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 nnand 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
Input
2 1 3
Output
3
Input
3 2 2
Output
1
Input
9 9 99
Output
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].
完全没有想到是dp，还和翔哥用深搜做个铲铲。
https://blog.csdn.net/c___c18/article/details/88251466
看了这个大哥做的懂了奥，
值得纪念的题目。
#pragma warning(disable:4996)
#include"stdio.h"
#include"algorithm"
#include"stack"
#include"string.h"
#include"vector"
#include"string"
#include"iostream"
#include"map"
#define INF 0x3f3f3f3f;
using namespace std;
const int MOD = 1000000007;
const int N = 200015;
int l, r,n;
long long dp[N][3];
int main()
{
while (~scanf("%d %d %d", &n,&l,&r))
{
int a, b, c;
a = b = c = (r - l + 1) / 3;
for (int i = 0; i < (r - l + 1) % 3; i++)
{
if ((i + l) % 3 == 0)
{
a++;
}
else if ((i + l) % 3 == 1)
{
b++;
}
else
{
c++;
}
}
dp[1][0] = a;
dp[1][1] = b;
dp[1][2] = c;
for (int i = 2; i <= n; i++)
{
dp[i][0] = (dp[i - 1][0] * a % MOD + dp[i - 1][1] * c % MOD + dp[i - 1][2] * b % MOD) % MOD;
dp[i][1]= (dp[i - 1][0] * b % MOD + dp[i - 1][1] * a % MOD + dp[i - 1][2] * c % MOD) % MOD;
dp[i][2]= (dp[i - 1][0] * c % MOD + dp[i - 1][1] * b % MOD + dp[i - 1][2] * a % MOD) % MOD;
}
printf("%lld\n", dp[n][0]);
}
return 0;
}
/*
aabbcabbs
*/