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題目

Sasha has an array of integers a₁,?a₂,?...,?a_n. 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 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

The next line contains n integers a₁,?a₂,?...,?a_n (1?≤?a_i?≤?10⁹).

Then follow m lines with queries descriptions. Each of them contains integers tp_i, l_i, r_i and may be x_i (1?≤?tp_i?≤?2, 1?≤?l_i?≤?r

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.

題目大意

給你一個長度為n的數列an，有兩種操作

1、將L到R的ai加上X

2、詢問L到R之間，f(aL)+f(aL+1)+……+f(aR)的和

f是斐波那契函數

分析

我們可以將斐波那契數轉化成它所對應的矩陣，對於每一次加x就是給原來矩陣乘上斐波那契矩陣的x次方。將為賦值的矩陣全部初始化為單位矩陣，然後進行樸素的線段樹為何兩節點之和即可。

代碼

#pragma G++ optimize (2)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define rri register int
const int mod=1e9+7;
struct mat {
      int g[5][5];
};
mat d[440000],one,per;
mat add[440000];
inline mat operator * (const mat a,const mat b){
      mat c;
      c.g[1][1]=c.g[1][2]=c.g[2][1]=c.g[2][2]=0;
      for(rri i=1;i<=2;++i)
         for(rri k=1;k<=2;++k)
            for(rri j=1;j<=2;++j)            
               c.g[i][j]=(c.g[i][j]+(long long)a.g[i][k]*b.g[k][j]%mod)%mod;
      return c;
}
inline mat operator + (const mat a,const mat b){
      mat c;
      for(rri i=1;i<=2;++i)
         for(rri j=1;j<=2;++j)
            c.g[i][j]=(a.g[i][j]+b.g[i][j])%mod;
      return c;
}
inline mat pw(mat a,int p){
      mat res=a;
      p-=1;
      while(p){
          if(p&1)res=res*a;
          a=a*a;
          p>>=1;
      }
      return res;
}
inline int read(){
      int x=0,f=1;char s=getchar();
      while(s<‘0‘||s>‘9‘){if(s==‘-‘)f=-1;s=getchar();}
      while(s>=‘0‘&&s<=‘9‘){x=(x<<3)+(x<<1)+(s-‘0‘);s=getchar();}
      return f*x;
}
inline void build(int le,int ri,int pl,mat k,int wh){
      add[wh]=per;
      if(le==ri){
        d[wh]=k;
        return;
      }
      int mid=(le+ri)>>1;
      if(mid>=pl)build(le,mid,pl,k,wh<<1);
        else build(mid+1,ri,pl,k,wh<<1|1);
      d[wh]=d[wh<<1]+d[wh<<1|1];
}
inline void pd(int wh){
      if(add[wh].g[1][1]!=1||add[wh].g[2][2]!=1||
         add[wh].g[1][2]!=0||add[wh].g[2][1]!=0){
        add[wh<<1]=add[wh<<1]*add[wh];
        add[wh<<1|1]=add[wh<<1|1]*add[wh];
        d[wh<<1]=d[wh<<1]*add[wh];
        d[wh<<1|1]=d[wh<<1|1]*add[wh];
        add[wh]=per;
      }
}
inline void update(int le,int ri,int x,int y,mat k,int wh){
      if(le>=x&&ri<=y){
          add[wh]=add[wh]*k;
          d[wh]=d[wh]*k;
          return;
      }
      int mid=(le+ri)>>1;
      pd(wh);
      if(mid>=x)update(le,mid,x,y,k,wh<<1);
      if(mid<y)update(mid+1,ri,x,y,k,wh<<1|1);
      d[wh]=d[wh<<1]+d[wh<<1|1];
}
inline int q(int le,int ri,int x,int y,int wh){
      if(le>=x&&ri<=y)return d[wh].g[1][2]%mod;
      int mid=(le+ri)>>1,ans=0;
      pd(wh);
      if(mid>=x)ans=(ans+q(le,mid,x,y,wh<<1))%mod;
      if(mid<y)ans=(ans+q(mid+1,ri,x,y,wh<<1|1))%mod;
      d[wh]=d[wh<<1]+d[wh<<1|1];
      return ans%mod;
}
int main()
{     int n,m,x,l,r,k;
      one.g[1][1]=0,one.g[1][2]=one.g[2][1]=one.g[2][2]=1;
      per.g[1][1]=per.g[2][2]=1,per.g[1][2]=per.g[2][1]=0;
      n=read(),m=read();
      for(rri i=1;i<=n;++i){
          x=read();
          build(1,n,i,pw(one,x),1);
      }
      for(rri i=1;i<=m;++i){
          k=read();
          if(k==1){
              l=read(),r=read(),x=read();
              update(1,n,l,r,pw(one,x),1);
          }else {
              l=read(),r=read();
              printf("%d\n",q(1,n,l,r,1)%mod);
          }
      }
      return 0;
}

718C Sasha and Array