【codeforces 508D】The Maths lecture

【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;
}