The Maths Lecture - CodeForces 507 D dp

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.

Sample test(s)

input

1 2 1000

output

4

input

2 2 1000

output

45

input

5 3 1103

output

590

题意：有多少个n位数，它的后缀可以被k整除。

思路：dp[i][j]表示i位数中可以被k整除为j的数有多少（已经去除掉之其中后缀能被k整除的情况）。因为如果i-1位能被k整除的数前面不管加什么都可以。每次转移的时候只需要添加前面的数字即可。注意要特殊处理后面全是0的情况。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;typedef long long ll;ll dp[1010][110],MOD,num[1010],ans,pow[1010];int n,m;int main(){    int i,j,k;    scanf("%d%d%I64d",&n,&m,&MOD);    for(i=1;i<=9;i++)       dp[1][i%m]++;    num[0]=1;num[1]=9%MOD;pow[1]=10%m;    for(i=2;i<=n;i++)       num[i]=num[i-1]*10%MOD;    for(i=2;i<=n;i++)       pow[i]=pow[i-1]*10%m;    for(i=1;i<=n;i++)    {        ans=(ans+dp[i][0]*num[n-i]%MOD)%MOD;        for(j=1;j<m;j++)          for(k=0;k<=9;k++)             dp[i+1][(j+k*pow[i]%m)%m]=(dp[i+1][(j+k*pow[i]%m)%m]+dp[i][j])%MOD;        for(k=1;k<=9;k++)           dp[i+1][k*pow[i]%m]++;    }    printf("%I64d\n",ans);}