Codeforces Round #287 (Div. 2) - D. The Maths Lecture （数位DP）

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本文介绍了一个使用数位动态规划（数位DP）解决特定数学问题的方法。该问题要求找出有多少个n位正整数，使得这些数的后缀（包括自身）能被k整除，并给出答案对m取模的结果。通过构建状态转移方程，文章提供了一种高效的求解策略。

Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.

First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:

Decimal representation of x (without leading zeroes) consists of exactly n digits;
There exists some integer y > 0 such that:
- $\text{[math]}$ ;
- decimal representation of y is a suffix of decimal representation of x.

As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.

Can you help Amr escape this embarrassing situation?

Input

Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).

Output

Print the required number modulo m.

Examples
inputCopy
1 2 1000
outputCopy
4
inputCopy
2 2 1000
outputCopy
45
inputCopy
5 3 1103
outputCopy
590

Note

A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.

题意：

找有几个n位数，他的后缀(包括自身)可以被k整除。答案对m取模

POINT：

数位DP，dp[x][y][z]记录当前x位数对k取模等于y，z代表有没有后缀已经被k整除了。即y==0。

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const LL maxn = 1234 + 10;
const int inf = 0x3f3f3f3f;

int shi[1234];

int dp[1111][111][2];
int n,k,m;

int dfs(int now,int mod,bool flag)
{
	if(dp[now][mod][flag]!=-1) return dp[now][mod][flag];
	if(now==n+1) return flag;
	int limit=now==n?1:0;
	int ans=0;
	for(int i=limit;i<=9;i++){
		ans+=dfs(now+1,(mod+shi[now-1]*i)%k,flag||(i>0&&(mod+shi[now-1]*i)%k==0));
		ans=ans%m;
	}
	return dp[now][mod][flag]=ans;
}

int main()
{
	memset(dp,-1,sizeof dp);

	scanf("%d%d%d",&n,&k,&m);
	shi[0]=1;
	for(int i=1;i<=1000;i++){
		shi[i]=shi[i-1]*10%k;
	}
	printf("%d\n",dfs(1,0,false)%m);


}