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

原题链接

D. The Maths Lecture

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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:
  • ;
  • 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

input

1 2 1000

output

4

input

2 2 1000

output

45

input

5 3 1103

output

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.

用dp[i][j]表示i位数中有多少个数MOD K 等于 j(j != 0),dp[i][0]表示i位数中有多少个数的后缀MOD %k == 0

#include 
#define maxn 200005
#define MOD 1000000007
using namespace std;
typedef long long ll;

ll dp[1005][105];
ll d[1005];
int main(){

    //freopen("in.txt", "r", stdin);
	int n, k;
	ll m;
	
	scanf("%d%d%I64d", &n, &k, &m);
	d[1] = 1;
	for(int i = 2; i <= n; i++)
	 (d[i] = d[i-1] * 10) %= k;
	for(int i = 1; i <= 9; i++){
		(dp[1][i%k] += 1) %= m;
	}
	
	for(int i = 2; i <= n; i++){
		for(int j = 0; j <= 9; j++){
			 if(j)
			  (dp[i][d[i]*j%k] += 1) %= m;
			for(int h = 0; h < k; h++){
				if(h == 0){
					(dp[i][0] += dp[i-1][h]) %= m;
				}
				else{
				    (dp[i][(d[i]*j+h)%k] += dp[i-1][h]) %= m;
				}
			}
		}
    }
	ll ans = (dp[n][0] - dp[n-1][0] + m) % m;
	printf("%I64d\n", ans);
	 
	return 0;
}