(L, R, rson))%M;
  PushUp(rt);
  return ret;
}

//debug用
void print(int l,int r,int rt){  
  if(l==r){  
    for(int i=0;i<2;i++){  
      for(int j=0;j<2;j++)  
        printf("%I64d ",sum[rt].v[i][j]);  
      printf("\n");  
    }  
    printf("\n");  
    return;  
  }  
  int m=(l+r)>>1;  
  print(lson);  
  print(rson);  
}  

int  main(){
  #ifndef ONLINE_JUDGE
  freopen("in.txt","r",stdin);
  #endif

  int n,m;
  init(A);
  read(n);read(m); 
  build(1, n, 1);
  int op,a,b;
  while(m--){
    read(op);read(a);read(b);
    if(op==1){
      int x;read(x);
      update(a, b, mtPow(A, x), 1, n, 1);
    }
    else{
      printf("%I64d\n", query(a, b, 1, n, 1));
    }
  }
  return 0;
}