【递推】Ayoub and Lost Array

题目：Ayoub had an array aa of integers of size nn and this array had two interesting properties:

All the integers in the array were between ll and rr (inclusive).
The sum of all the elements was divisible by 33.
Unfortunately, Ayoub has lost his array, but he remembers the size of the array nn and the numbers ll and rr, so he asked you to find the number of ways to restore the array.

Since the answer could be very large, print it modulo 109+7109+7 (i.e. the remainder when dividing by 109+7109+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 00.

Input
The first and only line contains three integers nn, ll and rr (1≤n≤2⋅105,1≤l≤r≤1091≤n≤2⋅105,1≤l≤r≤109) — the size of the lost array and the range of numbers in the array.

Output
Print the remainder when dividing by 109+7109+7 the number of ways to restore the array.

Examples
inputCopy
2 1 3
outputCopy
3
inputCopy
3 2 2
outputCopy
1
inputCopy
9 9 99
outputCopy
711426616
Note
In the first example, the possible arrays are : [1,2],[2,1],[3,3][1,2],[2,1],[3,3].

In the second example, the only possible array is [2,2,2][2,2,2].

解决方法：首先可以先将[l,r]范围内的整数根据其模3的到的余数分为三类，分别将其数量存在num[0],num[1],num[2]中，而dp[i] [j]代表在放置第i个数时使当前数组总和%3的值为j的情况数，由此可以推出
$dp[i][j]=dp[i-1][0]*num[j]+dp[i-1][1]*num[(j-1+3)\%3]+dp[i-1][2]*num[(j-2+3)\%3]$dp[i][j]=dp[i−1][0]∗num[j]+dp[i−1][1]∗num[(j−1+3)%3]+dp[i−1][2]∗num[(j−2+3)%3]

#include <iostream>
#include <cstdio>
using namespace std; 
typedef long long ll;
const int mod=1e9+7;
ll dp[200050][3];
ll num[3];
int main() 
{
    ios::sync_with_stdio(0),cin.tie(0);
    int n,l,r;
    cin>>n>>l>>r;
    num[0]=r/3-(l-1)/3;
    num[1]=num[0]-1;if(l%3==1) num[1]++;if(r%3==1||r%3==2) num[1]++;
    num[2]=num[0]-1;if(l%3==1||l%3==2) num[2]++;if(r%3==2) num[2]++;
    //cout<<num[0]<<" "<<num[1]<<" "<<num[2]<<endl;
    dp[1][0]=num[0];dp[1][1]=num[1];dp[1][2]=num[2];
    for(int i=2;i<=n;i++) {
    	dp[i][0]=((dp[i-1][0]*num[0])%mod+(dp[i-1][1]*num[2])%mod+(dp[i-1][2]*num[1])%mod)%mod;
    	dp[i][1]=((dp[i-1][0]*num[1])%mod+(dp[i-1][1]*num[0])%mod+(dp[i-1][2]*num[2])%mod)%mod;
    	dp[i][2]=((dp[i-1][0]*num[2])%mod+(dp[i-1][1]*num[1])%mod+(dp[i-1][2]*num[0])%mod)%mod;
    }
    cout<<dp[n][0]<<endl;
    return 0;
}