HDU3652 B-number(数位DP)

B-number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4565 Accepted Submission(s): 2617

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
13
100
200
1000

Sample Output
1
1
2
2

Author
wqb0039

Source
2010 Asia Regional Chengdu Site —— Online Contest

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数位DP模板题

#include "cstring"#include "cstdio"#include "string.h"#include "iostream"#include "cmath"using namespace std;int num[20];int dp[20][20][20][2];//0 don't have 13, 1 have 13int dfs(int pos,int mod,int pre,int have,int over){    if(pos<0)        return (mod==0&&have);    if(dp[pos][mod][pre][have]!=-1&&!over)        return dp[pos][mod][pre][have];    int last=over?num[pos]:9;    int ans=0;    for(int i=0;i<=last;i++)    {        int temp=10*mod;        temp+=i;        temp%=13;        if(pre==1&&i==3)            ans+=dfs(pos-1,temp,i,true,over&&i==last);        else            ans+=dfs(pos-1,temp,i,have,over&&i==last);    }    if(!over)        dp[pos][mod][pre][have]=ans;    return ans;}int solve(int n){    int len=0;    while(n)    {        num[len++]=n%10;        n/=10;    }    return dfs(len-1,0,0,false,true);}int main(){    int n;    memset(dp,-1,sizeof(dp));    while(~scanf("%d",&n))    {        printf("%d\n",solve(n));    }    return 0;}