HDU 3652 B-number(数位dp)

B-number

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

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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

1122

Author

wqb0039

Source

2010 Asia Regional Chengdu Site —— Online Contest

这道题的意思是给出一个数，问从1到这个数之间有多少数既有子串13，又能被13整除。

简单的数位dp题

#include <iostream>#include<string>#include<cstring>#include<set>#include<cstdio>using namespace std;typedef long long ll;ll dp[25][15][3];ll a[25];ll dfs(int pos,bool is,int state2,int state,bool limit){    if(pos==-1&&state==1&&state2==0) return 1;    else if(pos==-1) return 0;    if(!limit&&dp[pos][state2][state]!=-1) return dp[pos][state2][state];    int up=limit?a[pos]:9;    ll temp=0;    for(int i=0;i<=up;i++)    {        if((is&&i==3)||state==1)        temp+=dfs(pos-1,i==1,(state2*10+i)%13,1,limit&&a[pos]==i);        else if(i==1) temp+=dfs(pos-1,i==1,(state2*10+i)%13,2,limit&&a[pos]==i);        else temp+=dfs(pos-1,i==1,(state2*10+i)%13,0,limit&&a[pos]==i);    }    if(!limit) dp[pos][state2][state]=temp;    return temp;}ll solve(ll x){    memset(dp,-1,sizeof(dp));    int pos=0;    while(x)    {        a[pos++]=x%10;        x/=10;    }    return dfs(pos-1,0,0,0,true);}int main(){    ll x;    while(~scanf("%lld",&x))    {        cout<<solve(x)<<endl;    }    return 0;}