HDU

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B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6128 Accepted Submission(s): 3528

Problem Description

A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Input

Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

Output

Print each answer in a single line.

Sample Input

131002001000

Sample Output

Author

wqb0039

Source

2010 Asia Regional Chengdu Site —— Online Contest

题意是求1~n里有多少个数，数位上有13存在且这个数取余13==0

数位dp

state：1,：前一位是1 2：前面有13出现否则0

当时不知道怎么满足取余13等于0，其实按位数的权值一位一位的加，并膜13，只保留余数即可

这样：dp【位数】【state】【余数】

核心是想到记录余数这个状态，只有前面的余数相同，后面能取的个数才和之前求并记忆的一样

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>#define eps 1e-9#define PI 3.141592653589793#define bs 1000000007#define bsize 256#define MEM(a) memset(a,0,sizeof(a))#include<iostream>typedef long long ll;using namespace std;int dp[100][3][15];int a[100];int dfs(int pos,int state,int limit,int yu,long long t){if(pos==-1)    {        return state==2&&yu==0;    }if(!limit&&dp[pos][state][yu]!=-1)return dp[pos][state][yu];int up=limit?a[pos]:9;int ans=0;int now;for(int i=0;i<=up;i++){        if(state==2)            now=2;    else if(state==1&&i==3)            now=2;        else if(i==1)            now=1;        else            now=0;ans+=dfs(pos-1,now,limit&&i==a[pos],(yu+i*t)%13,t/10);}if(!limit&&dp[pos][state][yu]==-1)    {        dp[pos][state][yu]=ans;    }    return ans; } int solve(int x) {     int pos=0;     long long t=1;     while(x)     {         a[pos++]=x%10;         x/=10;         t*=10;     }     return dfs(pos-1,0,1,0,t/10); }int main(){    int n;    memset(dp,-1,sizeof(dp));    while(~scanf("%d",&n))    {        printf("%d\n",solve(n));    }    return 0; }

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