HDU 3652

http://acm.hdu.edu.cn/showproblem.php?pid=3652

数位DP，用了4维来记录，dp[vis][fg][k][index]，维数分别表示前面是否有13，上一位数是否为1，当前的余数，位数深度。

#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define int64 __int64
#define ll long long
#define mod 1000000007
#define M 20

int num[M] , dp[2][2][M][M];

int Dfs(int index , int vis , int fg , int k , int limit)
{
    if (!index)
    {
        return vis && !k;
    }
    if (!limit && dp[vis][fg][k][index] != -1)return  dp[vis][fg][k][index];
    int i , ret = 0 , up = limit ? num[index] : 9;
    for (i = 0 ; i <= up ; i++)
    {
        ret += Dfs(index-1 , vis||(fg&&i==3) , i==1 , (i+k*10)%13 , limit&&i==up);
    }
    if (!limit) dp[vis][fg][k][index] = ret;
    return ret;
}

int Solve(int k)
{
    int len = 0;
    while (k)
    {
        num[++len] = k%10;
         k /= 10;
    }
    return Dfs(len , 0 , 0 , 0 , 1);
}

int main()
{
    int n;
    memset(dp , -1 , sizeof dp);
    while (~scanf("%d",&n))
    {
        printf("%d\n",Solve(n));
    }
    return 0;
}