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HDU 3652 B-number (数位DP)

求区间内的包含’13’并且能被13整除的数的个数。

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<iomanip>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
//#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double Pi = acos(-1.0);
const double eps = 1e-9;
const int INF = 0x3f3f3f3f;
const int MOD = 2520;
const int MAXN = 320000+10;

LL dp[20][11][15][3];
int dig[20];

LL dfs(int len,int num,int mod,bool state,bool mxl)
{
    if(!len)
    {
        return state && (mod==0);
    }
    if(!mxl && dp[len][num][mod][state]!=-1)
    {
        return dp[len][num][mod][state];
    }
    LL res=0;
    int maxlen=mxl?dig[len]:9;
    for(int i=0; i<=maxlen; ++i)
    {
        int nmod=(mod*10+i)%13;
        if( state || (num==1 && i==3))
        {
            res+=dfs(len-1,i,nmod,1,mxl && i==maxlen);
        }
        else
        {
            res+=dfs(len-1,i,nmod,0,mxl && i==maxlen);
        }

    }
    if(!mxl )
    {
        dp[len][num][mod][state]=res;
    }
    return res;
}

LL solve(LL x)
{
    memset(dig,0,sizeof(dig));
    int len=0;
    while(x)
    {
        dig[++len]=x%10;
        x/=10;
    }
    return dfs(len,0,0,0,true);
}

int main()
{
    LL n;
    memset(dp,-1,sizeof(dp));
    while(cin>>n)
    {
        cout<<(solve(n))<<endl;
    }

    return 0;
}