HDU 3652

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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 20int 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;}