HDU3652 B-number（数位DP）题解

思路：

这里的状态分为3种，无13和末尾的1，无13且末尾为1，有13，然后DFS

等我搞清楚数位DP就来更新Orz

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<vector>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
#define ll long long
using namespace std;
const int N = 1000000000+5;
int dp[12][15][3],a[20];    //dp[pos][mod][status] //status:0无13和末尾1，1末尾为1，2有13
int dfs(int pos,int mod,int sta,bool limit){
    if(pos == -1) return mod == 0 && sta == 2;
    if(!limit && dp[pos][mod][sta] != -1)   return dp[pos][mod][sta];
    int top = limit? a[pos] : 9;
    int sta2;
    ll ret = 0;
    for(int i = 0;i <= top;i++){
        sta2 = sta;
        if(sta == 0 && i == 1) sta2 = 1;
        else if(sta2 == 1 && i == 3) sta2 = 2;
        else if(sta2 == 1 && i != 1) sta2 = 0;
        ret += dfs(pos-1,(mod*10+i)%13,sta2,limit && i == top);
    }
    if(!limit) dp[pos][mod][sta] = ret;
    return ret;
}
ll solve(int x){
    int pos = 0;
    while(x){
        a[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos-1,0,0,true);
}
int main(){
    int n;
    memset(dp,-1,sizeof(dp));
    while(~scanf("%d",&n)){

        printf("%lld\n",solve(n));
    }
    return 0;
}