Ayoub and Lost Array
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
- 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
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.
Print the remainder when dividing by 109+7109+7 the number of ways to restore the array.
Examples input Copy2 1 3output Copy
3input Copy
3 2 2output Copy
1input Copy
9 9 99output Copy
711426616Note
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].
#include<bits/stdc++.h> #define REP(i, a, b) for(int i = (a); i <= (b); ++ i) #define REP(j, a, b) for(int j = (a); j <= (b); ++ j) #define PER(i, a, b) for(int i = (a); i >= (b); -- i) using namespace std; const int maxn=2e5+5; const int mod=1e9+7; template <class T> inline void rd(T &ret){ char c; ret = 0; while ((c = getchar()) < ‘0‘ || c > ‘9‘); while (c >= ‘0‘ && c <= ‘9‘){ ret = ret * 10 + (c - ‘0‘), c = getchar(); } } int n,l,r; long long dp[maxn][5]; int main() { rd(n),rd(l),rd(r); int u=r/3-(l-1)/3; int v=(r+2)/3-(l-1+2)/3; int w=(r+1)/3-(l-1+1)/3; dp[1][0]=u,dp[1][1]=v,dp[1][2]=w; REP(i,2,n){ dp[i][0]=(dp[i-1][2]*v%mod+dp[i-1][1]*w%mod+dp[i-1][0]*u%mod)%mod; dp[i][1]=(dp[i-1][0]*v%mod+dp[i-1][1]*u%mod+dp[i-1][2]*w%mod)%mod; dp[i][2]=(dp[i-1][0]*w%mod+dp[i-1][1]*v%mod+dp[i-1][2]*u%mod)%mod; } cout<<dp[n][0]<<endl; }
Ayoub and Lost Array