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

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本文详细解析了Codeforces竞赛中C题“Ayoub and Lost Array”的解决方案，通过数位DP算法，计算在特定条件下的数组恢复可能性。考虑到数组元素范围及总和特性，通过预处理余数分布，实现高效计算。

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题目链接：

codeforces.com/contest/1105/problem/C

C. Ayoub and Lost Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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].

//数位dp的题目：

需要预处理出余数0，1，2的个数

This is the code

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x7f7f7f7f      //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]= {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]= {-1, 1, 0, 0, -1, 1, -1, 1};
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int mod=1e9+7;
LL dp[250000][3];
int main()
{
    LL n,l,r;
    scanf("%lld%lld%lld",&n,&l,&r);
    LL num=r-l+1;
    LL yu=num%3;
    num/=3;
    LL a,b,c;
    a=b=c=num;
    //求出余0，1，2的个数
    if(l%3==0)
    {
        if(yu==1)
            a+=1;
        if(yu==2)
            a+=1,b+=1;
    }
    else if(l%3==1)
    {
        if(yu==1)
            b+=1;
        if(yu==2)
            b+=1,c+=1;
    }
    else
    {
        if(yu==1)
            c+=1;
        if(yu==2)
            c+=1,a+=1;
    }
    dp[1][0]=a;//余数为零
    dp[1][1]=b;//余数为一
    dp[1][2]=c;//余数为二
    for(int i=2; i<=n; ++i)
    {
        dp[i][0]=(dp[i-1][0]*a+dp[i-1][1]*c+dp[i-1][2]*b)%mod;//累加余数为0
        dp[i][1]=(dp[i-1][0]*b+dp[i-1][1]*a+dp[i-1][2]*c)%mod;//累加余数为1
        dp[i][2]=(dp[i-1][0]*c+dp[i-1][1]*b+dp[i-1][2]*a)%mod;//累加余数为2
    }
    cout<<dp[n][0]<<endl;
    return 0;
}