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 3output
3input
3 2 2output
1input
9 9 99output
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].
//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0dp[i][1]=(dp[i-1][0]*b+dp[i-1][1]*a+dp[i-1][2]*c)%mod;//累加余数为1dp[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;
}