Codeforces Round #533 (Div. 2)C. Ayoub and Lost Array

Posted 2021-02-04 hazelnut

 

C. Ayoub and Lost Array

Ayoub had an array ?? of integers of size ?? and this array had two interesting properties:

All the integers in the array were between ?? and ?? (inclusive).
The sum of all the elements was divisible by 3.

Unfortunately, Ayoub has lost his array, but he remembers the size of the array ?? and the numbers ?? and ??, 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+7 (i.e. the remainder when dividing by 109+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 0.

Input
The first and only line contains three integers ??, ?? and ?? (1≤??≤2⋅105,1≤??≤??≤109) — the size of the lost array and the range of numbers in the array.

Output
Print the remainder when dividing by 109+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].

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

dp第一维枚举已经取的位数，第二维表示模3的余数：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define PI acos(-1)
#define pn printf("
");
using namespace std;

int mod = 1e9+7;
const int maxn = 2e5+5;
ll dp[maxn][3];
int n;

void solve(ll c0, ll c1, ll c2)
{
    dp[1][0] = c0;
    dp[1][1] = c1;
    dp[1][2] = c2;
    for(int i = 2; i <= n ;i++)
    {
        dp[i][0] = (dp[i-1][1] * c2 % mod + dp[i-1][2] * c1 % mod + dp[i-1][0] * c0 % mod) % mod;
        dp[i][1] = (dp[i-1][1] * c0 % mod + dp[i-1][2] * c2 % mod + dp[i-1][0] * c1 % mod) % mod;
        dp[i][2] = (dp[i-1][1] * c1 % mod + dp[i-1][2] * c0 % mod + dp[i-1][0] * c2 % mod) % mod;
    }
    
}

int main()
{
    ll l, r;
    scanf("%d%lld%lld", &n, &l, &r);
    ll uni = (r - l + 1) / 3;
    ll res = (r - l + 1) % 3;
    ll c0 = 0, c1 = 0, c2 = 0;
    if(l % 3 == 0)
    {
        c0 = uni + (res >= 1);
        c1 = uni + (res >= 2);
        c2 = uni;
    }
    else if(l % 3 == 1)
    {
        c0 = uni;
        c1 = uni + (res >= 1);
        c2 = uni + (res >= 2);
    }
    else
    {
        c0 = uni + (res >= 2);
        c1 = uni;
        c2 = uni + (res >= 1);
    }
    solve(c0, c1, c2);
    printf("%lld
", dp[n][0]);
}