【题目链接】:http://codeforces.com/problemset/problem/507/D

【题意】



让你找符合这样数字的数的个数:

1.有n个数码

2.某个后缀%k的值为0

3.大于0

【题解】



数位DP;

设f[i][j][0]和f[i][j][1]分别表示;

(把最后的数字倒过来)

前i个数字组成的数%k的结果为j;

且前i-1个数字没有出现过%k结果为0;

前i-1个数字有出现过%k结果为0的方案数;

枚举新添加的一位;

看看新的取余值是啥;

然后想一想转移方程就好;

dp[0][0][0]=1;

(一开始那个不算取余值为0)

(第一位必须为正数!)

最后累计∑f[n][0..k-1][1];

输出答案;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 1100;

const int K = 110;

LL n,k,m,dp[N][K][2],pre[N];

//0不含%k==0

//1含%k==0

int main()

{

    //Open();

    Close();//scanf,puts,printf not use

    //init??????

    cin >> n >> k >> m;

    pre[0] = 1%k;

    rep1(i,1,1000)

        pre[i] = (pre[i-1]*10)%k;

    dp[0][0][0] = 1;

    rep1(i,0,n-1)

        rep1(j,0,k-1)

            rep1(num,(i==(n-1)?1:0),9)

            {

                LL now = (j+num*pre[i])%k;

                if (now==0 && num!=0)

                {

                    dp[i+1][now][1] = (dp[i+1][now][1] + dp[i][j][0])%m;

                }

                else

                //now!=0 || (now==0 && num==0)

                    dp[i+1][now][0] = (dp[i+1][now][0] + dp[i][j][0])%m;

                dp[i+1][now][1] = (dp[i+1][now][1] + dp[i][j][1])%m;

            }

    LL ans = 0;

    rep1(i,0,k-1)

        ans = (ans + dp[n][i][1])%m;

    cout << ans << endl;

    return 0;

}